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I need to solve $\displaystyle \sum_{j=0}^\infty \sqrt{\frac{j!}{j^j}}$

Does this converge or diverge?

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    $\begingroup$ We saw your question the first time, no need to re-post. The problem is you are on the wrong side of town - this is about Mathematica, not about Mathematics. $\endgroup$
    – Sektor
    Jun 15, 2014 at 19:50

3 Answers 3

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Using Stirling's Formula, we have $$ j!\sim\sqrt{2\pi j}\left(\frac{j}{e}\right)^j $$ Since $j\le e^j$ for all $j$, we have $$ \begin{align} \sqrt{\frac{j!}{j^j}} &\sim\frac{(2\pi j)^{1/4}}{e^{j/2}}\\ &\le\frac{(2\pi)^{1/4}}{e^{j/4}} \end{align} $$ Therefore, $$ \sum_{j=0}^\infty\sqrt{\frac{j!}{j^j}} $$ converges.

Using the ratio test, we have $$ \frac{\displaystyle\sqrt{\frac{(j+1)!}{(j+1)^{j+1}}}}{\displaystyle\sqrt{\frac{j!}{j^j}}}=\frac1{\sqrt{\left(1+\frac1j\right)^j}}\to\frac1{\sqrt{e}}\lt1 $$ thus, the series converges.

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x = Sum[Sqrt[j!/j^j], {j, #}] & /@ Range@30

ListLinePlot@x

enter image description here

Now decide for yourself :)

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  • $\begingroup$ NSum[Sqrt[j!/j^j], {j, 1, Infinity}] produces $3.01278 $. $\endgroup$
    – user64494
    Jun 15, 2014 at 21:14
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Pushing a bit further along user64494 's track: NSum[Sqrt[j!/j^j], {j, 1, Infinity}, WorkingPrecision -> 96]
(* 3.01277702345391 *) Maybe even further along:
Sum[Sqrt[j!/j^j],{j,1,31}]; N[%,24]
(*3.01277593270324582978550*)
and the rest from j=32 to infinity can be approximated by:
NSum[Normal[Series[Sqrt[j!/j^j],{j,\[Infinity],12}]], {j, 32, \[Infinity]}, WorkingPrecision -> 96, AccuracyGoal -> 48]
(*1.09075065225228693*10^-6*)
to give a total of 3.01277702345389808207242
As to your question of convergence, remark that each term is majored by Sqrt[2^(j-1)] whose sum converges to 2+Sqrt[2]

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