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I have plugged the sum into mathematica and know that the sum diverges but I cannot prove it. When I tried the root test I was able to simplify it to $\lim_{n\to\infty}\frac{n}{(n!)^{\frac{1}{n}}e}$ which I also had to plug into mathematica and got 1 which means the test is inconclusive. The same issue arose when I tried the root test. The only thing left that I can think to do is the comparison test which would require finding something smaller than $\frac{n^n}{n!e^n}$ that diverges but I haven't had any luck.

Martin Sleziak
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Clade
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    Have you heard about Stirling's approximation? – Arthur Dec 15 '16 at 10:45
  • I had not but it seems to be smaller than n! so even though $\sum \frac{n^n}{\sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^n}=\sum \frac{1}{\sqrt{2\pi n}}$ diverges it still can't be used in the comparison test. Am I missing something? – Clade Dec 15 '16 at 10:56
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    Sorry, my last comment was wrong, I am apparently a bit distracted at the moment. Use the comparison test with the series $\sum \frac{1}{2\sqrt{2\pi n}}$, the terms of which quickly become smaller than your series. – Arthur Dec 15 '16 at 11:11
  • This is related. – J. M. ain't a mathematician Dec 15 '16 at 11:11
  • $\sum \frac2{\sqrt{2\pi n}}$ is eventually significantly greater than $\sum \frac{n^n}{n!e^n}$ therefore when I say that $\sum \frac2{\sqrt{2\pi n}}$ diverges I am not entitled to say that $\sum \frac{n^n}{n!e^n}$ also diverges. What comparison am I to make? – Clade Dec 15 '16 at 11:13
  • Ah that makes more sense thank you very much! – Clade Dec 15 '16 at 11:14

2 Answers2

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Lemma: $\left\{\left(1+\frac{1}{n}\right)^{n+1/2}\right\}_{n\geq 1}$ is a decreasing sequence converging to $e$.

Consequence: since we have $n=\prod_\limits{k=1}^{n-1}\left(1+\frac{1}{k}\right)$ for any $n\geq 2$, $$ n! = \frac{n^{n}}{\prod_\limits{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}=\frac{n^{n+1/2}}{\prod_\limits{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{k+1/2}}\leq\frac{n^{n+1/2}}{e^{n-1}} $$ and: $$ \frac{n^n}{n! e^n} \geq \frac{1}{e\sqrt{n}} $$ trivially implying that $\sum_\limits{n\geq 1}\frac{n^n}{n! e^n}$ is divergent.

Jose M Serra
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Jack D'Aurizio
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Since $\displaystyle\lim_{n\to\infty}\frac{n^n}{n!e^n}\div\frac{1}{n}=\lim_{n\to\infty}\frac{n^{n+1}}{n!e^n}=\infty\;$ (as shown below) and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$ diverges,

the series $\displaystyle\sum_{n=1}^{\infty}\frac{n^n}{n!e^n}$ diverges by the Limit Comparison Test.

Since $f(x)=\ln x$ is concave down, the Trapezoidal rule estimate gives that

$\displaystyle\int_1^n\ln x \,dx>\ln n!-\frac{1}{2}\ln n,\;$ so $\;\displaystyle\ln n!<n\ln n-n+1+\frac{1}{2}\ln n$ and therefore

$\displaystyle\ln\left(\frac{n^{n+1}}{n!e^n}\right)=(n+1)\ln n-\ln n!-n>(n+1)\ln n-[n\ln n-n+1+\frac{1}{2}\ln n]-n=\frac{1}{2}\ln n-1$.

Then $\displaystyle\lim_{n\to\infty}\ln\left(\frac{n^{n+1}}{n!e^n}\right)=\infty,\;$so $\displaystyle\lim_{n\to\infty}\frac{n^{n+1}}{n!e^n}=\infty$.

user84413
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