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Chapter 9: Problem 24

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

Short Answer

Expert verified
$$\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln ^{2} k}$$ Answer: Yes, the given alternating series converges.

Step by step solution

01

Determine the type of series

Before determining if our series converges, we first need to identify its type. Since the series is given by terms that alternate in signs, it is an alternating series.
02

Remember the Alternating Series Test

Now that we know that it's an alternating series, we need to apply the Alternating Series Test (AST). The AST states that an alternating series converges if the following two conditions are met: 1. The serie's terms decrease in magnitude, which means: \(a_{k+1} \leq a_k\) for all k. 2. The limit of the serie's terms goes to zero, which means: \(\lim\limits_{k \to \infty} a_k = 0\).
03

Check the first condition

For our series, the terms are given by \(a_k = \frac{1}{k \ln^2 k}\). We need to check if the terms decrease in magnitude: $$a_{k+1} = \frac{1}{(k+1) \ln^2 (k+1)} \leq \frac{1}{k \ln^2 k} = a_k$$ We can clearly see that as k increases, both the denominator of \(a_{k+1}\) and \(a_k\) increase, causing the fraction to decrease. Therefore, the first condition is met.
04

Check the second condition

We need to check if the limit of the terms goes to zero: $$\lim\limits_{k \to \infty} a_k = \lim\limits_{k \to \infty} \frac{1}{k \ln^2 k}.$$ Both k and \(\ln^2 k\) go to infinity as k goes to infinity, hence the whole fraction goes to 0: $$\lim\limits_{k \to \infty} \frac{1}{k \ln^2 k} = 0.$$ So, the second condition is met as well.
05

Conclude the convergence

Since our alternating series meets both conditions of the Alternating Series Test, we can conclude that the series converges: $$\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln ^{2} k}$$ The given series converges.

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Most popular questions from this chapter

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)^{\prime}}$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ).

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=0}^{\infty} x^{k}$$

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

Evaluate the limit of the following sequences. $$a_{n}=\frac{4^{n}+5 n !}{n !+2^{n}}$$

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0}\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G\). a. Show that \(a_{n}>b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).
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