Question Number 215315 by depressiveshrek last updated on 02/Jan/25
$$\mathrm{Evaluate}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}+...+\frac{\mathrm{1}}{\mathrm{4}{n}}\right) \\ $$
Answered by mr W last updated on 02/Jan/25
![lim_(n→∞) Σ_(k=1) ^(2n) (1/(2n+k)) =lim_(n→∞) Σ_(k=1) ^(2n) (1/(2n(1+(k/(2n))))) =∫_0 ^1 (dx/(1+x)) =[ln (1+x)]_0 ^1 =ln 2](Q215317.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+{k}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}\left(\mathrm{1}+\frac{{k}}{\mathrm{2}{n}}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}} \\ $$$$=\left[\mathrm{ln}\:\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{ln}\:\mathrm{2} \\ $$
Commented by depressiveshrek last updated on 02/Jan/25
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by MrGaster last updated on 02/Jan/25
![lim_(n→∞) Σ_(k=1) ^(2n) (1/(2n+k)) =lim_(n→∞) (1/n)Σ_(k=1) ^(2n) (1/(2+(k/n))) =∫_0 ^2 (1/(2+x))dx =[ln∣2+x∣]_0 ^2 =ln∣4∣−ln∣2∣ =ln((4/2)) =ln 2](Q215321.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+{k}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}+\frac{{k}}{{n}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}+{x}}{dx} \\ $$$$=\left[\mathrm{ln}\mid\mathrm{2}+{x}\mid\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\mathrm{ln}\mid\mathrm{4}\mid−\mathrm{ln}\mid\mathrm{2}\mid \\ $$$$=\mathrm{ln}\left(\frac{\mathrm{4}}{\mathrm{2}}\right) \\ $$$$=\mathrm{ln}\:\mathrm{2} \\ $$