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I want to determine whether the series $ \sum_{n=1}^{\infty}\frac{n!}{n^{n}\sqrt{n}}e^{n} $ and the series $ \sum_{n=1}^{\infty}\frac{n!}{n^{n}\sqrt{n}}\left(-e\right)^{n} $ converge. This was a part of the question: find the convergence segment of $\sum_{n=1}^{\infty}\frac{n!}{n^{n}\sqrt{n}}x^{n} $ and I found that the convergence radius is $e$, but I'm not sure how to determine for $x=e$ or $x=-e$ because the root test cannot determine when the limit is 1. Any ideas would help. Note that we cannot use Stirling's approximation because we haven't learned it yet. So I'm looking for another way.

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    $\begingroup$ The first one is divergent since $$e^n = 1 + n + \cdots + \frac{{n^n }}{{n!}} + \cdots \ge \frac{{n^n }}{{n!}} \Rightarrow \frac{{n!}}{{n^n \sqrt n }}e^n \ge \frac{1}{{\sqrt n }}. $$ $\endgroup$
    – Gary
    May 5 '20 at 18:27
  • $\begingroup$ You should make this an answer. $\endgroup$ May 6 '20 at 6:24
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Let $u_n=\dfrac{n!}{\sqrt{n}}\left(\dfrac en\right)^n\quad $ and $\quad v_n=\ln(u_n)$

$v_{n+1}-v_n=\ln\left(\frac{u_{n+1}}{u_n}\right)=\ln\left(e\cdot\left(\frac{n}{n+1}\right)^{n+\frac 12}\right)=1-(n+\frac 12)\ln(1+\frac 1n)=\cdots=O\left(\frac 1{n^2}\right)$

Since $\sum\frac 1{n^2}$ converges, then $(v_n)_n$ converges too (by telescoping sum) and so does $u_n\to c\neq 0$ ($c$ is not zero, because it is an exponential of a finite number).

Thus the series with or without the minus sign cannot converge, since the general term does not tend to zero.

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    $\begingroup$ (+1) All too easy $\endgroup$
    – Mark Viola
    May 5 '20 at 20:48

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