Question

$\underset{n\square \infty }{\lim }\left\{\frac{1}{2n+1}+\frac{1}{2n+2}+...+\frac{1}{2n+n}\right\}$ is equal to

(a) $\ln \left(\frac{1}{3}\right)$

(b) $\ln \left(\frac{2}{3}\right)$

(c) $\ln \left(\frac{3}{2}\right)$

(d) $\ln \left(\frac{4}{3}\right)$

Answer 1

(c) ln(3/2)

$$\begin{gathered} \underset{n\to \infty }{\lim }\left\{\frac{1}{2n+1}+\frac{1}{2n+2}+..........+\frac{1}{2n+n}\right\} \\ =\underset{n\to \infty }{\lim }{\sum }_{r=1}^n\frac{1}{2n+r} \\ =\underset{n\to \infty }{lim}\frac{1}{n}{\sum }_{r=1}^n\frac{1}{2+{\displaystyle \frac{r}{n}}} \\ let\frac{r}{n}=x \\ ={\int }_0^{\infty }\frac{1}{2+x}dx \\ ={\left[\log \left(2+x\right)\right]}_0^{\infty } \\ =\log 3-\log 2 \\ =log\frac{3}{2} \\ =\ln \left(\frac{3}{2}\right) \end{gathered}$$