Question

If $S_n=[\frac{1}{1+\sqrt{n}}+\frac{1}{2+\sqrt{2n}}+...+\frac{1}{n+\sqrt{n^2}}],$ then the value of $\underset{n\to \infty }{lim}{S}_n$ is equal to

• A. $\ln 2$
• B. $\ln 4$
• C. $\ln 8$
• D. $\ln 3$

Answer 1

The correct option is B $\ln 4$

$\underset{n\to \infty }{lim}{S}_n=\underset{n\to \infty }{lim}[\frac{1}{1+\sqrt{n}}+\frac{1}{2+\sqrt{2n}}+\cdots +\frac{1}{n+\sqrt{n^2}}]$
$=\underset{n\to \infty }{lim}\frac{1}{n}[\frac{1}{\frac{1}{n}+\frac{1}{\sqrt{n}}}+\frac{1}{\frac{2}{n}+\sqrt{\frac{2}{n}}}+\cdots +\frac{1}{\frac{n}{n}+\sqrt{\frac{n}{n}}}]$
$=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{n}\left(\frac{1}{\frac{r}{n}+\sqrt{\frac{r}{n}}}\right)$
$=\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{x}(\sqrt{x}+1)}$
Put $\sqrt{x}=z\Rightarrow \frac{1}{2\sqrt{x}}dx=dz$
$\Rightarrow \underset{n\to \infty }{lim}{S}_n=\underset{0}{\overset{1}{\int }}\frac{2dz}{z+1}=2{[\ln |z+1|]}_0^1=2(\ln 2-\ln 1)=\ln 4$