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  • Evaluting <img alt="\mathrm{\lim _{k \rightarrow \infty}\left\{\frac{\frac{1}{k} e^{\frac{1}{k}}+\frac{2}{k} e^{\frac{2}{k}}+\frac{3}{k} e^{\frac{3}{k}}+\cdots+\frac{k}{k} e^{\frac{k}{k}}}{k}\

Evaluting \mathrm{\lim _{k \rightarrow \infty}\left\{\frac{\frac{1}{k} e^{\frac{1}{k}}+\frac{2}{k} e^{\frac{2}{k}}+\frac{3}{k} e^{\frac{3}{k}}+\cdots+\frac{k}{k} e^{\frac{k}{k}}}{k}\right\}}

Option: 1

1


Option: 2

0


Option: 3

-1


Option: 4

2


Answers (1)

best_answer

\mathrm{L} is a Riemann sum. It is helpful to rewrite \mathrm{L} as
                \mathrm{ L=\lim _{k \rightarrow \infty}\left\{\frac{\frac{1}{k} e^{\frac{1}{k}}+\frac{2}{k} e^{\frac{2}{k}}+\frac{3}{k} e^{\frac{3}{k}}+\cdots+\frac{k}{k} e^{\frac{k}{k}}}{k}\right\} }
by moving one division by $k$ to the numerator. Now, letting $f(x)=x e^x$, we have
                \mathrm{ L=\lim _{k \rightarrow \infty} \sum_{i=1}^k \frac{f\left(\frac{i}{k}\right)}{k}=\int_0^1 f(x) d x . }
By integration by parts, \mathrm{\int x e^x d x=(x-1) e^x+C}, so \mathrm{L=0 e^1-(-1) e^0=1.}

Posted by

sudhir kumar

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