Question Number 155630 by SANOGO last updated on 03/Oct/21
Answered by Ar Brandon last updated on 03/Oct/21
![L=lim_(n→∞) nΣ_(k=1) ^n ((exp(−(n/k)))/k^2 )=lim_(n→∞) (1/n)Σ_(k=1) ^n (n^2 /k^2 )e^(−(n/k)) =∫_0 ^1 (e^(−(1/x)) /x^2 )dx=∫_0 ^1 e^(−(1/x)) d(−(1/x))=[e^(−(1/x)) ]_0 ^1 =(1/e)](https://www.tinkutara.com/question/Q155633.png)
$$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{exp}\left(−\frac{{n}}{{k}}\right)}{{k}^{\mathrm{2}} }=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}^{\mathrm{2}} }{{k}^{\mathrm{2}} }{e}^{−\frac{{n}}{{k}}} \\ $$$$\:\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−\frac{\mathrm{1}}{{x}}} }{{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−\frac{\mathrm{1}}{{x}}} {d}\left(−\frac{\mathrm{1}}{{x}}\right)=\left[{e}^{−\frac{\mathrm{1}}{{x}}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{{e}} \\ $$
Commented by SANOGO last updated on 03/Oct/21
$${merci} \\ $$