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Question Number 127085 by mnjuly1970 last updated on 26/Dec/20

...nice calculus... prove that:: lim_( n→∞) ( Σ_(k=1) ^n ((k/n))^n )=(e/(e−1))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:{prove}\:{that}:: \\ $$$$\:\:\:{lim}_{\:{n}\rightarrow\infty} \left(\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}}{{n}}\right)^{{n}} \right)=\frac{{e}}{{e}−\mathrm{1}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 27/Dec/20

thanks alot...

$$\:\:\:\:{thanks}\:\:{alot}... \\ $$

Answered by mindispower last updated on 27/Dec/20

Σ_(k≥1) e^(nln((k/n))) ln(1−(k/n))≤−(k/n)...E⇒e^(nln((k/n))) ≤e^(−k) (1/(1−x))>1..∀x∈[0,1[⇒∫_0 ^t (1/(1−x))dx>∫_0 ^t dx,t∈[0,1[ for E ⇔ln(1−t)<−t ⇒0≤Σ_(k=1) ^n ((k/n))^n =Σ_(k=0) ^(n−1) (((n−k)/n))^n ≤Σ_(k=1) ^n e^(−k) =(e/(e−1))... we have bounded sesuences by dominate cv Theorem lim_(n→∞) Σ_(k=1) ^(n−1) (1−(k/n))^n =Σlim_(n→∞) (1−(k/n))^n =Σ_0 ^∞ e^(−k) =(e/(e−1)) lim_(n→∞) ((k/n))^n =(e/(e−1))

$$\underset{{k}\geqslant\mathrm{1}} {\sum}{e}^{{nln}\left(\frac{{k}}{{n}}\right)} \\ $$$${ln}\left(\mathrm{1}−\frac{{k}}{{n}}\right)\leqslant−\frac{{k}}{{n}}...{E}\Rightarrow{e}^{{nln}\left(\frac{{k}}{{n}}\right)} \leqslant{e}^{−{k}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}}>\mathrm{1}..\forall{x}\in\left[\mathrm{0},\mathrm{1}\left[\Rightarrow\int_{\mathrm{0}} ^{{t}} \frac{\mathrm{1}}{\mathrm{1}−{x}}{dx}>\int_{\mathrm{0}} ^{{t}} {dx},{t}\in\left[\mathrm{0},\mathrm{1}\left[\right.\right.\right.\right. \\ $$$${for}\:{E}\: \\ $$$$\Leftrightarrow{ln}\left(\mathrm{1}−{t}\right)<−{t} \\ $$$$\Rightarrow\mathrm{0}\leqslant\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\frac{{k}}{{n}}\right)^{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\frac{{n}−{k}}{{n}}\right)^{{n}} \leqslant\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{−{k}} =\frac{{e}}{{e}−\mathrm{1}}... \\ $$$${we}\:{have}\:\:{bounded}\:{sesuences} \\ $$$${by}\:{dominate}\:{cv}\:{Theorem}\: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\left(\mathrm{1}−\frac{{k}}{{n}}\right)^{{n}} =\Sigma\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{{k}}{{n}}\right)^{{n}} =\underset{\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{k}} \\ $$$$=\frac{{e}}{{e}−\mathrm{1}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{k}}{{n}}\right)^{{n}} =\frac{{e}}{{e}−\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$

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