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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$e^2+1$

Vedclass · Answer

(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} = $

Similar Questions

The value of $\lim _{n \rightarrow \infty} \left[ \frac{n}{n^2+1^2} + \frac{n}{n^2+2^2} + \dots + \frac{n}{n^2+n^2} \right]$ is

$\lim _{n \rightarrow \infty} \left( \frac{\sqrt{1} + 2 \sqrt{2} + 3 \sqrt{3} + \ldots + n \sqrt{n}}{n^{5/2}} \right) = $

$\mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{\left( {n + 1} \right)\left( {n + 2} \right) \ldots \left( {3n} \right)}}{{{n^{2n}}}}} \right)^{\frac{1}{n}}} = $

If $f(n) = \frac{1}{n} [(n+1)(n+2)(n+3) \ldots (2n)]^{\frac{1}{n}}$,then $\lim_{n \rightarrow \infty} f(n) =$

Evaluate the following definite integral as the limit of a sum: $\int_{-1}^{1} e^{x} dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module