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Show that this series $$\sum_{n=1}^{\infty}\dfrac{e^n\cdot n!}{n^n}$$ is divergent.
My try: since $$u_{n}=\dfrac{e^n\cdot n!}{n^n},\Longrightarrow \dfrac{u_{n+1}}{u_{n}}=\dfrac{e^{n+1}\cdot(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^n}{e^n\cdot n!}=\dfrac{e}{\left(1+\dfrac{1}{n}\right)^n}$$ then $$\lim_{n\to\infty}\dfrac{u_{n+1}}{u_{n}}=\lim_{n\to\infty}\dfrac{e}{\left(1+\dfrac{1}{n}\right)^n}=1$$ so this limit is $1$, therefore I can't prove it by using the Ratio test.

From sos440 suggestion: by using Stirling's approximation, we have $$\dfrac{e^n\cdot n!}{n^n}\approx \sqrt{2n\pi}\to\infty.$$

Maybe this problem requires other methods.

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    $\begingroup$ If you are allowed to use Stirling's formula, then it is straightforward. $\endgroup$
    – Sangchul Lee
    Mar 8, 2014 at 4:06
  • $\begingroup$ The terms themselves do not approach $0$. Try to show that. $\endgroup$
    – 2'5 9'2
    Mar 8, 2014 at 4:07

2 Answers 2

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Since $\left(1+\dfrac{1}{n}\right)^n \nearrow e$, the succesive quotients are always greater or equal to $1$. This means that $a_{n+1}\geqslant a_n$ for each $n$. In particular $a_n\geqslant a_1$ for each $n$. Since $a_1$ is positive, the sequence doesn't go to $0$.

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  • $\begingroup$ Hello,then? Thank you $\endgroup$
    – math110
    Mar 8, 2014 at 4:11
  • $\begingroup$ @chinamath Does $a_n\to 0$? $\endgroup$
    – Pedro ♦
    Mar 8, 2014 at 4:25
  • $\begingroup$ But $n\to \infty$ $\endgroup$
    – math110
    Mar 8, 2014 at 4:38
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Using Logarithmic test..

$$ \lim_{n\to\infty}nlog\dfrac{u_{n}}{u_{n+1}}=\lim_{n\to\infty}-n+n^2log(1+\dfrac{1}{n})$$
expand $log(1+\frac{1}{n})$ into its series and u get limit as -1/2 < 1
So by log test series diverge

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