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Does the series$$\sum_{n=1}^{\infty}\frac{\ln{n!}}{n^3}$$ converge or diverge? I initially thought about using the ratio test but then I got the ratio is $1$, so the test is inconclusive here.

I was thinking maybe using the comparison test? But I am not too sure which series to compare with. Any help would be appreciated.

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    $\begingroup$ Do you know Stirling's Approximation for $\ln (n!)$? $\endgroup$
    – Clayton
    Oct 2, 2018 at 1:13
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    $\begingroup$ Stirling is unnecessarily fancy! $\endgroup$ Oct 2, 2018 at 1:41

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notice $\ln n! = \ln 1 + \ln 2 + \ln 3 + .. + \ln n $

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    $\begingroup$ So, in particular, $\ln n! \le n\ln n$. $\endgroup$ Oct 2, 2018 at 1:41
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You could use Raabe's test

Given $\sum_{n=1}^\infty\dfrac{\ln(n!)}{n^3}$

$$\lim_{n\rightarrow\infty} \left(n\left(\dfrac{\dfrac{\ln(n!)}{n^3}}{\dfrac{\ln((n+1)!)}{(n+1)^3}}-1\right)\right)=2$$

Since $L>1$ the series converges.

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  • $\begingroup$ Have u used stirling's here? $\endgroup$ Sep 29, 2021 at 9:42

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