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Chapter 9: Problem 13

Use the Ratio Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} k e^{-k}$$

Short Answer

Expert verified
Based on the Ratio Test, the given series $$\sum_{k=1}^{\infty} k e^{-k}$$ converges since the limit of the ratio of consecutive terms, which is $$e^{-1}$$, is less than 1.

Step by step solution

01

Write down the series term and the general formula for the ratio of consecutive terms

The given series is: $$\sum_{k=1}^{\infty} k e^{-k}$$ Let \(a_k = ke^{-k}\) represent the general term of the series. To apply the Ratio Test, we need to find the limit of the ratio \(\frac{a_{k+1}}{a_k}\), which is given by: $$\lim_{k \rightarrow \infty} \frac{a_{k+1}}{a_k} = \lim_{k \rightarrow \infty} \frac{(k+1)e^{-(k+1)}}{ke^{-k}}$$
02

Simplify the expression of the ratio

Before finding the limit, let's simplify the expression: $$\lim_{k \rightarrow \infty} \frac{(k+1)e^{-(k+1)}}{ke^{-k}} = \lim_{k \rightarrow \infty} \frac{(k+1)e^{-k}e^{-1}}{ke^{-k}}$$ Now cancelling out \(e^{-k}\) from both the numerator and the denominator, we get: $$\lim_{k \rightarrow \infty} \frac{(k+1)e^{-1}}{k}$$
03

Find the limit of the simplified expression

Now we evaluate the limit as k tends to infinity: $$\lim_{k \rightarrow \infty} \frac{(k+1)e^{-1}}{k}$$ We have two terms in the numerator with k: k and 1. Divide both terms and the denominator by k: $$\lim_{k \rightarrow \infty} \frac{(1+\frac{1}{k})e^{-1}}{1}$$ Now apply the limit: $$\lim_{k \rightarrow \infty} \frac{(1+\frac{1}{k})e^{-1}}{1} = \frac{(1+0)e^{-1}}{1} = e^{-1}$$
04

Apply the Ratio Test and conclude the convergence of the series

The limit of the ratio is \(e^{-1}\), which is approximately 0.37. Since the limit is less than 1, the Ratio Test tells us that the series converges. In conclusion, the given series $$\sum_{k=1}^{\infty} k e^{-k}$$ converges according to the Ratio Test.

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Most popular questions from this chapter

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}.\) (Assume the result of Exercise 63.)

Consider the following infinite series. a. Write out the first four terms of the sequence of partial sums. b. Estimate the limit of \(\left\\{S_{n}\right\\}\) or state that it does not exist. $$\sum_{k=1}^{\infty} \frac{3}{10^{k}}$$

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. When a biologist begins a study, a colony of prairie dogs has a population of \(250 .\) Regular measurements reveal that each month the prairie dog population increases by \(3 \%\) Let \(p_{n}\) be the population (rounded to whole numbers) at the end of the \(n\) th month, where the initial population is \(p_{0}=250\).

Reciprocals of odd squares Assume that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}\)

Determine whether the following series converge or diverge. $$\sum_{k=0}^{\infty} \frac{10}{k^{2}+9}$$
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