0
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$\sum n!e^{1/n}$

I'm not sure how to prove whether this converges or diverges. I have a hunch that it converges, due to the factorial growing more slowly than an nth root, but I can't figure out how to prove it. The factorial keeps getting in the way.

I did try to take the natural log, but again I wasn't sure how to proceed with the ln. Additionally, I feel like there's a better, more elegant solution. How can I determine this series' convergence?

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    $\begingroup$ The terms (spectacularly) don't go to $0$. Automatic divergence. $\endgroup$ – André Nicolas May 10 '16 at 23:27
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    $\begingroup$ $n!$ growing more slowly than the $n$th root? You need to rethink this. $\endgroup$ – zhw. May 10 '16 at 23:29
  • $\begingroup$ My thought process was that this was similar to indeterminate cases like 0*infinity, where we don't know just by looking at it which way it will go. But I see now that I was (for whatever reason) computing this as e-->0, and this is why I thought it might be indeterminate. My mistake, now that it's pointed out I feel rather foolish for not noticing. $\endgroup$ – Alex G May 10 '16 at 23:37
  • $\begingroup$ Note: e --> 0 makes no sense $\endgroup$ – zhw. May 10 '16 at 23:39
5
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$e^{x}>1$ for any $x>0$, so $$n!e^{1/n}>n!$$

Not only the series does not converge, even the sequence of which it is the sum diverges as a factorial.

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4
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Hint: $e^{1/n} > 1.\,\,\,\,$

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