Question Number 126886 by SOMEDAVONG last updated on 25/Dec/20
$$\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{n}+\mathrm{3}}\:+….+\:\frac{\mathrm{1}}{\mathrm{2n}}\right) \\ $$
Answered by Dwaipayan Shikari last updated on 25/Dec/20
$$\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{n}}}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}}{{n}}}+….+\frac{\mathrm{1}}{\mathrm{1}+\frac{{n}}{{n}}}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\frac{{r}}{{n}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}{dx}={log}\left(\mathrm{2}\right) \\ $$