lim _{n rightarrow infty} frac{1}{n^2} sum_{k | vedclass

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(B) The given expression is $S = \lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{2n} k e^{k/n}$.This can be rewritten as $S = \lim _{n \rightar

Vedclass · Accepted Answer

$e^2+1$

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(B) The given expression is $S = \lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{2n} k e^{k/n}$.This can be rewritten as $S = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{2n} \left(\frac{k}{n}\right) e^{k/n}$.Let $x = \frac{k}{n}$, then $dx = \frac{1}{n}$. As $k=1, x \rightarrow 0$ and as $k=2n, x \rightarrow 2$.The sum becomes a definite integral: $S = \int_{0}^{2} x e^x dx$.Using integration by parts, $\int u dv = uv - \int v du$, where $u=x$ and $dv=e^x dx$:$S = [x e^x]_{0}^{2} - \int_{0}^{2} e^x dx$.$S = (2 e^2 - 0) - [e^x]_{0}^{2}$.$S = 2 e^2 - (e^2 - e^0) = 2 e^2 - e^2 + 1 = e^2 + 1$.

$\lim _{n \rightarrow \infty} \frac{1}{n^2} \sum_{k=1}^{2n} k e^{k/n} = $

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