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I need to solve

$$\sum_{j = 1}^{\infty} \sqrt{\frac{j!}{j^j}}$$

Does this converge or diverge and why?

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closed as off-topic by heropup, vonbrand, qwr, Davide Giraudo, Hakim Jun 15 at 21:07

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Welcome to math.SE! Can you share what you've tried, and what you're having trouble with? Do you know the ratio test, or any related tests for convergence? –  T. Bongers Jun 15 at 20:01
    
I know very little about the topic, if you could please give me some information about it as well as the answer to this it would be appreciated –  NotFreddy Jun 15 at 20:04
    
@NotFreddy Read about the ratio test, either on Wikipedia or in your textbook. –  T. Bongers Jun 15 at 20:05
    
Thank you folk! –  NotFreddy Jun 15 at 20:07
    
Do you know the Stirling approximation for the factorial? That should make this problem nearly trivial... –  Steven Stadnicki Jun 15 at 20:08

4 Answers 4

up vote 2 down vote accepted

You can use d'Alembert's ratio test

$\sqrt{\frac{(j+1)!}{(j+1)^{j+1}}}\sqrt{\frac{j^j}{j!}} = \sqrt{\frac{(j+1)!}{j!}\frac{j^j}{(j+1)^{j+1}}} = \sqrt{\frac{j^{j+1}+j^j}{(j+1)^{j+1}}}$

You can now show that this quantity converges to some real number smaller than one in order to conclude.

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Right, the ratio is $\lt1$ for every $j$ (not only for $j$ large enough)... and so what? Note that the approach in this answer would also "prove" that $\sum\limits_n\frac1n$ converges since $\frac1{n+1}\frac{n}1\lt1$ for every $n$. @upvoters: Why the upvote? –  Did Jun 17 at 8:59
    
OP: You accepted the only answer which happens to be wrong. –  Did Jun 17 at 9:00
    
Actually you are right the argument is not complete enough and if go until the end then I come on the answer of Strants.. –  Surb Jun 18 at 8:43

Note for $j>4$ $$ {j!\over j^j}= \underbrace{ {j\over j}\cdot{j-1\over j}\cdot{j-2\over j}\cdots\cdot{4\over j}}_{<1}\cdot{3\over j}\cdot{2\over j}\cdot{1\over j}<{6\over j^3}. $$ So for $j>4$, we have $$\sqrt{j!\over j^j}<{\sqrt 6\over j^{3/2}}.$$

If you know that the $p$-series $\sum\limits_{j=1}^\infty{1\over j^{3/2}}$ converges (it does), then you can use the Comparison Test to deduce that your series converges.

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Stirling's formula states that that $$n!\sim n^ne^{-n}\sqrt{2n\pi}.$$

Thus, $$\lim\limits_{n\to\infty}\left( \frac{n!}{n^n}\right)^{\frac{1}{2n}}=\frac{1}{\sqrt{e}}$$ and the series converges.

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We can observe that $$\frac{a_{j+1}}{a_j} = \frac{\sqrt{\frac{(j+1)!}{(j+1)^{j+1}}}}{\sqrt{\frac{j!}{j^j}}} = \sqrt{\frac{(j+1)!j^j}{j!(j+1)^{j+1}}} = \sqrt{\frac{j^j}{(j+1)^j}} = \left(\left(1+\frac{1}{j}\right)^j\right)^{-\frac{1}{2}}.$$

Thus, $$L = \lim_{j\to\infty} \left|\frac{a_{j+1}}{a_j}\right| = \left(\lim_{j\to\infty}\left(1 + \frac{1}{j}\right)^j\right)^{-\frac{1}{2}} = e^{-\frac{1}{2}} < 1,$$

so the sequence converges by the ratio test.

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