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$$\sum_{n=2}^{\infty}\frac{(n+1)!(n+1)^{n-1}}{n^{2n}}$$ I used the Cauchy test and it lead me to $\frac{\sqrt[n]{n!}}{n^2}$. But I can't tell what is the limit of this. I tried the Squeeze theorem: $$\frac{\sqrt[n]{n}}{n^2} < \frac{\sqrt[n]{n!}}{n^2} < ??? $$ And I'm stuck.

Martin Sleziak
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miko
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  • You should try this: http://en.wikipedia.org/wiki/Stirling's_approximation – Siminore Jan 06 '15 at 09:29
  • $\displaystyle \frac{(n+1)!(n+1)^{n-1}}{n^{2n}} = \frac{n!}{n^n}\left(1+\frac{1}{n}\right)^n$, we have $\displaystyle n! = n\prod\limits_{k=1}^{n-1}(k(n-k))^{1/2} < n\prod\limits_{k=1}^{n-1}\frac{n}{2} = \frac{n^n}{2^{n-1}}$ and $\displaystyle \left(1+\frac{1}{n}\right)^n \le e$, the convergence should follow from Comparison test now. – r9m Jan 06 '15 at 09:38
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    Stirling's approximation leaded me to $\frac{\sqrt[2n]{2\pi n}}{en}$ and its limit is clearly $0$. Am I right? Is this a proper answer? – miko Jan 06 '15 at 09:41
  • @miko yes that does it too. You are right. – r9m Jan 06 '15 at 09:42
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    If there is $n$ instead of $n^2$ in the denominator, the limit is more complicated: http://math.stackexchange.com/questions/201906/showing-that-frac-sqrtnnn-rightarrow-frac1e – Martin Sleziak Jan 06 '15 at 10:41
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    I think the ratio test also works on this one. – paw88789 Jan 06 '15 at 10:54
  • I believe maybe you wanted n in the denominator instead of $n^2$. – user84413 Jan 07 '15 at 16:06

3 Answers3

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You could use AM-GM inequality. You get: $$\sqrt[n]{n!} = \sqrt{1\cdot2\cdots n} \le \frac{1+2+\dots+n}n=\frac{n+1}2$$ which means $$0\le \frac{\sqrt[n]{n!}}{n^2} \le \frac{n+1}{2n^2}$$ and thus $$\lim_{n\to\infty} \frac{\sqrt[n]{n!}}{n^2}=0.$$

Martin Sleziak
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You could just use the straight-forward ratio test: $$ a_{n+1}/a_n = \frac{(n+2)! (n+2)^n n^{2n}}{(n+1)! (n+1)^{n-1} (n+1)^{2n+2}} = \left(\frac{n+2}{n+1}\right)^{n+1} \left(\frac{n}{n+1}\right)^{2n} $$ Since $\left(\frac{n+2}{n+1}\right)^{n+1} = \left(1+\frac{1}{n+1}\right)^{n+1} \to \mathrm{e}$ and $\left(\frac{n}{n+1}\right)^n = \left(1-\frac{1}{n+1}\right)^n \to \mathrm{e}^{-1}$, we have $$a_{n+1}/a_n \to (\mathrm{e}) \left(\mathrm{e}^{-1}\right)^2 = \mathrm{e}^{-1} < 1. $$

Jason
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Using the root test,

$\displaystyle\lim_{n\to\infty}\frac{\left((n+1)!\right)^{1/n}\left(n+1\right)^{1-1/n}}{n^2}=\lim_{n\to\infty}\frac{\left((n+1)!\right)^{1/n}}{n}\cdot\frac{n+1}{n}\cdot\frac{1}{(n+1)^{1/n}}=\lim_{n\to\infty}\frac{\left((n+1)!\right)^{1/n}}{n}$

$\hspace{2.2 in}$since $\displaystyle\frac{n+1}{n}\to1$ and $\displaystyle (n+1)^{1/n}\to 1$.

If $c_n=\displaystyle\frac{(n+1)!}{n^n}$, then $\displaystyle\frac{c_{n+1}}{c_n}=\frac{(n+2)!}{(n+1)^{n+1}}\cdot\frac{n^n}{(n+1)!}=\frac{n+2}{n+1}\left(\frac{n}{n+1}\right)^n=\frac{n+2}{n+1}\frac{1}{(1+\frac{1}{n})^n}\to\frac{1}{e}$,

so $\displaystyle\lim_{n\to\infty}\frac{((n+1)!)^{1/n}}{n}=\lim_{n\to\infty}(c_n)^{1/n}=\frac{1}{e}<1$ and therefore the series converges.

user84413
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