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

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Question

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

Vedclass · Accepted Answer

$2 e^{\frac{\pi-4}{2}}$

Vedclass · Answer

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

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

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