If lim _{n} {rightarrow infty}left[left(1+fra | vedclass

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(C) Let $L = \lim _{n \rightarrow \infty}\left[\prod_{r=1}^n \left(1+\frac{r^2}{n^2}\right)\right]^{1 / n} = k$. Taking the natural logarithm on both

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$\log 2+\frac{\pi}{2}-2$

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(C) Let $L = \lim _{n ightarrow \infty}\left[\prod_{r=1}^n \left(1+\frac{r^2}{n^2}ight)ight]^{1 / n} = k$. Taking the natural logarithm on both sides: $\log k = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r^2}{n^2}ight)$. This is a Riemann sum,which can be expressed as the definite integral: $\log k = \int_0^1 \log(1+x^2) dx$. Using integration by parts,let $u = \log(1+x^2)$ and $dv = dx$: $\int \log(1+x^2) dx = x \log(1+x^2) - \int x \cdot \frac{2x}{1+x^2} dx$. $= x \log(1+x^2) - 2 \int \frac{x^2}{1+x^2} dx = x \log(1+x^2) - 2 \int \left(1 - \frac{1}{1+x^2}ight) dx$. $= x \log(1+x^2) - 2x + 2 	an^{-1}(x)$. Evaluating from $0$ to $1$: $\log k = [1 \cdot \log(2) - 2(1) + 2 	an^{-1}(1)] - [0 - 0 + 0]$. $\log k = \log 2 - 2 + 2 \left(\frac{\pi}{4}ight) = \log 2 + \frac{\pi}{2} - 2$.

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 $\log k=$

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