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Chapter 9: Problem 12

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

Short Answer

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

Step by step solution

01

Identify the Alternating Series

The given series is: $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}$$ This is an alternating series since the sign alternates between positive and negative due to the term \((-1)^{k}\).
02

Apply the Alternating Series Test

The Alternating Series Test states that if a series is of the form: $$\sum_{k=1}^{\infty} (-1)^{k}a_k$$ with \(a_k > 0\) for all \(k\), and \(a_{k+1} \leq a_k\) (the terms are decreasing) and \(\lim_{k\to\infty}a_k=0\), then the series converges. Here, \(a_k = \frac{1}{\sqrt{k}}\).
03

Check if the series' terms are decreasing

First, let's determine if the terms of the series are decreasing. Take the derivative of \(a_k\) with respect to k: $$\frac{d}{dk} a_k = \frac{d}{dk} (\frac{1}{\sqrt{k}}) = -\frac{1}{2k^{\frac{3}{2}}} < 0 $$ Since the derivative is negative, we can conclude that the terms are indeed decreasing.
04

Check if the limit of the terms is zero

Now, let's see if the limit of \(a_k\) as \(k\) approaches infinity is zero: $$\lim_{k\to\infty} a_k = \lim_{k\to\infty} \frac{1}{\sqrt{k}} = 0$$ Since the terms are decreasing, and the limit of the terms goes to zero, the Alternating Series Test tells us that the series converges. Therefore, the given series converges.

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