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Chapter 12: Problem 21

Show that the series diverges. $$\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k}$$

Short Answer

Expert verified
The given series is \(\sum_{k=1}^{\infty}\left(\frac{k+1}{k}\right)^{k}\). Using the Ratio Test, we compute the limit of the ratio of consecutive terms as \(k\to\infty\), and find it is inconclusive since the limit equals 1. We instead use the Comparison Test with the Integral Test on series \(\sum_{k=1}^{\infty} \left(1+\frac{1}{k}\right)^k\). Since the integral of this series converges for \(n>0\), we conclude that the given series also diverges.

Step by step solution

01

Find the Ratio of Consecutive Terms

Compute the ratio of consecutive terms by dividing the (k+1)-th term by the k-th term and taking the absolute value. \[ \frac{a_{k+1}}{a_k}=\frac{\left(\frac{k+2}{k+1}\right)^{k+1}}{\left(\frac{k+1}{k}\right)^k} \]
02

Simplify the Ratio Expression

Simplify the ratio by considering each part separately. First, we will simplify the terms inside parentheses. \[ \frac{\left(\frac{k+2}{k+1}\right)^{k+1}}{\left(\frac{k+1}{k}\right)^k}=\frac{\left(\frac{(k+1)+1}{(k+1)}\right)^{k+1}}{\left(\frac{k}{k}+\frac{1}{k}\right)^k} \] Now, let's simplify the expression further by cancelling the common terms. \[ \frac{\left(\frac{(k+1)+1}{(k+1)}\right)^{k+1}}{\left(\frac{k}{k}+\frac{1}{k}\right)^k}=\frac{\frac{k+2}{k+1}}{\frac{k+1}{k}} \]
03

Find the Limit of the Ratio as k goes to Infinity

Compute the limit as k approaches infinity for the simplified ratio expression. \[ \lim_{k\to\infty} \frac{\frac{k+2}{k+1}}{\frac{k+1}{k}} \] We can further simplify the given expression by multiplying the numerator and denominator by k(k+1). \[ \lim_{k\to\infty} \frac{(k+2)k}{(k+1)^2} \] Now, find the highest power of k in the numerator and denominator, which in this case is k^2. Divide each term by k^2. \[ \lim_{k\to\infty} \frac{1+\frac{2}{k}}{1+\frac{2}{k}+\frac{1}{k^2}} = \frac{1}{1} \] Since the limit of the ratio is equal to 1, the Ratio Test is inconclusive. We need another method.
04

Use Comparison Test with Integral Test

The given series can be compared to the series: \[ \sum_{k=1}^{\infty} \left(1+\frac{1}{k}\right)^k \] Since \(1 < (k+1)/k\), we are confident that the given series is greater than the second series. Now, we can perform the Integral Test on the second series, \(\left(1+\frac{1}{k}\right)^k\): \[ \int_{1}^{\infty} \left(1+\frac{1}{x}\right)^x \, dx \] Unfortunately, this integral does not have a closed-form solution, but we can use the fact that the integral converges for \(n > 0\): \[ \int_{n}^{\infty} \left(1+\frac{1}{x}\right)^x \, dx < \infty \] Since our second series is greater than the first one, and its integral converges, we can conclude that the first series also converges. Therefore, the series given in the exercise diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test
The Ratio Test is a popular tool to determine the convergence or divergence of an infinite series. For the series \( \sum_{k=1}^{\infty} a_k \), the ratio test involves the ratio of the \((k+1)\)-th term and the \(k\)-th term given by:
  • \( \frac{a_{k+1}}{a_k} \)

We then evaluate the limit as \( k \to \infty \):
  • \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \)

  • If \( L < 1 \), the series converges.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.

In our original series \( \left( \frac{k+1}{k} \right)^k \), applying the ratio yields \( L = 1 \). Hence, the ratio test is inconclusive by itself, necessitating the use of other tests.
Comparison Test
The Comparison Test is useful when analyzing a difficult series by comparing it with a known series. This test involves two primary considerations:
  • Find a series \( \sum b_k \) that is similar to \( \sum a_k \).
  • Ensure that \( a_k \leq b_k \) for all terms where the series \( \sum b_k \) converges or diverges.

For the given series \( \left( \frac{k+1}{k} \right)^k \), a handy comparison is \( \left( 1 + \frac{1}{k} \right)^k \), which is similar because of their shared fractions. Suppose \( \sum b_k \) diverges. Given that \( a_k \geq b_k \), the original series \( \sum a_k \) also diverges as it is not smaller than a diverging series.
This is crucial because it relates a complex series to a simpler, well-understood series.
Integral Test
The Integral Test validates the convergence or divergence of a series by comparing it to an integral. This technique is especially suitable for positive, decreasing functions \( f(x) \). To implement this:
  • Consider the function \( f(x) \) corresponding to \( a_k \).
  • Evaluate the integral \( \int_{1}^{\infty} f(x) \, dx \).

  • If the integral converges, the series \( \sum \) converges.
  • If the integral diverges, the series \( \sum \) diverges.

In the problem, although \( \int\left(1+\frac{1}{x}\right)^x \,dx \) does not settle easily, its behavior can be analyzed to show divergence, supporting the divergence of the original series \( \left( \frac{k+1}{k} \right)^k \).
The integral overlaps with the nature of the series, aiding us in solidifying conclusions about the series' behavior.
Limit of a Sequence
The limit of a sequence \( \{ a_n \} \) addresses the behavior of terms as \( n \to \infty \). For a sequence:
  • If \( \lim_{n \to \infty} a_n = L \), the sequence converges to limit \( L \).
  • If the limit does not exist or is infinite, the sequence diverges.

In this way, understanding sequences is crucial as they form the basis of series. Ensuring the limit of terms impacts the convergence of the entire series.
In this exercise, when considering limits, we observe that simplifications resolve towards \( \lim_{k \to \infty} \left( 1 + \frac{1}{k} \right)^k \), which converges to \( e \), a property reflected in the series' behavior. This limit insight assists in comparing series, contributing to understanding why some series diverge.

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Most popular questions from this chapter

Show that $$ \sum_{x=1}^{\infty} \ln \left(\frac{k+1}{k}\right) \quad \text { diverges } $$ although $$ \ln \left(\frac{k}{k}-\frac{1}{k}\right) \rightarrow 0 $$

Expand \(f(x)\) in powers of \(x\) $$f(x)=x^{3} e^{-x^{3}}$$

Expand \(f(x)\) in powers of \(x\) $$f(x)=x^{5}(\sin x+\cos 2 x)$$

Let \(r>0\) be arbitrary. Give an example of a power series \(\sum a_{k} x^{k}\) with radius of convergence \(r\)

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