Question Number 108474 by pticantor last updated on 17/Aug/20
$$\boldsymbol{{U}}_{\boldsymbol{{n}}} =\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+{k}}=? \\ $$$$\boldsymbol{{li}}\underset{\boldsymbol{{n}}>\infty} {\boldsymbol{{m}U}}_{{n}} =? \\ $$
Answered by Dwaipayan Shikari last updated on 17/Aug/20
![Σ_(k=0) ^(2n−1) (1/(n(2+(k/n))))=(1/n)lim_(n→∞ ) Σ_(k=1) ^(2n) (1/(2+(k/n)))=∫_0 ^2 (1/(2+x))dx=log[(x+2)]_0 ^2 =log(2)](https://www.tinkutara.com/question/Q108500.png)
$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{n}\left(\mathrm{2}+\frac{{k}}{{n}}\right)}=\frac{\mathrm{1}}{{n}}\underset{{n}\rightarrow\infty\:} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}+\frac{{k}}{{n}}}=\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}+{x}}{dx}={log}\left[\left({x}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} ={log}\left(\mathrm{2}\right) \\ $$