Telescoping Series Examples with Step-by-Step Explanations

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

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

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.