# Find  $\lim\limits _{n\rightarrow \infty} n^2 \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}$

I would like to find the limit

$\lim _{n\rightarrow \infty} n^2 \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}.$

This doesn't look like any limit I have seen before and I literally have no idea where to start.

We have
$L:=\lim _{n\rightarrow \infty} n^2 \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}=\lim _{n\rightarrow \infty} n^{\frac{2n}{n}} \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}$
$=\lim _{n\rightarrow \infty} \prod\limits_{k=1}^{n} (n^2)^{\frac{1}{n}} \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}=\lim _{n\rightarrow \infty} \prod\limits_{k=1}^{n} [n^2 (\frac{1}{k^2}+\frac{1}{n^2})]^{\frac{1}{n}}$
$=\lim _{n\rightarrow \infty} \prod\limits_{k=1}^{n} (\frac{n^2}{k^2}+1)^{\frac{1}{n}}=\lim _{n\rightarrow \infty} \prod\limits_{k=1}^{n} (\frac{1}{(\frac{k}{n})^2}+1)^{\frac{1}{n}}.$
Hence
$\ln L=\ln \left( \lim _{n\rightarrow \infty} \prod\limits_{k=1}^{n} (\frac{1}{(\frac{k}{n})^2}+1)^{\frac{1}{n}} \right)$
$= \lim _{n\rightarrow \infty} \ln \left( \prod\limits_{k=1}^{n} (\frac{1}{(\frac{k}{n})^2}+1)^{\frac{1}{n}} \right)= \lim _{n\rightarrow \infty} \sum_{k=1}^{n} \ln (\frac{1}{(\frac{k}{n})^2}+1)^{\frac{1}{n}}$
$= \lim _{n\rightarrow \infty}\frac{1}{n} \sum_{k=1}^{n} \ln (\frac{1}{(\frac{k}{n})^2}+1)=\int_{0}^{1}\ln(\frac{1}{x^2}+1)dx.$
Next we compute the above integral by integrating by parts. Let
$u=\ln(\frac{1}{x^2}+1) \text{and} dv=dx.$
Then
$du=\frac{-2x^{-3}}{1+x^{-2}} \text{and} v=x.$
So
$\ln L =\int_{0}^{1}\ln(\frac{1}{x^2}+1)dx= x\ln(\frac{1}{x^2}+1) \big|_0^{1}-\int_0^{1}x \frac{-2x^{-3}}{1+x^{-2}}dx$
$=\ln 2-\lim_{x \rightarrow 0} x\ln(\frac{1}{x^2}+1) +2\int_{0}^{1}\frac{1}{1+x^2}dx$
$=\ln 2-0 +2\arctan(x)\big|_0^{1}=\ln 2+2\arctan(1)-2\arctan(0)$
$=\ln 2+2\frac{\pi}{4}-0=\ln 2+\frac{\pi}{2}.$
Thus
$\ln L=\ln 2+\frac{\pi}{2} \Rightarrow L= e^{\ln 2+\frac{\pi}{2} }=2 e^{\frac{\pi}{2}}.$
Hence
$L=\lim _{n\rightarrow \infty} n^2 \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}=2 e^{\frac{\pi}{2}}.$

• Thank you so much Savin :)