Get the answer to "Use ana appropriate test to determine whether the..."

Clifland

Answered question

2020-11-05

Use ana appropriate test to determine whether the series converges.

\sum_{k = 1}^{\infty} (\frac{k!}{20^{k} k^{k}})

Answer & Explanation

Tasneem Almond

Skilled2020-11-06Added 91 answers

Given, the series is
$\sum_{k = 1}^{\infty} (\frac{k!}{20^{k} k^{k}})$
We have to check whether the series is convergent or divergent.
Use Ratio test,
If $\sum u_{k}$ is a series of positive terms such that
$lim_{k \to \infty} \frac{u_{k}}{u_{k + 1}} = l$ , then
$\sum u_{k}$ is convergent if $l > 1$
$\sum u_{k}$ is divergent if $l < 1$
Test fail when $l = 1$
$u_{k} = \frac{k!}{20^{k} k^{k}}$ and $u_{k + 1} = \frac{(k + 1)!}{20^{k + 1} (k + 1)^{k + 1}}$
$lim_{k \to \infty} \frac{u_{k}}{u_{k + 1}} = lim_{k \to \infty} \frac{\frac{k!}{20^{k} k^{k}}}{\frac{(k + 1)!}{20^{k + 1} (k + 1)^{k + 1}}}$
$= lim_{k \to \infty} \frac{k! 20^{k + 1} (k + 1)^{k + 1}}{20^{k} k^{k} (k + 1)!}$
$= lim_{k \to \infty} \frac{k! 20 (k + 1)^{k} (k + 1)}{k^{k} (k + 1) k!}$
$= lim_{k \to \infty} \frac{20 (k + 1)^{k}}{k^{k}}$
$= lim_{k \to \infty} 20 (1 + \frac{1}{k})^{k}$
$= 20 e > 1$
Hence the series is convergent by Ratio test.

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