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$$ \sum_{n=2}^{\infty}{1\over n\log(n!)} $$

I want to know whether the above converges...

I have no idea how to solve this question...

Please help...

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closed as off-topic by Namaste, projectilemotion, JMP, user91500, Claude Leibovici Mar 29 '17 at 7:11

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  • 1
    $\begingroup$ Lower bound $\log(n!)$ using Stirling's formula, then use a comparison test. $\endgroup$ – angryavian Mar 28 '17 at 16:54
  • $\begingroup$ You can bound $n! \geq (\frac{n}{2})^{n/2}$ then you should be able to compare your sum with $\sum_{i=1}^\infty 1/i^2$ $\endgroup$ – Zubzub Mar 28 '17 at 16:55
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Hint: Get an upper bound (for a direct comparison test) using the fact that $n! > e^n$ for sufficiently large $n$.

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Hint: $\ln n! =\sum_{k=1}^{n}\ln k.$ Now, roughly speaking, $n/2$  of those summands are $\ge \ln (n/2).$ Thus $$\ln n! \ge (n/2)\ln(n/2).$$

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  • $\begingroup$ (+1) for the 1 second victory. In a 100 meter dash, that is substantive. $\endgroup$ – Mark Viola Mar 28 '17 at 18:30
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HINT:

Note that $n!\ge \left(\frac n2\right)^{n/2}$. Hence, $\log(n!)\ge \frac{n}{2}\log(n/2)>\frac{n}{2}$.

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  • $\begingroup$ I beat you by 1 second. Let's leave both answers up. $\endgroup$ – zhw. Mar 28 '17 at 18:18
  • $\begingroup$ @zhw. It can't be a case of great minds here ... mine is so far from great that it is sad. $\endgroup$ – Mark Viola Mar 28 '17 at 18:29
  • $\begingroup$ Please let me know how I can improve my answer. I really want to given you the best answer I can. -Mark $\endgroup$ – Mark Viola Apr 12 '17 at 17:05

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