Prove that the series diverges: $$\sum_{n=1}^\infty \frac{e^nn!}{n^n}$$

I've tried using the ratio test but after all of my calculations I got that $\lim_{x\to\infty}\frac{a_{n+1}}{a_n} = 1$, which I can conclude nothing from.

The root test gave me the same result: $\lim_{x\to\infty}\sqrt[n]{a_n} = 1$

I'm instructed to first try and use Bernoulli's inequality to prove that the sequence $b_n = (1 + \frac{1}{n})^n$ is monotonically increasing.

I tried to follow the instructions, and I got that for each $n\geq1$, $b_n\geq2$. From here, I can conclude that the series $\sum_{n=1}^\infty b_n$ diverges.

from here, I've tried using the limit comparison test for non-negative series and the furthest I got was to the limit: $\lim_{x\to\infty} \frac{e^nn!}{(n+1)^n}$. I can't figure how to simplify that expression any more.

I'm really frustrated as I know I'm missing some minor yet important detail. I feel as though I've done anything I know and implemented any sort of test I've learned until now, with no results. Also, I can't really understand the direction the textbook wanted me to go as for using Bernoulli's inequality. Also, I am not allowed to use Sterling's approximation as I haven't learned nor proved it.

Any sort of help is greatly appreciated!

  • 2
    $\begingroup$ How did you compute the limit for the root test? $\endgroup$ – Kenny Lau May 19 '18 at 14:27
  • 1
    $\begingroup$ $\displaystyle{\mathrm{e}^{n}n! \over n^{n}} \sim \,\sqrt{\, 2\pi\,}\, \color{red}{n^{1/2}}$ as $\displaystyle n \to \infty$. $\endgroup$ – Felix Marin May 19 '18 at 16:11

If you can use Stirling's approximation, you can easily see that $a_n \to \infty$ as shown in the other (now removed) answer, implying divergence of the series.

Without Stirling, simple way is to notice $$ \frac{a_{n+1}}{a_n} = \frac{e n^n}{(n+1)^n}>1 $$ if and only if $$ \left(1+\frac{1}{n}\right)^{n}<e, $$ but the latter is a well known, since the left side is monotically increasing (you are instructed to show it using Bernoulli inequality) and has limit $e$. Thus $a_{n+1}>a_{n}$ and the sequence is monotically increasing, thus together with $a_1=e$, necessary condition $a_n \to 0$ is not satisfied. So the corresponding series diverges.

  • $\begingroup$ Thanks for the answer! Unfortunately I can't use Stirling's approximation, but your answer gave me great intuition about how I should approach this proof. :) $\endgroup$ – 0rka May 20 '18 at 14:44

Apply the logarithm to the $n$th term in the series to get

$$\tag 1 n + \sum_{k=1}^{n}\ln k - n\ln n.$$

Now $\sum_{k=1}^{n}\ln k \ge \int_1^n \ln x\, dx.$ That last integral equals

$$x\ln x - x \,\big|_1^n = n\ln n -n +1.$$

Thus $(1) \ge 1.$ Exponentiating back shows the $n$th term of the series is $\ge e.$ Thus the $n$th term does not go to $0,$ hence the series diverges.


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