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I tried to calculate following limit: (let me present my solution first, and then point errors or give hints how to finish)$$\lim_{n\rightarrow\infty}\frac{(n!)^n}{n^{{n}^{2}}}$$So my partly solution is: Let $a_{n}=\frac{(n!)^n}{n^{{n}^{2}}}$. Then $a_{n}=(\frac{(n!)}{n^{n}})^n$ and denoting $p_{n}=ln (a_{n})$ we have $p_{n}=ln((\frac{(n!)}{n^{n}})^n)$
$=n*ln\frac{(n!)}{n^{n}}=$
$n*(ln(n!)-n*ln(n)$
and now I use Stirling formula $n!\approx\sqrt{2\pi n}*(\frac{n}{e})^n$ and now
$p_{n}=n*(ln(\sqrt{2\pi n}*(\frac{n}{e})^n)-n*ln(n))$
$=n(ln(\sqrt{2\pi n)}+n*ln(n)-n-n*ln(n))=$
$=n((ln(\sqrt{2\pi n})+n))=$
$=n(ln(n\sqrt{{2n\pi n})})$
and that would imply that $\lim_{n\rightarrow\infty}a_{n}=exp(n\sqrt{2\pi n})^n$ what does not seem good... Where am I doing mistake, is it wrong way of acting? Any hints are desired, thank you in advance.

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2 Answers 2

up vote 1 down vote accepted

There is an error in your third step in the computation of $p_n$. \begin{align} p_n & = n (\ln(n!) - n \ln(n))\\ & \sim n (\ln(\sqrt{2 \pi n} \cdot (n/e)^n) - n \ln(n)) & (\because \text{Stirling's formula})\\ & = n (\ln(\sqrt{2 \pi n}) + n \ln n -n - n \ln n)\\ & = \color{blue}{\underbrace{n(\ln(\sqrt{2 \pi n}) - n) \to - \infty}_{n \to \infty}} \end{align} Hence, $\ln(a_n) \to -\infty$ as $n \to \infty$. Hence, what can you say about $a_n$?

EDIT

A simpler proof is as follows. Note that $n! \leq \dfrac{n^n}2$ for all $n > 1$. Hence, $$0 < \dfrac{(n!)^n}{n^{n^2}} = \left(\dfrac{n!}{n^n} \right)^n \leq \left(\dfrac12 \right)^n = \dfrac1{2^n}$$ Hence, $$0 \leq \lim_{n \to \infty} \dfrac{(n!)^n}{n^{n^2}} \leq \lim_{n \to \infty} \dfrac1{2^n} = 0$$

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OK, I see arithmetic mistake now, but... what further? –  fdhd Nov 7 '12 at 23:02
    
Stirling claims ownership of step two. –  Pedro Tamaroff Nov 7 '12 at 23:02
    
@PeterTamaroff Yes. Corrected. user46034: $\ln(a_n) \to -\infty$ as $n \to \infty$. Hence, what can you say about $a_n$? –  user17762 Nov 7 '12 at 23:04
    
I am blinded now.. seems to go to $0$ –  fdhd Nov 7 '12 at 23:05
1  
@user46034 No that is not sufficient. For instance, consider $\dfrac12 + \dfrac1{4n}$. It is bounded by $0$ and $1$ and is decreasing. But converges to $\dfrac12$. –  user17762 Nov 7 '12 at 23:09

You can see here that

$$\lim_{n\to\infty}\frac{n!}{n^n}=0$$

thus

$$(**)\;\;\;\;\;\;\;\;\;\;\;\;\;\sqrt[n]{\frac{(n!)^n}{n^{n^2}}}=\frac{n!}{n^n}\xrightarrow [n\to\infty]{} 0$$

From where it follows that the series

$$\sum_{n=1}^n\frac{(n!)^n}{n^{n^2}}$$

converges, since what we did in (**) above is merely apply Cauchy's Test of the n-th root, and thus the sequence's limit must be zero.

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Umm. I am afraid I do not fully understand Cauche's Test, could you explain, please? –  fdhd Nov 7 '12 at 23:17
    
Alright, used en.wikipedia.org/wiki/Cauchy%27s_root_test –  fdhd Nov 7 '12 at 23:19

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