Question Number 101582 by Rohit@Thakur last updated on 03/Jul/20
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{r}\:{e}^{{r}/{n}} \:= \\ $$
Answered by mathmax by abdo last updated on 03/Jul/20
![A_n =(1/n^2 )Σ_(k=1) ^(n ) k e^(k/n) ⇒A_n =(1/n) Σ_(k=1) ^n (k/n)e^(k/n) so A_n is a Rieman sum and lim_(n→+∞) A_n =∫_0 ^1 xe^x dx =[x e^x ]_0 ^1 −∫_0 ^1 e^x dx =e−(e−1) =1](https://www.tinkutara.com/question/Q101591.png)
$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}\:} \mathrm{k}\:\mathrm{e}^{\frac{\mathrm{k}}{\mathrm{n}}} \:\Rightarrow\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{k}}{\mathrm{n}}\mathrm{e}^{\frac{\mathrm{k}}{\mathrm{n}}} \:\:\mathrm{so}\:\mathrm{A}_{\mathrm{n}} \mathrm{is}\:\mathrm{a}\:\mathrm{Rieman}\:\mathrm{sum}\:\mathrm{and} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{xe}^{\mathrm{x}} \:\mathrm{dx}\:=\left[\mathrm{x}\:\mathrm{e}^{\mathrm{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{e}^{\mathrm{x}} \:\mathrm{dx} \\ $$$$=\mathrm{e}−\left(\mathrm{e}−\mathrm{1}\right)\:=\mathrm{1} \\ $$
Answered by Dwaipayan Shikari last updated on 03/Jul/20
![(1/n)lim_(n→∞) Σ_(r=1) ^n (r/n)e^(r/n) =∫_0 ^1 xe^x dx=[xe^x ]_0 ^1 −[e^x ]_0 ^1 =1](https://www.tinkutara.com/question/Q101586.png)
$$\frac{\mathrm{1}}{{n}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{r}}{{n}}{e}^{\frac{{r}}{{n}}} =\int_{\mathrm{0}} ^{\mathrm{1}} {xe}^{{x}} {dx}=\left[{xe}^{{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\left[{e}^{{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1} \\ $$
Commented by Rohit@Thakur last updated on 04/Jul/20
$${Thank}\:{you}\:{sir} \\ $$