Question

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

• A. $\log 2$
• B. $\log 4$
• C. $\log 8$
• D. $Noneofthese$

Answer 1

The correct option is B $\log 4$

${lim}_{n\to \infty }{S}_n={lim}_{n\to \infty }[\frac{1}{1+\sqrt{n}}+\frac{1}{2+\sqrt{2n}}+\cdots +\frac{1}{n+\sqrt{n^2}}]$

$={lim}_{n\to \infty }\frac{1}{n}[\frac{1}{\frac{1}{n}+\frac{1}{\sqrt{n}}}+\frac{1}{\frac{2}{n}+\sqrt{\frac{2}{n}}}+\cdots +\frac{1}{\frac{n}{n}+\sqrt{\frac{n}{n}}}]$

$={\int }_0^1\frac{dx}{\sqrt{x}(\sqrt{x}+1)}$

Put $\sqrt{x}=z$
$\therefore \frac{1}{2\sqrt{x}}dx=dz$

$\Rightarrow {lim}_{n\to \infty }{S}_n={\int }_0^1\frac{2dz}{z+1}=2|\log (z+1){|}_0^1$
$=2(\log 2-\log 1)$
$=2\log 2=\log 4$