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

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

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

Vedclass · Answer

(B) Let $V = \lim _{n ightarrow \infty}\left[\prod_{k=1}^{n} \left(1+\frac{k^2}{n^2}ight)ight]^{1 / n}$.Taking the natural logarithm on both sides:$\log V = \lim _{n ightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \log \left(1+\frac{k^2}{n^2}ight)$.This is a Riemann sum of the form $\int_{0}^{1} f(x) dx$ where $f(x) = \log(1+x^2)$:$\log V = \int_{0}^{1} \log(1+x^2) dx$.Using integration by parts,$\int \log(1+x^2) dx = x \log(1+x^2) - \int x \cdot \frac{2x}{1+x^2} dx = x \log(1+x^2) - 2 \int \frac{x^2+1-1}{1+x^2} dx = x \log(1+x^2) - 2x + 2 	an^{-1}(x)$.Evaluating from $0$ to $1$:$\log V = [1 \cdot \log(2) - 2(1) + 2 	an^{-1}(1)] - [0] = \log 2 - 2 + 2(\frac{\pi}{4}) = \log 2 - 2 + \frac{\pi}{2}$.Thus,$V = e^{\log 2 - 2 + \frac{\pi}{2}} = 2 e^{\frac{\pi}{2}-2}$.

$\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{4}{n^2}\right)\left(1+\frac{9}{n^2}\right) \ldots \left(1+\frac{n^2}{n^2}\right)\right]^{1 / n}=$

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