Divergence of the serie $\sum \frac{n^n}{n!}(\frac{1}{e})^n$

Question

Show that the serie

$$\sum \frac{n^n}{n!} \big(\frac{1}{e}\big)^n$$

Diverges.

The ratio test is inconclusive and the limit of the term is zero. So I think we should use the comparasion test. But I couldnt find any function to use, I've tried the harmonic ones, but doesnt work, since I cant calculate the limits. My guess is that we shpuld use $\frac{1}{n}$ Any hint?

Thanks!

If you know Stirling's approximation this should be straightforward; just pick a form that gives you an upper bound on $n!$. — Steven Stadnicki, Sep 25 '14 at 01:26

score 2 · Accepted Answer · answered Sep 25 '14 at 03:46

The trick here is to use Stirling's Approximation for $n!$ to provide a lower bound for the series. In particular, since $n!\lt e\sqrt{n}\left(\dfrac ne\right)^n$, then $\dfrac {n^n}{n!}\left(\dfrac1e\right)^n\gt \dfrac1{e\sqrt n}$. Can you do the rest given this?

RRL · Answer 2 · 2014-09-25T03:45:43.880

1

The series $\sum_{n=1}^{\infty}1/\sqrt{n}$ diverges (by comparison with the harmonic series).

Stirling's asymptotic formula is,

$$n! \sim \sqrt{2\pi n}\frac{n^n}{e^n}.$$

Hence

$$\lim_{n \rightarrow \infty}\frac{\frac{n^n}{e^nn!}}{\frac1{\sqrt{2\pi n}}} = 1.$$

By the limit comparison test the series $\sum_{n=1}^{\infty}\frac{n^n}{e^nn!}$ diverges.

edited Sep 25 '14 at 03:45

answered Sep 25 '14 at 01:22

RRL

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Divergence of the serie $\sum \frac{n^n}{n!}(\frac{1}{e})^n$

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