If $\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=k$ then $\log k=$

Question

Aicharya · Accepted Answer

Given, $ \begin{aligned} & k=\lim _{n ightarrow \infty}\left[\left(1+\frac{1}{n^2}ight)\left(1+\frac{2^2}{n^2}ight) \ldots\left(1+\frac{n^2}{n^2}ight)ight]^{1 / n} \ & \log k=\lim _{n ightarrow \infty} \log \left[\left(1+\frac{1}{n^2}ight)\left(1+\frac{2^2}{n^2}ight) \ldots\left(1+\frac{n^2}{n^2}ight)ight]^{1 / n} \ = & \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r^2}{n^2}ight)=\int_0^1 \log \left(1+x^2ight) d x \ = & {\left[\log \left(1+x^2ight) \cdot xight]_0^1-\int_0^1 \frac{2 x}{1+x^2} \cdot x d x } \ = & \log 2-2 \int_0^1 \frac{x^2}{1+x^2} d x=\log 2-2 \int_0^1\left(1-\frac{1}{1+x^2}ight) d x \ = & \log 2-2\left[x-	an ^{-1} xight]_0^1 \ = & \log 2-2\left[1-\frac{\pi}{4}ight] \ = & \log 2-2+\frac{\pi}{2} \ = & \log 2+\frac{\pi}{2}-2 \end{aligned} $

If $\lim _{n \rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right)…

Solution