# Use ana appropriate test to determine whether the series converges. sum_{k=1}^infty(frac{k!}{20^kk^k})

Question
Series
Use ana appropriate test to determine whether the series converges.
$$\displaystyle{\sum_{{{k}={1}}}^{\infty}}{\left({\frac{{{k}!}}{{{20}^{{k}}{k}^{{k}}}}}\right)}$$

2020-12-03
Given, the series is
$$\displaystyle{\sum_{{{k}={1}}}^{\infty}}{\left({\frac{{{k}!}}{{{20}^{{k}}{k}^{{k}}}}}\right)}$$
We have to check whether the series is convergent or divergent.
Use Ratio test,
If $$\displaystyle\sum{u}_{{k}}$$ is a series of positive terms such that
$$\displaystyle\lim_{{{k}\to\infty}}{\frac{{{u}_{{k}}}}{{{u}_{{{k}+{1}}}}}}={l}$$, then
$$\displaystyle\sum{u}_{{k}}$$ is convergent if $$\displaystyle{l}{>}{1}$$
$$\displaystyle\sum{u}_{{k}}$$ is divergent if$$\displaystyle{l}{<}{1}$$</span>
Test fail when $$\displaystyle{l}={1}$$
$$\displaystyle{u}_{{k}}={\frac{{{k}!}}{{{20}^{{k}}{k}^{{k}}}}}$$ and $$\displaystyle{u}_{{{k}+{1}}}={\frac{{{\left({k}+{1}\right)}!}}{{{20}^{{{k}+{1}}}{\left({k}+{1}\right)}^{{{k}+{1}}}}}}$$
$$\displaystyle\lim_{{{k}\to\infty}}{\frac{{{u}_{{k}}}}{{{u}_{{{k}+{1}}}}}}=\lim_{{{k}\to\infty}}{\frac{{{\frac{{{k}!}}{{{20}^{{k}}{k}^{{k}}}}}}}{{{\frac{{{\left({k}+{1}\right)}!}}{{{20}^{{{k}+{1}}}{\left({k}+{1}\right)}^{{{k}+{1}}}}}}}}}$$
$$\displaystyle=\lim_{{{k}\to\infty}}{\frac{{{k}!{20}^{{{k}+{1}}}{\left({k}+{1}\right)}^{{{k}+{1}}}}}{{{20}^{{k}}{k}^{{k}}{\left({k}+{1}\right)}!}}}$$
$$\displaystyle=\lim_{{{k}\to\infty}}{\frac{{{k}!{20}{\left({k}+{1}\right)}^{{{k}}}{\left({k}+{1}\right)}}}{{{k}^{{k}}{\left({k}+{1}\right)}{k}!}}}$$
$$\displaystyle=\lim_{{{k}\to\infty}}{\frac{{{20}{\left({k}+{1}\right)}^{{k}}}}{{{k}^{{k}}}}}$$
$$\displaystyle=\lim_{{{k}\to\infty}}{20}{\left({1}+{\frac{{{1}}}{{{k}}}}\right)}^{{k}}$$
$$\displaystyle={20}{e}{>}{1}$$
Hence the series is convergent by Ratio test.

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