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Question Number 193408 by cortano12 last updated on 13/Jun/23

$$\underset{―}{\underbrace{}}$$

Answered by MM42 last updated on 13/Jun/23

$${lim}_{n\to \mathrm{\infty }}\frac{1}{n}\times \left(\frac{1-{\left(e^{\frac{a}{n}}\right)}^n\times \frac{1}{e^{\frac{a}{n}}}}{1-e^{\frac{a}{n}}}\right)=$$$${lim}_{n\to \mathrm{\infty }}\frac{1}{n}\times \left(\frac{e^{\frac{a}{n}}-{\left(e^{\frac{a}{n}}\right)}^n}{1-e^{\frac{a}{n}}}\right)\times \frac{1}{e^{\frac{a}{n}}}=$$$${lim}_{n\to \mathrm{\infty }}\frac{1}{n}\times \frac{1}{e^{\frac{a}{n}}}\times \left(\frac{e^{\frac{a}{n}}-1+1-{\left(e^{\frac{a}{n}}\right)}^n}{1-e^{\frac{a}{n}}}\right)=$$$${lim}_{n\to \mathrm{\infty }}\frac{1}{n}\times \frac{1}{e^{\frac{a}{n}}}\left(\frac{e^{\frac{a}{n}}-1}{1-e^{\frac{a}{n}}}\right)+{lim}_{n\to \mathrm{\infty }}\frac{1}{n}\times \frac{1}{e^{\frac{a}{n}}}\times \left(\frac{1-{\left(e^{\frac{a}{n}}\right)}^n}{1-e^{\frac{a}{n}}}\right)=$$$${lim}_{n\to \mathrm{\infty }}\left(\frac{\frac{1}{n}}{1-e^{\frac{a}{n}}}\right)={lim}_{n\to \mathrm{\infty }}\left(\frac{-\frac{1}{n^2}}{\frac{a}{n^2}{e}^{\frac{a}{n}}}\right)=-\frac{1}{a}$$$$={lim}_{n\to \mathrm{\infty }}\frac{1}{n}\times \frac{1}{e^{\frac{a}{n}}}\times \left(\frac{1-{\left(e^{\frac{a}{n}}\right)}^n}{1-e^{\frac{a}{n}}}\right)=$$$${lim}_{n\to \mathrm{\infty }}\frac{1}{n}(1+e^{\frac{a}{n}}+e^{\frac{2a}{n}}+...+e^{\frac{na}{n}})=$$$${lim}_{n\to \mathrm{\infty }}\underset{i=1}{\sum ^n}{e}^{\frac{i}{n}a}\times \frac{1}{n}={\int }_0^1{e}^{ax}dx=\frac{e^a-1}{a}$$$$\Rightarrow ans=\frac{e^a-2}{a}✓$$

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