Is the following series convergent?
$$\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}$$
I treid the Ratio and Root tests, but both of them failed.
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Hint: Stirling tells us that $n!\sim \left( \frac ne \right)^n\sqrt {2\pi n}$ – lulu Feb 26 '17 at 15:16
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No. One may prove this using the Term Test. – projectilemotion Feb 26 '17 at 15:21
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No, your series is not convergent. Note that $$e^{n}=\sum_{k=0}^{\infty} \frac{n^k}{k!}>\frac{n^{n}}{n!} \implies e^{n}>\frac{n^{n}}{n!}$$ From the series expansion of $e^x$. Multiplying $\dfrac{n!}{n^n}$ on each side, we have that $$\frac{n!e^n}{n^n}>1 $$ Thus, by comparison test we have that the series diverges.
S.C.B.
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@soodehMehboodi Each terms are larger than $1$......But $\sum_{n=1}^{\infty}1 $ diverges... – S.C.B. Feb 26 '17 at 15:36
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How do you check the necessary condition for the series to converge? – soodehMehboodi Feb 26 '17 at 18:03
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1@soodehMehboodi I don't think that relates to my answer. If $a_{n}>1$ for all $n$, is a sufficient condition for $\sum\limits_{n=1}^{\infty}a_{n}$ to diverge. So, we just do comparison. We don't need the necessary condition. – S.C.B. Feb 27 '17 at 03:17
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