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

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 
\[L=\lim _{n\rightarrow \infty} n^2 \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}=2 e^{\frac{\pi}{2}}.\]