Question Number 163318 by KONE last updated on 06/Jan/22
Answered by Mathspace last updated on 06/Jan/22
![=lim_(n→∞) (1/n)Σ_(k=1) ^n [e^(k/n) ] =∫_0 ^1 [e^x ] dx changement e^x =t give x=lnt ∫^1 _0 [e^x ]dx=∫_1 ^e [t](dt/t) =∫_1 ^2 (([t])/t)dt +∫_2 ^e (([t])/t)dt =∫_1 ^2 (dt/t) +2∫_2 ^e (dt/t) =ln2+2{1−ln2} =2−ln2](https://www.tinkutara.com/question/Q163404.png)
$$={lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \left[{e}^{\frac{{k}}{{n}}} \right] \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[{e}^{{x}} \right]\:{dx}\:\:\:{changement} \\ $$$${e}^{{x}} ={t}\:\:{give}\:{x}={lnt} \\ $$$$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \left[{e}^{{x}} \right]{dx}=\int_{\mathrm{1}} ^{{e}} \left[{t}\right]\frac{{dt}}{{t}} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\left[{t}\right]}{{t}}{dt}\:+\int_{\mathrm{2}} ^{{e}} \:\frac{\left[{t}\right]}{{t}}{dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{dt}}{{t}}\:+\mathrm{2}\int_{\mathrm{2}} ^{{e}} \:\frac{{dt}}{{t}} \\ $$$$={ln}\mathrm{2}+\mathrm{2}\left\{\mathrm{1}−{ln}\mathrm{2}\right\} \\ $$$$=\mathrm{2}−{ln}\mathrm{2} \\ $$