Question

solve it by using limit sum property​

Answer 1

$\sf \Large Solution!!$

$\sf \displaystyle \lim_{n\to \infty}\bigg[\bigg(1+\dfrac{1}{n^{2}}\bigg)\bigg(1+\dfrac{2^{2}}{n^{2}}\bigg)+\dots \bigg(1+\dfrac{n^{2}}{n^{2}}\bigg)\bigg]^{\dfrac{1}{n}}$

$\sf Let\:P=\displaystyle \lim_{n\to \infty}\bigg[\bigg(1+\dfrac{1}{n^{2}}\bigg)\bigg(1+\dfrac{2^{2}}{n^{2}}\bigg)+\dots \bigg(1+\dfrac{n^{2}}{n^{2}}\bigg)\bigg]^{\dfrac{1}{n}}$

$\sf \log P=\displaystyle \lim_{n\to \infty}\dfrac{1}{n}\bigg[\log\bigg(1+\dfrac{1}{n^{2}}\bigg)+\log\bigg(1+\dfrac{2^{2}}{n^{2}}\bigg)+\dots \log\bigg(1+\dfrac{n^{2}}{n^{2}}\bigg)\bigg]$

$\sf \log P=\displaystyle \lim_{n\to \infty}\dfrac{1}{n}\displaystyle \sum\limits_{r=1}^{n}\log\bigg(1+\dfrac{r^{2}}{n^{2}}\bigg)$

$\sf \dfrac{r}{n},\:r\to 1,\:\dfrac{r}{n}\to 0\\r\to n,\:\dfrac{r}{n}\to 1$

$\sf \log P=\displaystyle \int\limits_{0}^{1} \log(1+x^{2})\,dx$

$\sf \log P=\log(1+x^{2})\displaystyle \int\limits_{0}^{1}dx-\displaystyle \int\limits_{0}^{1}\bigg[\dfrac{2x}{1+x^{2}}\displaystyle \int dx\bigg]dx$

$\sf \log P=\log(1+x^{2})\left. x \right |_{0}^{1}-2\displaystyle \int\limits_{0}^{1}\dfrac{x^{2}}{1+x^{2}}dx$

$\sf \log P=\log 2-2\displaystyle \int\limits_{0}^{1}\dfrac{x^{2}+1-1}{x^{2}+1}dx$

$\sf \log P=\log 2-2\displaystyle \int\limits_{0}^{1}\dfrac{x^{2}+1}{x^{2}+1}dx+2\displaystyle \int\limits_{0}^{1}\dfrac{1}{1+x^{2}}dx$

$\sf \log P=\log 2-2+2\tan ^{-1}\left. x \right |_{0}^{1}=2\times \dfrac{\pi }{4}=\dfrac{\pi }{2}$

$\sf \log P=\log 2-2+\dfrac{\pi }{2}$

$\sf \log P-\log 2=\dfrac{\pi }{2}-2$

$\sf \log \dfrac{P}{2}=\dfrac{\pi }{2}-2$

$\sf \dfrac{P}{2}=e^{\frac{\pi }{2}-2}\implies P=2e^{\frac{\pi }{2}-2}$

$\sf \therefore \displaystyle \lim_{n\to \infty}\bigg[\bigg(1+\dfrac{1}{n^{2}}\bigg)\bigg(1+\dfrac{2^{2}}{n^{2}}\bigg)+\dots \bigg(1+\dfrac{n^{2}}{n^{2}}\bigg)\bigg]^{\dfrac{1}{n}}\\=2e^{\frac{\pi -4}{2}}$