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Chapter 9: Problem 63

Choose your test Use the test of your choice to determine whether the following series converge. $$\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Choose a series to compare with

We will compare our series with the convergent p-series: $$\sum_{k=2}^{\infty} \frac{1}{k^2}$$
02

Calculate the limit of the ratio between the terms of the given series and the chosen series

Set up the limit: $$\lim_{k\to\infty} \frac{\frac{1}{k^{\ln k}}}{\frac{1}{k^2}}$$ Simplify the limit expression: $$\lim_{k\to\infty} \frac{k^2}{k^{\ln k}}$$
03

Use logarithm properties to simplify the limit expression

Since both the numerator and the denominator have the same base, use logarithm properties to simplify the expression: $$\lim_{k\to\infty} k^{2-\ln k}$$
04

Calculate the limit

As k approaches infinity, the term \(k^{2-\ln k}\) approaches infinity, so we use L'H么pital's rule to find the limit as follows: $$\lim_{k\to\infty} \frac{2-\ln k}{1}$$ Now, as k approaches infinity, \(2-\ln k\) approaches negative infinity. This means that the limit of the ratio between the two series is 0.
05

Conclude convergence by the Limit Comparison Test

The Limit Comparison Test states that if \(\lim_{k\to\infty} \frac{a_k}{b_k} = 0\) or a finite number (not equal to 0), then both series either converge or diverge together. Since we have chosen a convergent p-series to compare with, we can conclude that our given series also converges. Thus, the series $$\sum_{k=2}^{\infty} \frac{1}{k^{\ln k}}$$ converges.

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Most popular questions from this chapter

An early limit Working in the early 1600 s, the mathematicians Wallis, Pascal, and Fermat were attempting to determine the area of the region under the curve \(y=x^{p}\) between \(x=0\) and \(x=1\) where \(p\) is a positive integer. Using arguments that predated the Fundamental Theorem of Calculus, they were able to prove that $$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^{p}=\frac{1}{p+1}$$ Use what you know about Riemann sums and integrals to verify this limit.

Convergence parameter Find the values of the parameter \(p>0\) for which the following series converge. $$\sum_{k=2}^{\infty}\left(\frac{\ln k}{k}\right)^{p}$$

Determine whether the following series converge or diverge. $$\sum_{k=2}^{\infty} \frac{4}{k \ln ^{2} k}$$

Given that \(\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90},\) show that \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{4}}=\frac{7 \pi^{4}}{720}.\) (Assume the result of Exercise 63.)

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5, n=0,1,2, \dots$$
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