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Question Number 144394 by SOMEDAVONG last updated on 25/Jun/21

$$\mathrm{L}=\underset{\mathrm{n}\to +\propto }{\lim }(\frac{1}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+2}+...+\frac{1}{\mathrm{n}+\mathrm{n}})=?$$

Commented by Canebulok last updated on 25/Jun/21

$$$$$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:$$$$\Rightarrow \underset{n\to +\mathrm{\infty }}{\lim }\left(\underset{k=1}{\sum ^n}\frac{1}{n+k}\right)=L$$$$\Rightarrow \underset{n\to +\mathrm{\infty }}{\lim }\left(\frac{1}{n}\right)\left(\underset{k=1}{\sum ^n}\frac{n}{n+k}\right)=L$$$$\Rightarrow \underset{n\to +\mathrm{\infty }}{\lim }\left(\frac{1}{n}\right)\left(\underset{k=1}{\sum ^n}\frac{1}{(1+\frac{k}{n})}\right)=L$$$$\therefore$$$$\Rightarrow {\int }_0^1\frac{1}{x+1}dx={\left[Ln(\mid x+1\mid )\right]}_0^1$$$$\Rightarrow L=Ln(\mid 2\mid )$$

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