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I am trying to find out whether the following series is convergent or divergent. I already tried the Root test and the Ratio test and all of them are inconclusive.

$$\sum_{n\geq1} \frac{n^n}{e^n * n!}$$

I also have a hint from my teacher that says

$$\lim_{n\to\infty}n(e - (1 + \frac{1}{n})^n) = \frac{e}{2} $$

Thank you!

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    $\begingroup$ Try to connect the quotient you get in the ratio test with the hint from your teacher to obtain the result. $\endgroup$ – Daniel Fischer Nov 2 '15 at 14:18
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    $\begingroup$ When I compute the quotient I get something like $$\frac{\lim_{n\to\infty}(1 + \frac{1}{n}) ^ n}{e}$$ And it's always 1.... $\endgroup$ – cr.rusucosmin Nov 2 '15 at 14:27
  • $\begingroup$ See also Stirling's approximation. $\endgroup$ – Lucian Nov 2 '15 at 19:47
  • $\begingroup$ I managed solved it (answer in the comments). Thank you so much! $\endgroup$ – cr.rusucosmin Nov 3 '15 at 8:52
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I managed to solve it myself. it's quite easy. We just have to use the Raabe's Test. By this, let $$R_n=n(\frac{x_{n}}{x_{n + 1}} - 1)$$ Where $$x_n = \frac{n^n}{e^n+n!}$$ $$\frac{x_{n + 1}}{x_{n}} = \frac{1}{e}(1 + \frac{1}{n})^n$$ $$\frac{x_{n}}{x_{n + 1}} = \frac{e}{(1 + \frac{1}{n})^n}$$ $$R_n = n(\frac{e}{(1 + \frac{1}{n})^n} - 1) = \frac{n(e - (1 + \frac{1}{n}^n)^n}{(1 + \frac{1}{n})^n}$$

We know by the hint that the numerator of $R_n$ tends to $\frac{e}{2}$ and the denominator tends to $e$. Thank gives us that $$\lim_{n\to\infty}R_n = \frac{1}{2}$$ Because $\lim_{n\to\infty}R_n \lt 0$, we conclude that the series $\sum_{x_n}$ is $divergent$.

The Kelenner's answer is too complex for me and this approach seems easier.

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You have as hint that

$$(1+\frac{1}{n})^n=e-\frac{e}{2n}+o(1/n)$$ Let $\alpha\in ]1/2,1[$. Recall that $(1+\frac{1}{n})^{\alpha}=1+\frac{\alpha}{n}+o(1/n)$, and now put $v_n=n^{\alpha}u_n$ with $u_n$ your series. We have then $$\frac{v_{n+1}}{v_n}=(1-\frac{1}{2n}+o(1/n))(1+\frac{\alpha}{n}+o(1/n))=1+(\alpha-\frac{1}{2})\frac{1}{n}+o(1/n)$$ This gives that there exists $N$ such that if $n\geq N$, we have $v_{n+1}\geq v_n$. Now as $v_n>0$ for all $n$, we get a $c>0$ such that $v_n\geq c$ for all $n$. This show that $\displaystyle u_n\geq \frac{c}{n^{\alpha}}$, and as $\alpha<1$, that $\displaystyle \sum u_n=+\infty$.

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  • $\begingroup$ I'm more than sure that what you have done is correct and true, but I can only use the convergence test as we are just at the beginning of the course... $\endgroup$ – cr.rusucosmin Nov 2 '15 at 15:25
  • $\begingroup$ I managed to solve it myself. Thank you. I will accept my answer as it's not so complex. Thank you! $\endgroup$ – cr.rusucosmin Nov 2 '15 at 16:09

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