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Chapter 9: Problem 16

Determine whether the following series converge. $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k^{2}+10}$$

Short Answer

Expert verified
Question: Determine if the following series converges or diverges: $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k^2 + 10}$$ Answer: Converges.

Step by step solution

01

Write the general term of the series

The general term of the series is given by: $$a_k = \frac{(-1)^k}{k^2 + 10}$$
02

Show that the absolute values of the terms are non-increasing

First, we'll look at the function \(f(k) = \frac{1}{k^2+10}\), which represents the absolute value of the term \(a_k\). We want to show that \(f(k+1) \leq f(k)\) for all \(k\). Now, let's find the first derivative of \(f(k)\): $$f'(k) = -\frac{2k}{(k^2+10)^2}$$ Since \(k^2+10>0\) and \((k^2+10)^2>0\) for all \(k\), we can see that \(f'(k) \leq 0\) for all \(k\). Hence, \(f(k)\) is non-increasing. This implies that the absolute values of the terms are non-increasing: $$|a_{k+1}| \leq |a_k|$$
03

Show that the limit of the absolute value of the terms approaches zero

Now, let's find the limit of the absolute value of \(a_k\) as \(k\) goes to infinity: $$\lim_{k \to \infty} |a_k| = \lim_{k \to \infty} \frac{1}{k^2+10}$$ As \(k\) goes to infinity, \(k^2+10\) also goes to infinity. Thus: $$\lim_{k \to \infty} \frac{1}{k^2+10} = 0$$
04

Conclude using the Alternating Series Test

Since both conditions of the Alternating Series Test are satisfied 鈥� the absolute values of the terms are non-increasing and the limit of the absolute value of the terms approaches zero as \(k\) goes to infinity 鈥� we can conclude that the given series converges: $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k^{2}+10}$$ converges.

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Most popular questions from this chapter

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. A material transmutes \(50 \%\) of its mass to another element every 10 years due to radioactive decay. Let \(M_{n}\) be the mass of the radioactive material at the end of the \(n\) th decade, where the initial mass of the material is \(M_{0}=20 \mathrm{g}.\)

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5 7, \(11,13, \ldots .\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

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

Give an argument, similar to that given in the text for the harmonic series, to show that \(\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}\) diverges.

Infinite products Use the ideas of Exercise 88 to evaluate the following infinite products. $$\text { a. } \prod_{k=0}^{\infty} e^{1 / 2^{k}}=e \cdot e^{1 / 2} \cdot e^{1 / 4} \cdot e^{1 / 8} \ldots$$ $$\text { b. } \prod_{k=2}^{\infty}\left(1-\frac{1}{k}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \dots$$
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