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Chapter 9: Problem 13

Determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k} k}{3 k+2}$$

Short Answer

Expert verified
Series: $$\sum_{k=1}^{\infty} \frac{(-1)^{k} k}{3 k+2}$$ Answer: The given series converges.

Step by step solution

01

Identify the sequence of terms

The terms of the given series can be represented by the sequence: $$a_k = \frac{k}{3k+2}$$
02

Check if the terms decrease in absolute value

To check if the terms decrease in absolute value, we will compare the \((k+1)\)-th term to the \(k\)-th term: $$|a_{k+1}| = \left|\frac{(k+1)}{3(k+1)+2}\right| \quad \text{and} \quad |a_k| = \left|\frac{k}{3k+2}\right|$$ Let's check if \(|a_{k+1}| < |a_k|\). We can ignore the absolute value signs because both sequences are always positive. $$\frac{k+1}{3(k+1)+2} < \frac{k}{3k+2}$$ $$k(3k+2) < (k+1)(3k+5)$$ $$k^2 -_k < 0$$ Since \(k \geq 1\), the inequality holds, and the terms decrease in absolute value.
03

Check if the terms approach zero

Now, let's take the limit of the sequence \(a_k\) as \(k\) approaches infinity: $$\lim_{k \to \infty} \frac{k}{3k+2} = \lim_{k \to \infty} \frac{1}{3+\frac{2}{k}}$$ Since the limit exists and equals 0, the terms approach zero.
04

Apply the Alternating Series Test

Using the Alternating Series Test, since the terms decrease in absolute value and approach zero, the alternating series converges. Therefore, the given series \(\sum_{k=1}^{\infty} \frac{(-1)^{k} k}{3 k+2}\) converges.

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Most popular questions from this chapter

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty}\left(-\frac{1}{3}\right)^{k}$$

Find the limit of the sequence $$\left\\{a_{n}\right\\}_{n=2}^{\infty}=\left\\{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \cdots\left(1-\frac{1}{n}\right)\right\\}.$$

For a positive real number \(p,\) how do you interpret \(p^{p^{p \cdot *}},\) where the tower of exponents continues indefinitely? As it stands, the expression is ambiguous. The tower could be built from the top or from the bottom; that is, it could be evaluated by the recurrence relations \(a_{n+1}=p^{a_{n}}\) (building from the bottom) or \(a_{n+1}=a_{n}^{p}(\text { building from the top })\) where \(a_{0}=p\) in either case. The two recurrence relations have very different behaviors that depend on the value of \(p\). a. Use computations with various values of \(p > 0\) to find the values of \(p\) such that the sequence defined by (2) has a limit. Estimate the maximum value of \(p\) for which the sequence has a limit. b. Show that the sequence defined by (1) has a limit for certain values of \(p\). Make a table showing the approximate value of the tower for various values of \(p .\) Estimate the maximum value of \(p\) for which the sequence has a limit.

For what values of \(x\) does the geometric series $$f(x)=\sum_{k=0}^{\infty}\left(\frac{1}{1+x}\right)^{k}$$ converge? Solve \(f(x)=3\)

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.)
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