Clifland

2020-11-05

Use ana appropriate test to determine whether the series converges.
$\sum _{k=1}^{\mathrm{\infty }}\left(\frac{k!}{{20}^{k}{k}^{k}}\right)$

Tasneem Almond

Given, the series is
$\sum _{k=1}^{\mathrm{\infty }}\left(\frac{k!}{{20}^{k}{k}^{k}}\right)$
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
$\underset{k\to \mathrm{\infty }}{lim}\frac{{u}_{k}}{{u}_{k+1}}=l$, 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 ${u}_{k+1}=\frac{\left(k+1\right)!}{{20}^{k+1}\left(k+1{\right)}^{k+1}}$
$\underset{k\to \mathrm{\infty }}{lim}\frac{{u}_{k}}{{u}_{k+1}}=\underset{k\to \mathrm{\infty }}{lim}\frac{\frac{k!}{{20}^{k}{k}^{k}}}{\frac{\left(k+1\right)!}{{20}^{k+1}\left(k+1{\right)}^{k+1}}}$
$=\underset{k\to \mathrm{\infty }}{lim}\frac{k!{20}^{k+1}\left(k+1{\right)}^{k+1}}{{20}^{k}{k}^{k}\left(k+1\right)!}$
$=\underset{k\to \mathrm{\infty }}{lim}\frac{k!20\left(k+1{\right)}^{k}\left(k+1\right)}{{k}^{k}\left(k+1\right)k!}$
$=\underset{k\to \mathrm{\infty }}{lim}\frac{20\left(k+1{\right)}^{k}}{{k}^{k}}$
$=\underset{k\to \mathrm{\infty }}{lim}20\left(1+\frac{1}{k}{\right)}^{k}$
$=20e>1$
Hence the series is convergent by Ratio test.

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