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Compute the following limit:

$$\lim_{n\to\infty}\frac{{\left(1^1 \cdot 2^2 \cdot3^3\cdots n^n\right)}^\frac{1}{n^2}}{\sqrt{n}} $$

I'm interested in almost any approaching way for this limit. Thanks.

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do you know the answer? take the limit equal to some k and then log on both the sides. –  sukunrt Jun 23 '12 at 20:28
+1. I find that all your questions, in general, to be interesting and intriguing. –  user17762 Jun 23 '12 at 20:34
I wonder who is that troll, diligently downvoting your questions... –  Norbert Jun 23 '12 at 20:40

2 Answers 2

up vote 22 down vote accepted

Let's begin $$ \lim\limits_{n\to\infty}\frac{\left(\prod\limits_{k=1}^n k^k\right)^{\frac{1}{n^2}}}{\sqrt{n}}= \lim\limits_{n\to\infty}\exp\left(\frac{1}{n^2}\sum\limits_{k=1}^n k\log k - \frac{1}{2}\log n\right)= $$ $$ \lim\limits_{n\to\infty}\exp\left(\frac{1}{n^2}\sum\limits_{k=1}^n k\log\left(\frac{k}{n}\right)+\frac{1}{n^2}\sum\limits_{k=1}^n k\log n - \frac{1}{2}\log n\right)= $$ $$ \lim\limits_{n\to\infty}\exp\left(\sum\limits_{k=1}^n \frac{k}{n}\log\left(\frac{k}{n}\right)\frac{1}{n}+\frac{1}{2}\log n\left(\frac{n^2+n}{n^2}-1\right)\right)= $$ $$ \exp\left(\lim\limits_{n\to\infty}\sum\limits_{k=1}^n \frac{k}{n}\log\left(\frac{k}{n}\right)\frac{1}{n}+\frac{1}{2}\lim\limits_{n\to\infty}\frac{\log n}{n}\right)= $$ $$ \exp\left(\int\limits_{0}^1 x\log x dx\right)=\exp\left(-1/4\right) $$ And now we are done!

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very fast and elegant at the same time! Thanks! –  Chris's sis Jun 23 '12 at 20:33
You are welcome! –  Norbert Jun 23 '12 at 20:35

$$ \frac1{n^2}\sum_{k=1}^nk\log(k)-\frac12\log(n)=\frac1{n}\sum_{k=1}^n\frac{k}n\log\left(\frac{k}n\right)+\frac12\frac{\log(n)}n=\int_0^1x\log(x)\mathrm dx+o(1) $$

Edit: Per request, a solution without integrals, using only the elementary version of Stirling's approximation. Let $s_n=\displaystyle\sum_{k=1}^nk\log(k)$. Then, $$ \sum_{k=1}^n\log(k!)=\sum_{k=1}^n\sum_{i=1}^k\log(i)=\sum_{i=1}^n(n-i+1)\log(i)=(n+1)\log(n!)-s_n. $$ On the other hand, Stirling's approximation in its simplest form is $\log(k!)=k\log(k)-k+r_k$ with $r_k=O(\log k)$, which yields $$ \sum_{k=1}^n\log(k!)=\sum_{k=1}^nk\log(k)-\sum_{k=1}^nk+\sum_{k=1}^nr_k=s_n-\frac12n(n+1)+t_n,\quad t_n=\sum_{k=1}^nr_k. $$ Comparing these two expressions and using once more $\log(n!)=n\log(n)-n+r_n$, one gets $$ 2s_n=n(n+1)\log n-n(n+1)+(n+1)r_n+\frac12n(n+1)-t_n, $$ hence $2s_n-n^2\log n=-\frac12n^2+u_n$ with $$ u_n=n\log(n)-\frac12n+(n+1)r_n-t_n. $$ Since $r_n=O(\log n)$, $t_n=O(n\log n)$ and each term in $u_n$ is $O(n\log n)$, hence $$ \frac{s_n}{n^2}-\frac{\log n}2=-\frac14+\frac{u_n}{2n^2}=-\frac14+O\left(\frac{\log n}{n}\right)\to-\frac14. $$

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thanks for your solution. –  Chris's sis Jun 23 '12 at 20:34

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