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.