Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, indicate whether the given series converges or diverges. If it converges, find its sum. Hint: It may help you to write out the first few terms of the series. $$ \sum_{k=1}^{\infty}\left(\frac{e}{\pi}\right)^{k+1} $$

Step-by-Step Solution

Answer

The series converges, and its sum is \( \frac{e^3/\pi^3}{1 - e/\pi} \).

Step 1: Write down the series

The series given is \( \sum_{k=1}^{\infty}\left(\frac{e}{\pi}\right)^{k+1} \). It can be rewritten as \( \sum_{k=1}^{\infty} a_{k} \) where each term \( a_{k} = \left( \frac{e}{\pi} \right)^{k+1} \).

Step 2: Rewrite the first few terms

Write out the first few terms of the series to observe the pattern: \( \left( \frac{e}{\pi} \right)^2, \left( \frac{e}{\pi} \right)^3, \left( \frac{e}{\pi} \right)^4, \ldots \).

Step 3: Simplify using a common factor

Notice that each term is of the form \( \left( \frac{e}{\pi} \right)^{k+1} \), which can be rewritten as \( \left( \frac{e}{\pi} \right) \cdot \left( \frac{e}{\pi} \right)^k \). Thus, factor out \( \left( \frac{e}{\pi} \right)^2 \), giving \( \left( \frac{e}{\pi} \right)^2 \sum_{k=1}^{\infty} \left( \frac{e}{\pi} \right)^{k} \).

Step 4: Identify the series type

The series \( \sum_{k=1}^{\infty} \left( \frac{e}{\pi} \right)^k \) is a geometric series. A geometric series has the form \( \sum_{k=0}^{\infty} ar^k \) where \( a \) is the initial term and \( r \) is the common ratio.

Step 5: Check for convergence

For a geometric series \( \sum_{k=0}^{\infty} ar^k \) to converge, the absolute value of the common ratio \( r \) must be less than 1. Here, \( r = \frac{e}{\pi} \). Since \( e \approx 2.718 < \pi \approx 3.141 \), we have \( \left| \frac{e}{\pi} \right| < 1 \). Thus, the series converges.

Step 6: Find the sum of the geometric series

The sum of a convergent geometric series \( \sum_{k=0}^{\infty} ar^k \) is \( \frac{a}{1-r} \). Our series starts at \( k=1 \), so the first term considered is \( ar \) and the sum of \( \sum_{k=1}^{\infty} r^k \) is \( \frac{r}{1-r} \). Substitute \( r = \frac{e}{\pi} \): \( \frac{\frac{e}{\pi}}{1-\frac{e}{\pi}} \).

Step 7: Calculate the sum of the original series

Since \( \left( \frac{e}{\pi} \right)^2 \) was factored out, multiply it by the sum found. The total sum is: \[ \left( \frac{e}{\pi} \right)^2 \cdot \frac{\frac{e}{\pi}}{1-\frac{e}{\pi}} = \frac{e^3/\pi^3}{1 - e/\pi}. \]

Key Concepts

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 constant, known as the common ratio. This series is represented as \( \sum_{k=0}^{\infty} ar^k \), where \( a \) is the first term and \( r \) is the common ratio.
Understanding geometric series is key, as they appear often in mathematics and physics. For our original problem, the series is identified as geometric since it follows the form \( \left( \frac{e}{\pi} \right)^k \), with \( a = 1 \) and the common ratio \( r = \frac{e}{\pi} \).
Geometric series have straightforward characteristics and are easy to identify and work with, once you recognize the repetitive multiplication pattern involving the common ratio. Given this context, identifying and understanding this pattern allows us to simplify and determine the sum or convergence of the series.

sum of series

The sum of an infinite geometric series is one of the simpler sums to compute, owing to its pattern of multiplication by the common ratio. When such a series is convergent, the sum formula comes into play.
For a converging geometric series \( \sum_{k=0}^{\infty} ar^k \), the sum can be calculated using the formula \( \frac{a}{1-r} \). The series must start at \( k=0 \), but in our case, since the series begins from \( k=1 \), we adapt: \( \sum_{k=1}^{\infty} ar^k \) results in \( \frac{ar}{1-r} \).
In our example with \( r = \frac{e}{\pi} \), the sum becomes \( \frac{\frac{e}{\pi}}{1-\frac{e}{\pi}} \). Always ensure that the series starts at the right index and adjust the formula accordingly.
Once you have this framework down, the sum of any convergent geometric series is at your fingertips.

convergence criteria

Convergence in the context of series, especially geometric ones, depends primarily on the common ratio \( r \). This is the central factor in determining if an infinite series will sum to a finite value.
For a geometric series \( \sum_{k=0}^{\infty} ar^k \) to converge, the absolute value of the common ratio must be less than 1. Mathematically, this is expressed as \( |r| < 1 \).
In the given exercise, this condition is satisfied because \( e \approx 2.718 \) and \( \pi \approx 3.141 \), ensuring that \( \frac{e}{\pi} < 1 \).
This simple test helps you quickly verify if the series will converge and if you can subsequently apply the sum formula. Determining the convergence guarantees not only that the series has a sum, but that mathematical operations to find it are well-founded.