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If I want to get the interval of convergence of $$\sum_{n=1}^{\infty}\frac{n!x^n}{n^n}$$ Applying Ratio test, I got the radius of convergence $= R = e$.

But How can I check the endpoints since the ratio test fails at the endpoints?

When I find factorials in a series, I use always ratio test .. How can we deal with factorials when ratio test fails ?!!

Note: I found similar question here, but answers were not helpful for me Find the radius of convergence for the series $\sum_{k=0}^{\infty}\frac{k!}{k^k}x^k$.

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    $\begingroup$ You might need Stirling formula. $\endgroup$ – hamam_Abdallah Dec 12 '17 at 19:42
  • $\begingroup$ The radius of convergence is the number $R$ for which the following holds: if $|x|<R$, then $\sum a_nx^n$ converges and if $|x|>R$, then $\sum a_nx^n$ diverges. In general, what happens at the boundary $|x|=R$ must be dealt in a case by case basis, and often on the complex plane different points on the circle $|z|=R$ may behave differently with respect to convergence. $\endgroup$ – Fimpellizieri Dec 12 '17 at 19:43
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For the $x=e$ case, you need knowledge about the factorial function, for example Stirling's approximation. But there is an easier inequality you can use: Solve by induction: $n!>(n/e)^n$ , which implies that the terms in your series don't tend to zero. This also handles the case $x=-e$, since the terms in your series won't tend to zero when $x=-e$ either.

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hint

The Stirling formula gives

$$\frac {n!e^n}{n^n}\sim \sqrt {2\pi n} $$

the series diverges at $x=e $.

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In this problem, Ratio test is right and the answer is 1/e that implies the radius is e. I think when the ration test fails we can set n!=(sqrt(2pin))(n/e)^n.

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Try with Raabe's test, which states that if $$\lim n \left(1-\frac{|a_n|}{|a_{n+1}|}\right)=R$$ and $R>1$, then $\sum a_n$ converges absolutely (then if $a_n>0$ then you can simply say that the sum converges), and if $R<1$ then the series $\sum|a_n|$ diverges (but you cannot say anything about $\sum a_n$ itself).

Be very careful with the differences between Raabe and ratio test, specially with the fact that convergence condition is the opposite to the usual one.

In fact for $x=e$ you will see that the series is divergent. But you can say nothing for $x=-e$ with this test, although you can analyse convergence with Leibnitz test (alternating series test).

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