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Chapter 9: Problem 34

Use the Comparison Test or Limit Comparison Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{1}{3^{k}-2^{k}}$$

Short Answer

Expert verified
Answer: Yes, the series converges.

Step by step solution

01

Write down the original series and the comparison series

Let's denote the original series by: $$a_k = \frac{1}{3^k - 2^k},$$ and we will compare this with the geometric series: $$b_k = \frac{1}{3^k}$$
02

Apply the Limit Comparison Test

According to the Limit Comparison Test, if: $$\lim_{k\to\infty} \frac{a_k}{b_k} = L$$ and \(0 < L < \infty\), then either both series converge or both diverge.
03

Calculate the limit of the ratio

Calculate the limit of the ratio of the terms as \(k\) approaches infinity: $$ \lim_{k\to\infty} \frac{a_k}{b_k} = \lim_{k\to\infty} \frac{\frac{1}{3^k - 2^k}}{\frac{1}{3^k}} = \lim_{k\to\infty} \frac{3^k}{3^k - 2^k}$$
04

Simplify the limit expression

Divide the numerator and the denominator by \(3^k\): $$ \lim_{k\to\infty} \frac{1}{1 - \frac{2^k}{3^k}}$$
05

Calculate the limit

The limit can now be computed easily: $$ \lim_{k\to\infty} \frac{1}{1 - \frac{2^k}{3^k}} = \frac{1}{1 - 0} = 1$$
06

Interpret the result

Since the limit of the ratio is 1, which is finite and greater than 0, according to the Limit Comparison Test, the original series converges if the comparison series converges. In this case, the comparison series \(\sum_{k=1}^{\infty} \frac{1}{3^k}\) is a convergent geometric series with a common ratio \(r = \frac{1}{3}\). Therefore, since the comparison series converges, the original series also converges: $$\sum_{k=1}^{\infty} \frac{1}{3^{k}-2^{k}}$$ converges.

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