Author: tom

Linearity Properties of Series

1. Addition of Series:

   Formula: \( \sum_{n=1}^{\infty} (a_n + b_n) = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n \)

   Explanation: You can add two series together by adding their corresponding terms. This property allows the addition operation to be distributed over the sum.

2. Subtraction of Series:

   Formula: \( \sum_{n=1}^{\infty} (a_n – b_n) = \sum_{n=1}^{\infty} a_n – \sum_{n=1}^{\infty} b_n \)

   Explanation: You can subtract one series from another by subtracting their corresponding terms. This property allows the subtraction operation to be distributed over the sum.

3. Constant Multiple of Series:

   Formula: \( \sum_{n=1}^{\infty} c \cdot a_n = c \cdot \sum_{n=1}^{\infty} a_n \), where \( c \) is a constant.

   Explanation: You can multiply every term of a series by a constant, and this can be done by multiplying the sum of the series by that value. This property allows the scalar multiplication to be distributed over the sum.

These properties are referred to as the “linearity of summation” because they demonstrate how summation behaves linearly with respect to addition, subtraction, and scalar multiplication. They are foundational rules in working with series and have broad applications in mathematics.

Example: Addition of Two Convergent Series

Let’s consider two convergent series:

• Series A: \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) (a convergent geometric series with sum \( \frac{1}{2} \div (1 – \frac{1}{2}) = 1 \))
• Series B: \( \sum_{n=1}^{\infty} \frac{1}{3^n} \) (another convergent geometric series with sum \( \frac{1}{3} \div (1 – \frac{1}{3}) = \frac{1}{2} \))

We want to find the sum of these two series:

\( \sum_{n=1}^{\infty} \left( \frac{1}{2^n} + \frac{1}{3^n} \right) \)

Using the property of addition of series, we can write this as:

\( \sum_{n=1}^{\infty} \frac{1}{2^n} + \sum_{n=1}^{\infty} \frac{1}{3^n} = 1 + \frac{1}{2} = \frac{3}{2} \)

This expression represents the sum of Series A and Series B, and the total sum is \( \frac{3}{2} \).

Note: The addition property is valid only when both series converge. In this example, both Series A and Series B are convergent geometric series, so we can apply the addition property to find their sum.

Explanation: The addition property allows us to add the corresponding terms of the two series together. In this example, we added the \( n \)-th term of Series A with the \( n \)-th term of Series B for each \( n \). This resulted in a new series that represents the sum of the original two series. The property of addition of series enables us to work with the sum of two series in a straightforward and systematic way, provided that both series converge.

Example: Difference of Two Convergent Series

Let’s consider the same two convergent series as before:

• Series A: \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) (a convergent geometric series with sum \( \frac{1}{2} \div (1 – \frac{1}{2}) = 1 \))
• Series B: \( \sum_{n=1}^{\infty} \frac{1}{3^n} \) (another convergent geometric series with sum \( \frac{1}{3} \div (1 – \frac{1}{3}) = \frac{1}{2} \))

We want to find the difference between these two series:

\( \sum_{n=1}^{\infty} \left( \frac{1}{2^n} – \frac{1}{3^n} \right) \)

Using the property of subtraction of series, we can write this as:

\( \sum_{n=1}^{\infty} \frac{1}{2^n} – \sum_{n=1}^{\infty} \frac{1}{3^n} = 1 – \frac{1}{2} = \frac{1}{2} \)

This expression represents the difference between Series A and Series B, and the total difference is \( \frac{1}{2} \).

Note: The difference property is valid only when both series converge. In this example, both Series A and Series B are convergent geometric series, so we can apply the difference property to find their difference.

Explanation: The difference property allows us to subtract the corresponding terms of the two series. In this example, we subtracted the \( n \)-th term of Series B from the \( n \)-th term of Series A for each \( n \). This resulted in a new series that represents the difference between the original two series. The property of subtraction of series enables us to work with the difference of two series in a systematic way, provided that both series converge.

Example: Constant Multiple of a Convergent Series

Consider a convergent geometric series:

• Series A: \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) (a convergent geometric series with sum \( \frac{1}{2} \div (1 – \frac{1}{2}) = 1 \))

We want to find the constant multiple of this series by a factor of 3:

\( 3 \cdot \sum_{n=1}^{\infty} \frac{1}{2^n} \)

Using the constant multiple property, we can distribute the constant inside the summation:

\( 3 \cdot \left( \sum_{n=1}^{\infty} \frac{1}{2^n} \right) = \sum_{n=1}^{\infty} \left( 3 \cdot \frac{1}{2^n} \right) \)

This means we multiply each term inside the summation by the constant 3:

\( \sum_{n=1}^{\infty} \left( 3 \cdot \frac{1}{2^n} \right) = 3 \cdot \frac{1}{2} + 3 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} + \ldots \)

The sum of this new series is:

\( 3 \cdot \frac{1}{2} \div (1 – \frac{1}{2}) = 3 \)

Note: The constant multiple property is valid only when the series converges. In this example, Series A is a convergent geometric series, so we can apply the constant multiple property to find the constant multiple of the series.

Explanation: The constant multiple property allows us to multiply each term of the series by a constant factor. In this example, we multiplied each term of Series A by 3. This resulted in a new series that represents the constant multiple of the original series by a factor of 3. The property of constant multiplication enables us to work with the constant multiple of a series in a systematic way, provided that the series converges.

Understanding Convergence and Divergence in Mathematical Series

The study of mathematical series, specifically convergence and divergence, is a fundamental concept in mathematics. This summary explores key ideas, including the proof of the divergence of the harmonic series, the definition of convergent series, the relationship between convergence and the behavior of individual terms, and an illustrative example using a geometric series.

Proof: The Harmonic Series Diverges

The harmonic series is a sequence that adds the reciprocals of natural numbers. Through a method of grouping terms and comparing with a known divergent geometric series, it’s proven that the harmonic series does not converge to a finite sum.

Definition of Convergent Series

A series is convergent if the sequence of its partial sums approaches a finite limit. If a series is convergent, then the individual terms must go to 0. This relationship is explored through mathematical reasoning and illustrated with a specific geometric series example.

Counterexample: The Harmonic Series

The harmonic series serves as a counterexample to the converse statement. While the terms of the series go to 0, the series itself does not converge. This highlights the importance of the rate at which the terms go to 0 in determining convergence.

Summary

The convergence and divergence of mathematical series are complex topics with deep implications. The harmonic series, in particular, provides a rich example that challenges intuition and underscores the need for careful analysis. Understanding these concepts is essential for various mathematical contexts and offers insight into the intricate nature of mathematical reasoning.

Proof: The Harmonic Series Diverges

The harmonic series is given by \( \sum_{n=1}^{\infty} \frac{1}{n} \). We will prove that it diverges by comparing it to a known divergent geometric series.

1. Grouping Terms:

   We can group the terms of the harmonic series to make it easier to compare with a geometric series: \[ 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \ldots \]

2. Comparison with a Known Divergent Series:

   We will compare the grouped quantities to the corresponding fractions on the right side: \[ 1 = 1, \quad \frac{1}{2} = \frac{1}{2}, \quad \left(\frac{1}{3} + \frac{1}{4}\right) > \frac{1}{4}, \quad \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) > \frac{1}{8}, \ldots \]

   Therefore, the harmonic series is larger than the divergent geometric series: \[ 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \ldots > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \]

3. Conclusion:

   The series on the right is a geometric series with a common ratio of \( \frac{1}{2} \), and it diverges. Since the harmonic series is larger, it must also diverge. This proves that the harmonic series does not converge, even though the terms go to 0.

Definition of Convergent Series

A series is said to be convergent if the sequence of its partial sums approaches a finite limit as the number of terms goes to infinity. In other words, as we add more and more terms, the sum gets closer to a specific number.

The Result: Terms Go to 0

If a series is convergent, then the individual terms must go to 0. Here’s why:

1. Understanding Consecutive Partial Sums:
   • We look at the limit of the partial sum as \( N \) goes to infinity: \[ \lim_{N \to \infty} S_N. \]
   • We also look at the limit of the previous partial sum as \( N \) goes to infinity: \[ \lim_{N \to \infty} S_{N-1}. \]
   • Since the series is convergent, both of these limits are equal to the same finite value \( L \): \[ \lim_{N \to \infty} S_N = \lim_{N \to \infty} S_{N-1} = L. \]
2. Expressing the Terms as Differences of Partial Sums: We express the terms as differences of partial sums. This forms a bridge between the terms and the partial sums. \[ S_N – S_{N-1} = a_N. \]
3. Applying the Limit to the Difference: We take the limit to understand what happens to the terms as we go to infinity. This step connects the terms to the convergence of the series. \[ \lim_{N \to \infty} (S_N – S_{N-1}) = \lim_{N \to \infty} a_N. \]
4. Using the Convergence of the Series: We use the fact that the series is convergent to simplify the expression. \[ L – L = \lim_{N \to \infty} a_N. \]
5. Concluding that the Terms Go to 0: We conclude that the terms must go to 0. \[ \lim_{N \to \infty} a_N = 0. \]

Summary

Taking the limit is not an arbitrary step. It’s a way to understand what happens to the terms as we go to infinity. It connects the behavior of the terms to the convergence of the series, allowing us to prove that the terms must go to 0 if the series is convergent. This understanding helps us work with series in various mathematical contexts.

Example: Geometric Series with Common Ratio \( \frac{1}{2} \)

We’ll use the geometric series \( \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n \) to illustrate the concept that the terms must go to 0 if the series is convergent.

1. Understanding the Series:

   This is a geometric series with a common ratio of \( \frac{1}{2} \), and it is known to be convergent. The sum of the series is \( \frac{1}{1 – \frac{1}{2}} = 2 \).

2. Looking at Consecutive Partial Sums:
   • The partial sum up to \( N \) terms: \[ S_N = 1 + \frac{1}{2} + \frac{1}{4} + \ldots + \left(\frac{1}{2}\right)^N. \]
   • The previous partial sum: \[ S_{N-1} = 1 + \frac{1}{2} + \frac{1}{4} + \ldots + \left(\frac{1}{2}\right)^{N-1}. \]
   • Both of these partial sums approach 2 as \( N \) goes to infinity: \[ \lim_{N \to \infty} S_N = \lim_{N \to \infty} S_{N-1} = 2. \]
3. Expressing the Terms as Differences of Partial Sums:

   The difference between consecutive partial sums gives the \( N \)-th term: \[ S_N – S_{N-1} = \left(\frac{1}{2}\right)^N. \]

4. Applying the Limit to the Difference:

   We take the limit of both sides: \[ \lim_{N \to \infty} (S_N – S_{N-1}) = \lim_{N \to \infty} \left(\frac{1}{2}\right)^N. \]

5. Using the Convergence of the Series:

   We use the fact that the series converges to 2: \[ 2 – 2 = \lim_{N \to \infty} \left(\frac{1}{2}\right)^N. \]

6. Concluding that the Terms Go to 0:

   We conclude that the terms must go to 0: \[ \lim_{N \to \infty} \left(\frac{1}{2}\right)^N = 0. \]

Summary

This example illustrates how the terms of a convergent series must go to 0. By looking at the partial sums and applying the limit, we can see that the terms of this geometric series approach 0 as the number of terms goes to infinity. This understanding is consistent with the general result for convergent series.

Counterexample: The Harmonic Series

The harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) serves as a counterexample to the converse statement. While the terms of the series go to 0, the series itself does not converge.

1. Terms Go to 0:

   The terms of the harmonic series are given by \( \frac{1}{n} \), and as \( n \) goes to infinity, the terms approach 0: \[ \lim_{n \to \infty} \frac{1}{n} = 0. \]

2. Partial Sums Grow Without Bound:

   Despite the terms going to 0, the partial sums of the harmonic series grow without bound. By grouping terms, we can see that the partial sums are larger than a divergent geometric series: \[ 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \ldots > 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \]

3. Comparison with a Known Divergent Series:

   The series on the right is a geometric series with a common ratio of \( \frac{1}{2} \), and it diverges. Since the harmonic series is larger, it must also diverge.

4. Conclusion:

   The harmonic series illustrates that even if the terms of a series go to 0, the series may still diverge. The convergence of a series requires more than just the terms going to 0; the rate at which they go to 0 is also crucial.

Summary

The harmonic series disproves the converse statement that if the terms of a series go to 0, then the series converges. While the terms of the harmonic series do go to 0, the series itself diverges. This example highlights the importance of careful analysis in understanding the behavior of infinite series.

Partial Fraction Decomposition of \(\frac{1}{n(n+1)}\)

In mathematics, partial fraction decomposition is a method used to break down complex rational expressions into simpler parts. It’s particularly useful in integration, simplifying complex expressions, and solving differential equations. Below, we’ll explore the step-by-step process of decomposing the expression \(\frac{1}{n(n+1)}\) into simpler fractions.

1. \(\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}\)    |    Break down the expression into two simpler fractions.
2. \(1 = A(n+1) + Bn\)    |    Multiply both sides by \(n(n+1)\) to clear the denominators.
3. \(1 = An + A + Bn\)    |    Distribute \(n\) inside the parentheses to expand \(A(n+1)\).
4. \(1 = (A + B)n + A\)    |    Combine like terms by adding the coefficients of \(n\); this means adding the coefficients of the same power of \(n\).
5. \(A = 1, A + B = 0\)    |    Equate the coefficients of the corresponding powers of \(n\) on both sides; this involves setting the coefficients of the same power of \(n\) on both sides equal to each other.
6. \(B = -1\)    |    Substitute \(A = 1\) into the equation for the coefficient of \(n\) to find \(B\).
7. \(\frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}\)    |    Substitute the values of \(A\) and \(B\) into the original expression.

Telescoping Series for \(\frac{1}{n(n+1)}\)

The partial fraction decomposition of \(\frac{1}{n(n+1)}\) leads to a telescoping series. A telescoping series is a series where most of the terms cancel out, and only a few terms remain after simplification. Here’s how the series unfolds:

1. Start with the expression: \(\frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}\)
2. Write the series: \(\sum_{n=1}^{\infty} \left(\frac{1}{n} – \frac{1}{n+1}\right)\)
3. Expand the series:
   • \(\frac{1}{1} – \frac{1}{2} + \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \ldots\)
   • The terms cancel out, leaving only the first term: \(1\)
4. The sum of the series is \(1\).

This telescoping series beautifully illustrates how complex expressions can simplify to a single value. It’s a powerful concept in mathematics, with applications in various fields.

Partial Fraction Decomposition of \(\frac{1}{n(n+2)}\)

\(\frac{1}{n(n+2)}\)	Start with the expression
\(\frac{A}{n} + \frac{B}{n+2}\)	Write it as a sum of two fractions
\(1 = A(n+2) + Bn\)	Multiply both sides by \(n(n+2)\)
\(1 = An + 2A + Bn\)	Expand
\(1 = (A + B)n + 2A\)	Group like terms
\(A + B = 0\), \(2A = 1\)	Compare coefficients
\(A = \frac{1}{2}\), \(B = -\frac{1}{2}\)	Solve for \(A\) and \(B\)
\(\frac{1}{n(n+2)} = \frac{1}{2n} – \frac{1}{2(n+2)}\)	Write the final expression

This decomposition allows us to write the expression as a difference of two simpler fractions, which can be useful in various mathematical contexts.

Telescoping Series for \(\frac{1}{n(n+2)}\)

Using the partial fraction decomposition, we can write the telescoping series for \(\frac{1}{n(n+2)}\):

1. Start with the expression: \(\frac{1}{n(n+2)} = \frac{1}{2n} – \frac{1}{2(n+2)}\)
2. Write the series: \(\sum_{n=1}^{\infty} \left(\frac{1}{2n} – \frac{1}{2(n+2)}\right)\)
3. Expand the series:
   • \(\frac{1}{2} – \frac{1}{4} + \frac{1}{4} – \frac{1}{6} + \frac{1}{6} – \frac{1}{8} + \ldots\)
   • The terms cancel out, leaving only the first term: \(\frac{1}{2}\)
4. The sum of the series is \(\frac{1}{2}\).

This example further illustrates the concept of telescoping series and how complex expressions can simplify to a single value.

Partial Fraction Decomposition of \(\frac{1}{(n+1)(n+2)}\)

\(\frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2}\)	Express the original expression as a sum of two unknown fractions, a common step in partial fraction decomposition.
\(1 = A(n+2) + B(n+1)\)	Multiply both sides by \((n+1)(n+2)\) to clear the denominators, using the fact that the original expression equals 1.
\(1 = An + 2A + Bn + B\)	Expand the right side by distributing \(A\) and \(B\) over the expressions inside the parentheses.
\(1 = (A + B)n + (2A + B)\)	Group like terms by collecting the coefficients of \(n\) and the constant terms together.
\(A + B = 0\), \(2A + B = 1\)	Set up a system of equations by comparing the coefficients of \(n\) and the constant terms on both sides.
\(A = 1\), \(B = -1\)	Solve the system of equations to find the values of \(A\) and \(B\).
\(\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} – \frac{1}{n+2}\)	Substitute the values of \(A\) and \(B\) back into the original expression to obtain the final result.

This decomposition breaks down the original expression into simpler fractions, making it easier to work with in various mathematical contexts, such as integration or series expansion.

Telescoping Series for \(\frac{1}{(n+1)(n+2)}\)

We have already found the partial fraction decomposition:

\(\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} – \frac{1}{n+2}\)

Now, we can write out the series by substituting successive integer values for \( n \):

\(S = \left(\frac{1}{1} – \cancel{\frac{1}{2}}\right) + \left(\cancel{\frac{1}{2}} – \cancel{\frac{1}{3}}\right) + \left(\cancel{\frac{1}{3}} – \cancel{\frac{1}{4}}\right) + \ldots\)

Notice that most of the terms cancel out:

\(S = 1 – \cancel{\frac{1}{2}} + \cancel{\frac{1}{2}} – \cancel{\frac{1}{3}} + \cancel{\frac{1}{3}} – \cancel{\frac{1}{4}} + \ldots = 1\)

The sum of the infinite series corresponding to \(\frac{1}{(n+1)(n+2)}\) is 1.

Understanding the Ellipsis in a Series

In mathematical notation, the ellipsis (“…”) is used to indicate that a pattern continues indefinitely. It represents a continuation of the terms in a sequence or series, following the established pattern.

For example, in the series:

\(1 + 2 + 3 + \ldots + n\)

The ellipsis indicates that the pattern of adding consecutive integers continues up to \( n \), where \( n \) is a specific integer.

In the context of the telescoping series we discussed earlier:

\(\left(\frac{1}{1} – \frac{1}{2}\right) + \left(\frac{1}{2} – \frac{1}{3}\right) + \left(\frac{1}{3} – \frac{1}{4}\right) + \ldots\)

The ellipsis signifies that the pattern of subtracting consecutive reciprocal integers continues indefinitely, following the established pattern of the series.

It’s a concise way to represent an infinite sequence of terms without having to write them all out. The understanding of the pattern and the context in which the ellipsis is used allows mathematicians to work with infinite series and sequences.

Understanding the Ellipsis in a Series

In mathematical notation, the ellipsis (“…”) is used to indicate that a pattern continues indefinitely. It represents a continuation of the terms in a sequence or series, following the established pattern.

For example, in the series:

\(1 + 2 + 3 + \ldots + n\)

The ellipsis indicates that the pattern of adding consecutive integers continues up to \( n \), where \( n \) is a specific integer.

In the context of the telescoping series we discussed earlier:

\(\left(\frac{1}{1} – \frac{1}{2}\right) + \left(\frac{1}{2} – \frac{1}{3}\right) + \left(\frac{1}{3} – \frac{1}{4}\right) + \ldots\)

The ellipsis signifies that the pattern of subtracting consecutive reciprocal integers continues indefinitely, following the established pattern of the series.

It’s a concise way to represent an infinite sequence of terms without having to write them all out. The understanding of the pattern and the context in which the ellipsis is used allows mathematicians to work with infinite series and sequences.

Understanding Telescoping Series: Term Breakdown

Let’s explore a telescoping series by carefully breaking down its terms:

\( S = \sum_{n=1}^{\infty} \left( \frac{1}{n} – \frac{1}{n+1} \right) \)

Breaking Down the Terms

As we expand the series, we get:

\( S = 1 – \frac{1}{2} + \frac{1}{2} – \frac{1}{3} + \frac{1}{3} – \frac{1}{4} + \ldots \)

Notice that each fraction in the series is paired with its negative counterpart from the next term. Let’s understand why this happens.

Understanding the Cancellations

When we group the positive and negative terms together, the cancellations become evident:

\( S = 1 + \left( -\frac{1}{2} + \frac{1}{2} \right) + \left( -\frac{1}{3} + \frac{1}{3} \right) + \ldots \)

Each pair of terms within the parentheses cancels each other out, leaving us with:

\( S = 1 \)

Understanding the Result

The telescoping series simplifies to a single term: 1. This might seem surprising, but it’s a result of the strategic pairing of terms.

Even though the individual terms vary, their differences lead to a remarkable cancellation pattern, resulting in a constant sum.

This breakdown illustrates how terms combine and cancel to achieve a seemingly unexpected outcome, shedding light on the nature of telescoping series.

Level 1 Telescoping Series

A level 1 telescoping series is a series where many terms cancel out, leaving only a few terms to determine the sum. A classic example is:

\( S = \sum_{n=1}^{\infty} \left(\frac{1}{n} – \frac{1}{n+1}\right) \)

This can be expanded as:

\( S = \left(1 – \frac{1}{2}\right) + \left(\frac{1}{2} – \frac{1}{3}\right) + \left(\frac{1}{3} – \frac{1}{4}\right) + \ldots \)

Now, let’s look at the cancellation process:

• The \(-\frac{1}{2}\) in the first term cancels with the \(+\frac{1}{2}\) in the second term.
• The \(-\frac{1}{3}\) in the second term cancels with the \(+\frac{1}{3}\) in the third term.
• The \(-\frac{1}{4}\) in the third term cancels with the \(+\frac{1}{4}\) in the fourth term.
• And so on, with each negative fraction canceling with the corresponding positive fraction in the next term.

After all the cancellations, we are left with the first term of each parenthesis:

\( S = 1 \)

The sum of this telescoping series is 1, as all other terms have canceled out.

Telescoping Series: Understanding the Summation Process

In a telescoping series, we often sum up to a finite number of terms (N) to understand the pattern and behavior of the series. Once we have the expression for the sum of the first N terms, we can then take the limit as N approaches infinity to find the sum of the infinite series.

Consider the finite sum of the first N terms:

\( S_N = \sum_{n=1}^{N} \left( \frac{1}{n} – \frac{1}{n+1} \right) \)

Expanding the series, we see:

\( S_N = \left( \frac{1}{1} – \frac{1}{2} \right) + \left( \frac{1}{2} – \frac{1}{3} \right) + \ldots + \left( \frac{1}{N-1} – \frac{1}{N} \right) + \left( \frac{1}{N} – \frac{1}{N+1} \right) \)

Now, we can see the cancellation pattern, and what’s left is:

\( S_N = 1 – \frac{1}{N} + \frac{1}{N} – \frac{1}{N+1} \)

The \( \frac{1}{N} \) and \( -\frac{1}{N} \) cancel each other out, leaving:

\( S_N = 1 – \frac{1}{N+1} \)

Now, to find the sum of the infinite series, we take the limit as N approaches infinity:

\( S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left( 1 – \frac{1}{N+1} \right) = 1 \)

The key to understanding the disappearance of the \( \frac{1}{N} \) part is to recognize the cancellation pattern and to see how the terms pair up to cancel each other out. By carefully tracking each term, we can see how the series collapses to a simple expression, and then we can find the limit as N goes to infinity.

Telescoping Series: A Detailed Breakdown

Consider the series:

\( S = \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} – \frac{1}{2n+1} \right) \)

We’ll break down the sum into the first N terms and then analyze what happens as N approaches infinity:

\( S_N = \sum_{n=1}^{N} \left( \frac{1}{2n-1} – \frac{1}{2n+1} \right) \)

Expanding the Sum

When we expand the sum, we notice that many terms cancel out:

\( S_N = 1 – \frac{1}{3} + \frac{1}{3} – \frac{1}{5} + \frac{1}{5} – \frac{1}{7} + \ldots – \frac{1}{2N+1} \)

The pattern continues, with every positive fraction being followed by the same negative fraction, except for the first term and the last term.

Understanding the Last Term

After all the cancellations, we’re left with:

\( S_N = 1 – \frac{1}{2N+1} \)

The last term, \(\frac{1}{2N+1}\), becomes smaller and smaller as N increases. It represents the remaining fraction after all the cancellations.

For example, when N=100, the last term is \(\frac{1}{201}\), and when N=1000, it’s \(\frac{1}{2001}\). As N goes to infinity, this term approaches zero.

Finding the Infinite Sum

To find the sum of the infinite series, we take the limit as N approaches infinity:

\( S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \left( 1 – \frac{1}{2N+1} \right) = 1 \)

The last term \(\frac{1}{2N+1}\) vanishes as N goes to infinity, so the sum of the infinite series is 1.

This detailed explanation helps to understand how the last term behaves and why it doesn’t affect the sum of the infinite series. The telescoping series collapses to a simple expression, and the last term becomes negligible as we consider more and more terms.

The geometric series ∑ xⁿ from n=0 to ∞ has different behaviors depending on the value of x. For -1 < x < 1, the series converges, and the sum is given by 1 / (1 - x). When x = 1, the series 1 + 1 + 1 + ... diverges, and there is no finite sum. When x = -1, the series 1 - 1 + 1 - 1 + ... also diverges, and the sum does not converge to a finite value. These special cases illustrate the importance of the condition -1 < x < 1 for the sum formula to be valid.

Understanding the Sum of \( x^n \): A Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed number called the common ratio. In this case, the common ratio is \( x \), and the series can be expressed using sigma notation:

\( S = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \ldots \)

This series represents a pattern where you start with 1 and then multiply by \( x \) again and again, adding up all the terms.

Finding the Sum

If the common ratio \( x \) is between -1 and 1 (excluding -1 and 1), the terms get smaller and smaller, and the sum approaches a specific value. We can find this value using a clever trick:

1. Let \( S \) be the sum of the series: \( S = 1 + x + x^2 + x^3 + \ldots \)

2. Multiply both sides by \( x \): \( xS = x + x^2 + x^3 + x^4 + \ldots \)

3. Subtract the second equation from the first: \( S – xS = 1 \Rightarrow S = \frac{1}{1 – x} \)

This formula gives us the sum of the series when \( x \) is between -1 and 1. It’s a powerful insight into how the individual terms of the series combine to create a finite sum, and it has many applications in mathematics and real life.

Special Case: \( x = 1 \) in the Geometric Series

When \( x = 1 \), the geometric series becomes:

\( S = \sum_{n=0}^{\infty} 1^n = 1 + 1 + 1 + 1 + \ldots \)

Since each term is equal to 1, the series does not converge to a finite value. Instead, the sum grows without bound as more and more terms are added. In mathematical terms, we say that the sum diverges.

Expression \( \frac{1}{1 – x} \)

The expression \( \frac{1}{1 – x} \) gives the sum of the geometric series when \( x \) is between -1 and 1 (excluding -1 and 1). When \( x = 1 \), the expression becomes:

\( \frac{1}{1 – 1} = \frac{1}{0} \)

This expression is undefined because division by zero is not allowed in mathematics. It reflects the fact that the sum of the series diverges when \( x = 1 \), and there is no finite value that represents the sum.

In summary, when \( x = 1 \), both the geometric series and the expression \( \frac{1}{1 – x} \) indicate that the sum does not converge to a finite value. It’s a special case that illustrates the importance of the condition \( -1 < x < 1 \) for the sum formula to be valid.

Special Case: \( x = -1 \) in the Geometric Series

When \( x = -1 \), the geometric series becomes:

\( S = \sum_{n=0}^{\infty} (-1)^n = 1 – 1 + 1 – 1 + 1 – 1 + \ldots \)

The series alternates between 1 and -1, and there is no consistent pattern towards a specific value. In mathematical terms, we say that the sum does not converge to a finite value, and it is considered divergent.

Expression \( \frac{1}{1 – x} \)

The expression \( \frac{1}{1 – x} \) gives the sum of the geometric series when \( x \) is between -1 and 1 (excluding -1 and 1). When \( x = -1 \), the expression becomes:

\( \frac{1}{1 – (-1)} = \frac{1}{2} \)

However, this value does not represent the sum of the series when \( x = -1 \), as the series does not converge. The expression \( \frac{1}{1 – x} \) is only valid for \( -1 < x < 1 \), and it does not provide a meaningful result for \( x = -1 \).

In summary, when \( x = -1 \), both the geometric series and the expression \( \frac{1}{1 – x} \) indicate that the sum does not converge to a finite value. It’s another special case that illustrates the importance of the condition \( -1 < x < 1 \) for the sum formula to be valid.

Geometric Series and Partial Sums

The plot above illustrates both the terms of the geometric series (2/3)ⁿ and the sequence of its partial sums:

• The blue dots represent the terms of the series (2/3)ⁿ, which decrease towards zero as n increases.
• The orange dots represent the partial sums of the series, which converge to a specific value.

This visualization helps to understand how the individual terms of the series decrease, while the sum of the series approaches a finite value.

Formation of Partial Sums

Here are some calculations showing the formation of the partial sums for a few samples:

Sum for n=1: (2/3)¹ = 2/3

Sum for n=2: (2/3)¹ + (2/3)² = 2/3 + 2/3 = 4/3

Sum for n=3: (2/3)¹ + (2/3)² + (2/3)³ = 2/3 + 2/3 + 2/3 = 2

Sum for n=4: (2/3)¹ + (2/3)² + (2/3)³ + (2/3)⁴ = 2/3 + 2/3 + 2/3 + 2/3 = 8/3

Actual Sum of the Infinite Series

The sum of the infinite geometric series with a common ratio of 2/3 and the first term 1 is given by:

Sum = 1 / (1 – 2/3) = 3

As n increases, the sum of the partial sums approaches this value, demonstrating the convergence of the series.

Geometric Series and Partial Sums

The plot above illustrates both the terms of the geometric series (4/3)ⁿ and the sequence of its partial sums:

• The blue dots represent the terms of the series (4/3)ⁿ, which increase as n increases.
• The orange dots represent the partial sums of the series, which diverge, as the series does not converge.

This visualization helps to understand how the individual terms of the series increase without bound, and the sum of the series does not approach a finite value.

Formation of Partial Sums

Here are some calculations showing the formation of the partial sums for a few samples:

Sum for n=1: (4/3)¹ = 4/3

Sum for n=2: (4/3)¹ + (4/3)² = 4/3 + 16/9 = 28/9

Sum for n=3: (4/3)¹ + (4/3)² + (4/3)³ = 4/3 + 16/9 + 64/27 = 100/27

Sum for n=4: (4/3)¹ + (4/3)² + (4/3)³ + (4/3)⁴ = 4/3 + 16/9 + 64/27 + 256/81 = 388/81

Actual Sum of the Infinite Series

The series with a common ratio of 4/3 does not converge, so the sum of the infinite series does not exist.

Series and Partial Sums of \((-1)^n/n\)

Formation of Partial Sums

Here are some calculations showing the formation of the partial sums for a few samples:

Sum for n=1: \((-1)^1/1 = -1\)

Sum for n=2: \((-1)^1/1 + (-1)^2/2 = -1 + 1/2 = -1/2\)

Sum for n=3: \((-1)^1/1 + (-1)^2/2 + (-1)^3/3 = -1 + 1/2 – 1/3 = -5/6\)

Sum for n=4: \((-1)^1/1 + (-1)^2/2 + (-1)^3/3 + (-1)^4/4 = -1 + 1/2 – 1/3 + 1/4 = -7/12\)

…

This visualization helps to understand how the individual terms of the series oscillate, while the sum of the series does not converge to a finite value.

The blue dots represent the terms of the series \((-1)^n\), which alternate between -1 and 1 as \(n\) increases.

The orange dots represent the partial sums of the series, which alternate between -1 and 0 as \(n\) increases.

Partial Sums of the Series (-1)ⁿ/ⁿ

The following are the first 8 partial sums for the series (-1)ⁿ/ⁿ:

1. (-1)¹/1 = -1 ≈ -1.000

2. (-1)¹/1 + (-1)²/2 = -1 + 1/2 = -1/2 ≈ -0.500

3. (-1)¹/1 + (-1)²/2 + (-1)³/3 = -1 + 1/2 – 1/3 = -11/6 ≈ -1.833

4. (-1)¹/1 + (-1)²/2 + (-1)³/3 + (-1)⁴/4 = -1 + 1/2 – 1/3 + 1/4 = -7/12 ≈ -0.583

5. (-1)¹/1 + (-1)²/2 + (-1)³/3 + (-1)⁴/4 + (-1)⁵/5 = -1 + 1/2 – 1/3 + 1/4 – 1/5 = -37/60 ≈ -0.617

6. (-1)¹/1 + (-1)²/2 + (-1)³/3 + (-1)⁴/4 + (-1)⁵/5 + (-1)⁶/6 = -1 + 1/2 – 1/3 + 1/4 – 1/5 + 1/6 = -49/60 ≈ -0.817

7. (-1)¹/1 + (-1)²/2 + (-1)³/3 + (-1)⁴/4 + (-1)⁵/5 + (-1)⁶/6 + (-1)⁷/7 = -1 + 1/2 – 1/3 + 1/4 – 1/5 + 1/6 – 1/7 = -363/420 ≈ -0.864

8. (-1)¹/1 + (-1)²/2 + (-1)³/3 + (-1)⁴/4 + (-1)⁵/5 + (-1)⁶/6 + (-1)⁷/7 + (-1)⁸/8 = -1 + 1/2 – 1/3 + 1/4 – 1/5 + 1/6 – 1/7 + 1/8 = -67/105 ≈ -0.638

This series is known as the alternating harmonic series, and these calculations represent the partial sums up to the 8th term.

Terms of the Sequence \((-1)^n/n\)

The blue dots represent the terms of the sequence \((-1)^n/n\), showing the alternating pattern as \(n\) increases.

Partial Sums of the Sequence \((-1)^n/n\)

The orange dots represent the partial sums of the sequence \((-1)^n/n\), showing how the sum fluctuates as \( n \) increases.

A geometric series is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This page provides a step-by-step guide to deriving the sum of an infinite geometric series, explaining how to identify the common ratio (r), write down the series (S = a + ar + ar² + …), multiply by the common ratio (rS = ar + ar² + …), subtract the two series (S – rS = a), and finally solve for the sum (S = a / (1 – r)). Examples include the geometric sum of 2ⁿ/3ⁿ⁻¹ from n=1 to n=∞, which converges to 9, the geometric sum of 2ⁿ⁺¹/3ⁿ⁻² from n=1 to n=∞, which converges to 54, and the geometric sum of 2²ⁿ * 3¹⁻ⁿ from n=1 to n=∞, which does not converge. Whether you’re a math enthusiast or a total beginner, you’ll find these explanations accessible and clear.

Deriving the Sum of an Infinite Geometric Series

A geometric series is a series where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. An infinite geometric series continues indefinitely.

Example: Geometric Series

Consider a geometric series with a common ratio \( r \) and the first term \( a \):

\[ S = \sum_{n=0}^{\infty} ar^n \]

We want to find the sum of this series, assuming it converges (i.e., the sum approaches a specific value).

Step 1: Identify the Common Ratio

The common ratio \( r \) is the factor by which each term is multiplied to get the next term. It must be between -1 and 1 for the series to converge.

Step 2: Write Down the Series

Write down the series as:

\[ S = a + ar + ar^2 + ar^3 + \ldots \]

Step 3: Multiply the Series by the Common Ratio

Multiply the entire series by \( r \):

\[ rS = ar + ar^2 + ar^3 + ar^4 + \ldots \]

Notice that we’ve simply shifted the terms one position to the right.

Step 4: Subtract the Two Series

Subtract the second series from the first:

\[ S – rS = (a + ar + ar^2 + ar^3 + \ldots) – (ar + ar^2 + ar^3 + ar^4 + \ldots) \]

The terms after the first in the series cancel out:

\[ S – rS = a \]

Step 5: Solve for the Sum

We now have a simple equation with one unknown, \( S \):

\[ S(1 – r) = a \]

Divide by \( (1 – r) \):

\[ S = \frac{a}{1 – r} \]

Conclusion

The sum of an infinite geometric series with a common ratio \( r \) (where \( -1 < r < 1 \)) and the first term \( a \) is given by:

\[ S = \frac{a}{1 – r} \]

This derivation shows how the sum of an infinite geometric series can be found by manipulating the series itself. By multiplying the series by the common ratio and then subtracting, we arrive at a simple expression for the sum. This method provides a powerful tool for finding the sum of a converging geometric series.

Geometric Sum of \(\frac{2^n}{3^{n-1}}\) from \(n=1\) to \(n=\infty\)

The formula for the sum of an infinite geometric series with common ratio \(r\) is given by:

\( \text{Sum} = \frac{a}{1 – r} \), where \(a\) is the first term and \(r\) is the common ratio.

\( a_n = \frac{2^n}{3^{n-1}} \)	Given general term
\( a_n = \frac{2^n}{3^n} \times \frac{3}{3} \)	Multiply by \(\frac{3}{3} = 1\) to align the exponents
\( a_n = \frac{2^n \times 3}{3^n \times 3} \)	Rewrite the expression
\( a_n = \frac{3 \times 2^n}{3^n} \)	Combine the numerators
\( a_n = 3 \times \frac{2^n}{3^n} \)	Rewrite as a product
\( a_n = 3 \times \left(\frac{2}{3}\right)^n \)	Express \(\frac{2^n}{3^n}\) as \(\left(\frac{2}{3}\right)^n\)
\( \text{Sum} = \frac{3}{1 – \frac{2}{3}} \)	Apply formula for sum of infinite geometric series
\( \text{Sum} = \frac{3}{\frac{1}{3}} \)	Simplify the denominator
\( \text{Sum} = 9 \)	Simplify the expression to find the sum

The sum of the geometric series \(\frac{2^n}{3^{n-1}}\) from \( n=1 \) to \( n=\infty \) converges to 9.

Geometric Sum of \(\frac{2^{n+1}}{3^{n-2}}\) from \(n=1\) to \(n=\infty\)

The formula for the sum of an infinite geometric series with common ratio \(r\) is given by:

\( \text{Sum} = \frac{a}{1 – r} \), where \(a\) is the first term and \(r\) is the common ratio.

\( a_n = \frac{2^{n+1}}{3^{n-2}} \)	Given general term
\( a_n = \frac{2^{n+1} \cdot 2^{-1} \cdot 3^2}{3^{n-2} \cdot 2^{-1} \cdot 3^2} \)	Multiply by \(\frac{2^{-1}}{2^{-1}} \cdot \frac{3^2}{3^2}\), which is equivalent to multiplying by 1. This specific form of 1 helps to align the exponents of 2 and 3 in the numerator and denominator.
\( a_n = \frac{2^n \cdot 2 \cdot 9}{3^n} \)	Simplify the expression by combining like terms. The multiplication by our specific form of 1 has allowed us to express the term in a more recognizable geometric form.
\( a_n = 18 \times \left(\frac{2}{3}\right)^n \)	Combine constants and express \(\frac{2^n}{3^n}\) as \(\left(\frac{2}{3}\right)^n\). This is now in the standard form for a geometric series with first term 18 and common ratio \(\frac{2}{3}\).
\( \text{Sum} = \frac{18}{1 – \frac{2}{3}} \)	Apply formula for sum of infinite geometric series, using the values we found for the first term and common ratio.
\( \text{Sum} = 54 \)	Simplify the expression to find the sum

The sum of the geometric series \(\frac{2^{n+1}}{3^{n-2}}\) from \( n=1 \) to \( n=\infty \) converges to 54. The key step was multiplying by a specific form of 1 that allowed us to align the exponents and express the term in standard geometric form.

Geometric Sum of \(2^{2n} \cdot 3^{1-n}\) from \(n=1\) to \(n=\infty\)

We’ll start by expressing the given series in a more recognizable form, and then we’ll determine if it converges.

\( a_n = 2^{2n} \cdot 3^{1-n} \)	Given general term
\( a_n = (2^2)^n \cdot 3 \cdot 3^{-n} \)	Expressing \(2^{2n}\) as \((2^2)^n\) and splitting the 3’s
\( a_n = 4^n \cdot 3 \cdot \frac{1}{3^n} \)	Expressing \(3^{-n}\) as \(\frac{1}{3^n}\) and \(2^2\) as 4
\( a_n = 3 \cdot (4/3)^n \)	Combining terms to get the expression into the form \(a \cdot r^n\)
\( r = \frac{4}{3} \)	Common ratio

Since the common ratio \( r = \frac{4}{3} \) is greater than 1, the series does not converge, and the sum does not exist.

This detailed breakdown shows each step of the simplification process, using valid algebraic operations and careful manipulation of the expression. The key was recognizing how to multiply by various forms of 1 to combine terms and reveal the common ratio.

Introduction to Series

A series is the sum of the terms of a sequence. It’s a way to represent and analyze the sum of infinitely many terms. While a sequence is a list of numbers, a series is the sum of those numbers.

A series is denoted by the summation symbol (Σ) and is defined as S = Σ (from n=1 to ∞) aₙ, where aₙ is the nth term of the sequence.

A partial sum is the sum of the first n terms of a series, denoted as Sₙ = Σ (from i=1 to n) aᵢ. The sequence of partial sums (Sₙ) provides insight into the behavior of the series and is a fundamental concept in understanding series.

Examples include the series defined by aₙ = n, aₙ = 1/n, and an oscillating series where aₙ alternates between -1 and 1. The total sum of a series is the limit of the sequence of partial sums, and this relationship helps us understand whether the series converges or diverges.

An important example is the geometric series, represented as Σ (from n=0 to N-1) arⁿ or Σ (from n=1 to ∞) ar^(n-1). Both representations produce the same terms, and the sum of a geometric series can be derived using specific formulas.

Shifting the index in summation notation involves changing the starting and ending values of the index and adjusting the expression inside the summation. For example, Σ (from n=0 to N-1) arⁿ can be shifted to Σ (from n=1 to N) ar^(n-1).

This introduction covers the essential concepts of series, including definitions, partial sums, examples, and techniques for working with series.

Introduction to Series

A series is the sum of the terms of a sequence. It’s a way to represent and analyze the sum of infinitely many terms. While a sequence is a list of numbers, a series is the sum of those numbers.

Definition of a Series

A series is denoted by the summation symbol and is defined as:

\[ S = \sum_{n=1}^{\infty} a_n \]

where \( a_n \) is the nth term of the sequence.

Partial Sums

A partial sum is the sum of the first \( n \) terms of a series. It is denoted as:

\[ S_n = \sum_{i=1}^{n} a_i \]

The sequence of partial sums \( (S_n) \) provides insight into the behavior of the series and is a fundamental concept in understanding series.

Example of the Series Defined by aₙ = n

Consider the series defined by the sequence aₙ = n:

S = Σ (from n=1 to ∞) n = 1 + 2 + 3 + 4 + 5 + …

We can compute several partial sums to see how the series behaves:

• S₁ = Σ (from i=1 to 1) i = 1
• S₂ = Σ (from i=1 to 2) i = 1 + 2 = 3
• S₃ = Σ (from i=1 to 3) i = 1 + 2 + 3 = 6
• S₄ = Σ (from i=1 to 4) i = 1 + 2 + 3 + 4 = 10
• S₅ = Σ (from i=1 to 5) i = 1 + 2 + 3 + 4 + 5 = 15

The sequence of partial sums is a new sequence {S₁, S₂, S₃, …}, and the total sum of the series is the limit of this sequence of partial sums.

In this case, the sequence of partial sums grows without bound, so the limit does not exist. Therefore, the series is divergent, meaning that it does not converge to a specific value.

The relationship between the total sum of a series and the limit of the sequence of partial sums is a fundamental concept in the study of series. It helps us understand how the individual terms contribute to the overall sum and whether the series converges or diverges.

Example of the Series Defined by aₙ = 1/n

Consider the series defined by the sequence aₙ = 1/n:

S = Σ (from n=1 to ∞) 1/n = 1 + 1/2 + 1/3 + 1/4 + 1/5 + …

We can compute several partial sums to see how the series behaves:

• S₁ = Σ (from i=1 to 1) 1/i = 1
• S₂ = Σ (from i=1 to 2) 1/i = 1 + 1/2 = 1.5
• S₃ = Σ (from i=1 to 3) 1/i = 1 + 1/2 + 1/3 ≈ 1.833
• S₄ = Σ (from i=1 to 4) 1/i = 1 + 1/2 + 1/3 + 1/4 ≈ 2.083
• S₅ = Σ (from i=1 to 5) 1/i = 1 + 1/2 + 1/3 + 1/4 + 1/5 ≈ 2.283

The sequence of partial sums is a new sequence {S₁, S₂, S₃, …}, and the total sum of the series is the limit of this sequence of partial sums.

In this case, the sequence of partial sums continues to grow without bound as we add more terms, even though the individual terms are getting smaller. Therefore, the limit does not exist, and the series is divergent.

This series is known as the harmonic series, and its divergence is a classic result in mathematics. It illustrates how a series can diverge even if the individual terms are getting smaller and smaller.

Example of an Oscillating Series

Consider the series defined by the sequence \( a_n = (-1)^n \), where each term alternates between -1 and 1:

\[ S = \sum_{n=0}^{\infty} (-1)^n = 1 – 1 + 1 – 1 + 1 – 1 + \ldots \]

We can compute several partial sums to see how the series behaves:

• \( S_1 = \sum_{i=0}^{0} (-1)^i = 1 \)
• \( S_2 = \sum_{i=0}^{1} (-1)^i = 1 – 1 = 0 \)
• \( S_3 = \sum_{i=0}^{2} (-1)^i = 1 – 1 + 1 = 1 \)
• \( S_4 = \sum_{i=0}^{3} (-1)^i = 1 – 1 + 1 – 1 = 0 \)
• \( S_5 = \sum_{i=0}^{4} (-1)^i = 1 – 1 + 1 – 1 + 1 = 1 \)

As we can see, the partial sums oscillate between 0 and 1, depending on whether the number of terms is even or odd. This pattern continues indefinitely, and the series does not settle down to a specific value.

This series is an example of a divergent series, meaning that it does not converge to a specific value. The oscillating behavior illustrates the importance of carefully considering the convergence or divergence of a series, especially when dealing with infinite sums.

Sum of a Series with Given Limit of Partial Sums

Suppose we have a series defined by the sequence aₙ, and we know that the limit of the partial sums (the sum of the first n terms) is given by:

lim (n → ∞) sₙ = 3n / (4n + 7)

Then the sum of the series, if it converges, is the limit of this expression as n approaches infinity:

Sum of series = lim (n → ∞) 3n / (4n + 7)

When dealing with limits as n approaches infinity, constants and lower-order terms often become insignificant relative to the leading terms. In this case, the constant 7 in the denominator becomes insignificant compared to the 4n term as n grows large.

So we can focus on the leading terms and write:

Sum of series ≈ lim (n → ∞) 3n / 4n

Now we can cancel the n’s:

Sum of series = lim (n → ∞) 3 / 4 = 3 / 4

So the sum of the series converges to 3/4.

This example illustrates how the limit of the sequence of partial sums can be used to find the sum of the series, and it highlights the idea that constants and lower-order terms can often be dropped when dealing with limits at infinity.

Two Representations of a Geometric Series

A geometric series can be represented in two different ways using summation notation, but both produce the same terms.

1. Starting from \( n=0 \) to \( N-1 \):

\( \sum_{n=0}^{N-1} ar^n = a + ar + ar^2 + ar^3 + \ldots \)

This representation starts with \( n=0 \) and goes up to \( N-1 \), where \( N \) is the number of terms in the partial sum.

2. Starting from \( n=1 \) to \( \infty \):

\( \sum_{n=1}^{\infty} ar^{(n-1)} = a + ar + ar^2 + ar^3 + \ldots \)

This representation starts with \( n=1 \) and goes up to \( \infty \), and uses \( ar^{(n-1)} \) to produce the same terms as the first representation.

Comparison:

Both representations produce the same terms:

• When \( n=0 \) in the first representation, the term is \( ar^0 = a \).
• When \( n=1 \) in the second representation, the term is \( ar^{(1-1)} = ar^0 = a \).
• Both continue with the same pattern: \( ar, ar^2, ar^3, \ldots \)

These two representations show that the choice of indexing in summation notation can be flexible, and different choices can lead to the same series.

Shifting the Index in Summation Notation

Shifting the index in summation notation involves changing the starting and ending values of the index, as well as adjusting the expression inside the summation. Here’s how it works:

Original Representation (starting from \( n=0 \)):

\( \sum_{n=0}^{N-1} ar^n \)

Shifting the Index (starting from \( n=1 \)):

To shift the index from starting at \( n=0 \) to starting at \( n=1 \), we need to make two adjustments:

1. Adjust the Expression: Since \( n \) goes from \( 0 \) to \( 1 \), the expression has to decrease by \( 1 \). So, we replace \( n \) with \( n-1 \) inside the expression: \( ar^n \) becomes \( ar^{(n-1)} \).
2. Adjust the Bounds: The starting value of \( n \) changes from \( 0 \) to \( 1 \), and the ending value changes from \( N-1 \) to \( N \): \( \sum_{n=1}^{N} ar^{(n-1)} \).

The result is a new representation that produces the same terms as the original:

\( \sum_{n=1}^{N} ar^{(n-1)} = \sum_{n=0}^{N-1} ar^n \)

This example illustrates that shifting the index is a flexible technique that can be used to change the appearance of a series without changing its value.