2
$\begingroup$

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!

$\endgroup$
  • 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.