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I have already found the endpoints of the interval of convergence of $$\sum_{n=1}^{\infty}\frac{n^{n}x^{n}}{e^{n}(n+2)!}$$ to be $x=\pm 1$. How do I prove that the series diverges/converges at these points?

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    $\begingroup$ This is one of those times I would recommend using Stirling's approximation on the factorial (after first rewriting it as $(n+2)(n+1)n!$ so it comes out nicer). $\endgroup$ Mar 30, 2020 at 1:27
  • $\begingroup$ @BrianMoehring I'm not sure what you are trying to say. $\endgroup$ Mar 30, 2020 at 1:30
  • $\begingroup$ @SimplyBeautifulArt Basically, I'm saying that you seem to be using $\frac{a_n}{b_n} \to 1$ means that $\sum a_n$ converges if and only if $\sum b_n$ does. I might be mis-remembering, but I thought this needs absolute convergence, or something to that effect. (deleted my previous comment, since this supercedes it) $\endgroup$ Mar 30, 2020 at 1:33
  • $\begingroup$ @BrianMoehring It does require $b_n$ to be positive and it does only work for proving absolute convergence. What of it? $\endgroup$ Mar 30, 2020 at 1:35
  • $\begingroup$ @SimplyBeautifulArt Huh... I had to actually go through the approximation, but I guess that $(n+2)$ is what saves us. If it had come out differently (e.g. replacing $(n+2)!$ with $(n+1)! / \sqrt{n}$) I don't know how we'd deal with it. In any case, my concern certainly doesn't matter here. $\endgroup$ Mar 30, 2020 at 1:45

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We can derive weak Stirling bounds by noticing that

\begin{align}\ln(n!)&=\sum_{k=1}^n\ln(k)\\&\ge\int_1^n\ln(t)~\mathrm dt\\&=n\ln(n)-n+1\\n!&\ge e\left(\frac ne\right)^n\end{align}

which gives us a direct comparison with

\begin{align}S&=\sum_{n=1}^\infty\frac{n^n}{e^n(n+2)!}\\&=\sum_{n=1}^\infty\frac{(n/e)^n}{(n+2)(n+1)n!}\\&\le\sum_{n=1}^\infty\frac1{e(n+2)(n+1)}\\&=\frac1e\sum_{n=1}^\infty\left(\frac1{n+1}-\frac1{n+2}\right)\\&=\frac1{2e}\end{align}

and hence it converges absolutely at both endpoints.

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