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My task is to calculate limit: $$\lim_{n \rightarrow \infty} \sqrt[n^2]{ \frac{(n+1)^{n+1}(n+2)^{n+2}\cdots(n+n)^{n+n}}{n^{n+1}n^{n+2}\cdots n^{n+n}} }$$I denoted that limit as $a_n$. So: $$\log a_n=\frac{1}{n^2} \left ( (n+1)\log \left (1+\frac{1}{n} \right )+\cdots+(n+n)\log \left (1+\frac{n}{n} \right )\right )=$$$$=\cdots=\frac{1}{n} \left ( \sum_{k=1}^n\log(1+\frac{k}{n})+\sum_{k=1}^n\frac{k}{n}\log(1+\frac{k}{n}) \right )$$ The only (quite crucial however) problem I've got is the term $\frac{1}{n}$ before the above. I know how to calculate parenthesis:$$ \lim_{n \rightarrow \infty}\left ( \sum_{k=1}^n\log(1+\frac{k}{n})+\sum_{k=1}^n\frac{k}{n}\log(1+\frac{k}{n}) \right )=$$$$\int_0^1\log(1+x)dx+\int_0^1x\log(1+x)dx=\cdots$$But I have no idea what $\frac{1}{n} $does. Any hints? Thanks in advance.

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    $\begingroup$ If $f:[0,1]\to\Bbb R$ is Riemann integrable, then $\int_0^1f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^nf(\frac{k}{n})$. $\endgroup$
    – 23rd
    Commented Jun 9, 2013 at 10:46
  • $\begingroup$ aaa.... so the answer is just the sum of two integrals? $\endgroup$
    – fdhd
    Commented Jun 9, 2013 at 10:50
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    $\begingroup$ Yes, and please do not forget to take exponential to go back to the original limit. You may post an answer by yourself as you wish. $\endgroup$
    – 23rd
    Commented Jun 9, 2013 at 10:54

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We will use the fact that $\int_0^1f(x)dx=\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=1}^nf(\frac{k}{n})$ So:$$\lim_{n \rightarrow \infty} \frac{1}{n}\left ( \sum_{k=1}^n\log(1+\frac{k}{n})+\sum_{k=1}^n\frac{k}{n}\log(1+\frac{k}{n}) \right )=$$$$\int_0^1\log(1+x)dx+\int_0^1x\log(1+x)dx= \cdots=\log4-\frac{3}{4}$$ Going back to the original limit:$$\lim_{n \rightarrow \infty}a_n=\exp(\log4-\frac{3}{4})$$

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  • $\begingroup$ Integration is easy, everyone can do it on his own, I hope it is now correct :) $\endgroup$
    – fdhd
    Commented Jun 9, 2013 at 11:07
  • $\begingroup$ It looks good to me. +1. $\endgroup$
    – 23rd
    Commented Jun 9, 2013 at 11:24

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