Question Number 29510 by abdo imad last updated on 09/Feb/18
$${let}\:\:\:{w}_{{n}} =\:\frac{\mathrm{1}}{{n}}\left(\:\mathrm{1}\:+{e}^{\frac{\mathrm{1}}{{n}}} \:+{e}^{\frac{\mathrm{2}}{{n}}} \:+\:...\:{e}^{\frac{{n}−\mathrm{1}}{{n}}} \right)\:{find}\:{lim}_{{n}\rightarrow+\infty} {w}_{{n}} . \\ $$
Commented by abdo imad last updated on 11/Feb/18
![we have w_n = (1/n)Σ_(k=0) ^(n−1) e^(k/n) ⇒ lim_(n→∞) w_n =lim_(n→∞) ((1−0)/n)Σ_(k=0) ^(n−1) e^((k(1−o))/n) = ∫_0 ^1 e^x dx (Reiman sum) = [e^x ]_0 ^1 =e−1 .](Q29678.png)
$${we}\:{have}\:{w}_{{n}} =\:\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{\frac{{k}}{{n}}} \:\Rightarrow\:{lim}_{{n}\rightarrow\infty} {w}_{{n}} \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}−\mathrm{0}}{{n}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{e}^{\frac{{k}\left(\mathrm{1}−{o}\right)}{{n}}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{e}^{{x}} {dx}\:\:\:\:\:\left({Reiman}\:{sum}\right) \\ $$$$=\:\left[{e}^{{x}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:\:={e}−\mathrm{1}\:. \\ $$