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Find the limit:
$ \lim\limits_{n\to\infty}\sqrt[n]{\left(\dfrac{1 + n}{n^2} \right)\left(\dfrac{4 + 2n}{n^2} \right)...\left(\dfrac{n^2 + n^2}{n^2} \right)} $

I tried simplifying this limit and the one I get to is:

$ \lim\limits_{n\to\infty}\dfrac{1}{n^2}\bigg(\big(2n\big)!\bigg)^{\dfrac{1}{n}} $

I have an instruction to write the limit as a definite integral and then calculate its value. I think that there should be a way to represent the last limit as a Riemann sum and then calculate it with the integral. But I'm not sure how to get to the Riemann sum.

Looking forward for any ideas!

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    $\begingroup$ +1 and welcome to our community! However, don't you think you can make the title of this question a bit better? $\endgroup$ – Nike Dattani Apr 21 at 1:34
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    $\begingroup$ Considered your advice! $\endgroup$ – A. Mason Apr 21 at 12:08
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The logarithm of the initial expression is $$ \frac 1n \sum_{k=1}^n \ln\left(\frac{k^2+kn}{n^2}\right) = \frac 1n \sum_{k=1}^n \ln\left(\left(\frac kn \right)^2 + \frac kn \right) $$ which is a Riemann sum for $$ \int_0^1 \ln(x^2 + x) \, dx = \int_0^2 \ln x \, dx = 2 \ln 2 - 2 \, . $$

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Let $\;a_n=\dfrac{(2n)!}{n^{2n}}>0\quad$ for all $\;n\in\mathbb{N}\;.$

Since

$\exists\lim\limits_{n\to\infty}\dfrac{a_{n+1}}{a_n}=\lim\limits_{n\to\infty}\left[\dfrac{(2n+2)!}{(n+1)^{2n+2}}\cdot\dfrac{n^{2n}}{(2n)!}\right]=$

$=\lim\limits_{n\to\infty}\left[\dfrac{(2n+1)(2n+2)}{(n+1)^2}\cdot\dfrac{n^{2n}}{(n+1)^{2n}}\right]=$

$=\lim\limits_{n\to\infty}\left[\dfrac{4n^2+6n+2}{n^2+2n+1}\cdot\dfrac1{\left(1+\frac1n\right)^{2n}}\right]=\dfrac4{e^2}$

by using an application of Stolz-Cesàro theorem, we get that

$\exists\lim\limits_{n\to\infty}\dfrac{1}{n^2}\bigg(\big(2n\big)!\bigg)^{\dfrac{1}{n}}=\lim\limits_{n\to\infty}\sqrt[n]{\dfrac{(2n)!}{n^{2n}}}=\lim\limits_{n\to\infty}\sqrt[n]{a_n}=\dfrac4{e^2}\;.$

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  • $\begingroup$ Nice answer. I don't understand the use of the existential quantifier $\exists$ here. $\endgroup$ – Taladris Apr 21 at 0:57
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    $\begingroup$ It means that the limit exists, indeed there are limits that do not exist, for example: $\;\lim\limits_{n\to\infty}\sin n\;.$ $\endgroup$ – Angelo Apr 21 at 12:37
  • $\begingroup$ I know, but to me, it is putting the cart before the horse, since when writing it, you don't know yet that the limit exists. $\endgroup$ – Taladris Apr 21 at 13:36
  • $\begingroup$ I know it because I have already calculated it, moreover if a limit does not exist, then it does not make sense to write that it is equal to something else. For example, it does not make sense to write: $\;\lim\limits_{n\to\infty}\sin n=\lim\limits_{n\to\infty}(-1)^n\;.$ $\endgroup$ – Angelo Apr 21 at 14:22

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