# Ratio Test for Convergence of an Infinite Series

## Key Questions

• It is not always clear-cut, but if a series contains exponential functions or/and factorials, then Ratio Test is probably a good way to go.

I hope that this was helpful.

• By Ratio Test, the posted series converges absolutely.

By Ratio Test:

${\lim}_{n \to \infty} | {a}_{n + 1} / {a}_{n} | = {\lim}_{n \to \infty} | \frac{{\left(- 10\right)}^{n + 1}}{{4}^{2 n + 3} \left(n + 2\right)} \cdot \frac{{4}^{2 n + 1} \left(n + 1\right)}{{\left(- 10\right)}^{n}} |$

By canceling out common factors:

$= {\lim}_{n \to \infty} | \frac{- 10 \left(n + 1\right)}{{4}^{2} \left(n + 2\right)} |$

since $| \frac{- 10}{4} ^ 2 | = \frac{5}{8}$, we have:

$= \frac{5}{8} {\lim}_{n \to \infty} \frac{n + 1}{n + 2}$

by dividing the numerator and the denominator by $n$,

$= \frac{5}{8} {\lim}_{n \to \infty} \frac{1 + \frac{1}{n}}{1 + \frac{2}{n}} = \frac{5}{8} \cdot 1 = \frac{5}{8} < 1$

Hence, ${\sum}_{n = 1}^{\infty} \frac{{\left(- 10\right)}^{n}}{{4}^{2 n + 1} \left(n + 1\right)}$ is absolutely convergent.