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Question Number 50405 by Abdo msup. last updated on 16/Dec/18

$$let{V}_n=\frac{1}{2n+1}+\frac{1}{2n+3}+...+\frac{1}{4n-1}$$$$determine{lim}_{n\to +\mathrm{\infty }}{V}_n$$

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

$$V_n=\frac{1}{2n+1}+\frac{1}{2n+3}+...+\frac{1}{2n+2n-1}$$$$V_n=\frac{\frac{1}{n}}{2+\frac{1}{n}}+\frac{\frac{1}{n}}{2+\frac{3}{n}}+...+\frac{\frac{1}{n}}{2+\frac{2n-1}{n}}$$$$li\underset{n\to \mathrm{\infty }}{m}\frac{1}{n}\underset{r=1}{\sum ^n}\frac{1}{1+\left(\frac{2r-1}{n}\right)}=li\underset{n\to \mathrm{\infty }}{m}\frac{1}{n}\underset{r=1}{\sum ^n}(\frac{1}{1+2\left(\frac{r}{n}\right)-\frac{1}{n}}$$$${\int }_0^1\frac{dx}{1+2x}=\frac{1}{2}{\int }_0^1\frac{dx}{\frac{1}{2}+x}=\frac{1}{2}\mid ln(\frac{1}{2}+x){\mid }_0^1$$$$=\frac{1}{2}[ln\left(\frac{3}{2}\right)-ln\left(\frac{1}{2}\right)]=\frac{1}{2}ln3$$$$$$$$plscheckisitcorrect...$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

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