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Chapter 9: Problem 54

Determine whether the following series converge or diverge. $$\sum_{k=0}^{\infty} \frac{10}{k^{2}+9}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Identify test for convergence

We will use the limit comparison test to determine the convergence of the given series.
02

Set up the series for comparison

We will compare the series \(\sum_{k=0}^{\infty} \frac{10}{k^2+9}\) to the convergent p-series \(\sum_{k=1}^{\infty} \frac{1}{k^2}\)
03

Compute limit

Compute the limit of the ratio of the terms: $$\lim_{k \to \infty} \frac{\frac{10}{k^2 + 9}}{\frac{1}{k^2}}$$
04

Simplify the limit expression

We obtain: $$\lim_{k \to \infty} \frac{10}{k^2 + 9} \cdot k^2$$
05

Evaluate the limit

As \(k\) goes to infinity, we have: $$\lim_{k \to \infty} \frac{10k^2}{k^2 + 9} = \lim_{k \to \infty} \frac{10}{1 + \frac{9}{k^2}} = \frac{10}{1 + 0} = 10$$
06

Apply the limit comparison test

Since the limit obtained is a finite positive number (10), the given series converges or diverges with the compared series. Since the series \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) is convergent, our given series also converges. Conclusion: The series \(\sum_{k=0}^{\infty} \frac{10}{k^{2}+9}\) converges.

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Most popular questions from this chapter

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5, n=0,1,2, \dots$$

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