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mattgondek4

Answered question

2020-12-16

Use the Limit Comparison Test to determine the convergence or divergence of the series.

\sum_{n = 1}^{\infty} \frac{5}{4^{n} + 1}

Answer & Explanation

Lacey-May Snyder

Skilled2020-12-17Added 88 answers

Given series, $\sum_{n = 1}^{\infty} \frac{5}{4^{n} + 1}$
we have to determine the convergence or divergence of the given series.
Consider $a_{n} = \frac{5}{4^{n} + 1}$
Since $4^{n} + 1 > 4^{n}$
$\Rightarrow \frac{1}{4^{n} + 1} < \frac{1}{4^{n}}$
$\Rightarrow \frac{5}{4^{n} + 1} < \frac{5}{4^{n}}$
so, $0 < a_{n} < \frac{5}{4^{n}}$
Thus, we can compare the given series $\sum_{n = 1}^{\infty} \frac{5}{4^{n} + 1}$ with the geometric series
$\sum_{n = 1}^{\infty} \frac{5}{4^{n}} = 5 \sum_{n = 1}^{\infty} \frac{1}{4^{n}}$
this geometric series converges since $| \frac{1}{4} | < 1$
so the comparison test tells us that $\sum_{n = 1}^{\infty} \frac{5}{4^{n} + 1}$ also converges.

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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