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Chapter 7: Problem 52

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{2}{3^{n}-5} $$

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} \frac{2}{3^{n}-5} \) diverges.

Step by step solution

01

Identify a reference series

To compare with the given series, choose the series \( \sum_{n=1}^{\infty} \frac{1}{n} \). The series \( \frac{2}{3^n-5} \) behaves similar to \( \frac{1}{n} \) as \( n \) approaches infinity.
02

Calculate the limit

The limit comparison test calculates the limit of the ratio of the terms of the two series as \( n \) approaches infinity. If this limit is positive and finite, both series either converge or diverge. Calculate the limit:\( \lim_{n \rightarrow \infty} \frac{\frac{2}{3^n-5}}{\frac{1}{n}} = \lim_{n \rightarrow \infty} \frac{2n}{3^n-5} \)
03

Examine the limit

This limit can be examined by using L'hopital rule as it has the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). After applying L'hopital rule it becomes:\( \lim_{n \rightarrow \infty} \frac{2}{\ln(3) \cdot 3^n} \)So, \( \lim_{n \rightarrow \infty} \frac{2n}{3^n-5} = 0 \)
04

Convergence/Divergence of the series

Since the limit is 0 and finite, according to the limit comparison test, if the reference series converges or diverges, so does the given series. The reference series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is a harmonic series which is a known divergent series. Therefore, the given series \( \sum_{n=1}^{\infty} \frac{2}{3^{n}-5} \) also diverges.

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