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Question Number 155630 by SANOGO last updated on 03/Oct/21

Answered by Ar Brandon last updated on 03/Oct/21

$$\mathcal{L}=\underset{n\to \mathrm{\infty }}{\lim }n\underset{k=1}{\sum ^n}\frac{\exp (-\frac{n}{k})}{k^2}=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n}\underset{k=1}{\sum ^n}\frac{n^2}{k^2}{e}^{-\frac{n}{k}}$$$$={\int }_0^1\frac{e^{-\frac{1}{x}}}{x^2}dx={\int }_0^1{e}^{-\frac{1}{x}}d(-\frac{1}{x})={\left[e^{-\frac{1}{x}}\right]}_0^1=\frac{1}{e}$$

Commented by SANOGO last updated on 03/Oct/21

$$merci$$

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