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Chapter 9: Problem 27

Use the Comparison Test or Limit Comparison Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{1}{k^{2}+4}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Applying Limit Comparison Test

In order to apply the Limit Comparison Test, define the following two series: $$a_k = \frac{1}{k^2+4} \text{ and } b_k = \frac{1}{k^2}$$ Now, calculate the limit of the ratio \(\frac{a_k}{b_k}\) as \(k\to\infty\).
02

Calculate the limit of the ratio

We will find the limit of \(\frac{a_k}{b_k}\) as \(k\to\infty\): $$\lim_{k\to\infty} \frac{a_k}{b_k} = \lim_{k\to\infty} \frac{\frac{1}{k^2+4}}{\frac{1}{k^2}}$$ Multiply the denominator and numerator by \(k^2(k^2+4)\) to simplify the expression: $$\lim_{k\to\infty} \frac{k^2}{k^2+4}$$ Next, divide each term by \(k^2\): $$\lim_{k\to\infty} \frac{1}{1+\frac{4}{k^2}}$$
03

Evaluate the limit

As we let \(k\to\infty\), \(\frac{4}{k^2}\) will approach 0. Therefore, the limit converges to 1: $$\lim_{k\to\infty} \frac{1}{1+\frac{4}{k^2}} = \frac{1}{1+0} = 1$$
04

Apply the Limit Comparison Test result

Since the limit \(\frac{a_k}{b_k}\) converges to 1, a finite positive number, we can conclude from the Limit Comparison Test that the series \(\sum_{k=1}^{\infty} \frac{1}{k^{2}+4}\) has the same convergence behavior as the series \(\sum_{k=1}^{\infty} \frac{1}{k^2}\). Moreover, we know that the latter series converges (it's a convergent p-series with p=2). Therefore, the given series \(\sum_{k=1}^{\infty} \frac{1}{k^{2}+4}\) converges.

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