Question

$={lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots +\frac{1}{\sqrt{n^2+(n-1)n}}]$ is equal to [RPET 2000]

• A. $2+2\sqrt{2}$
• B. $2\sqrt{2}-2$
• C. $2\sqrt{2}$
• D. $2$

Answer 1

The correct option is B $2\sqrt{2}-2$

$y={lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{\sqrt{n^2+n}}+\cdots +\frac{1}{\sqrt{n^2+(n-1)n}}]$
$\Rightarrow y={lim}_{n\to \infty }[\frac{1}{n}+\frac{1}{n\sqrt{1+\frac{1}{n}}}+\cdots +\frac{1}{n\sqrt{1+\frac{(n-1)}{n}}}]$
$\Rightarrow y=\frac{1}{n}{lim}_{n\to \infty }[1+\frac{1}{\sqrt{1+\frac{1}{n}}}+\cdots +\frac{1}{\sqrt{1+\frac{(n-1)}{n}}}]$
$y={lim}_{n\to \infty }\frac{1}{n}{\sum }_n^{k=1}\frac{1}{\sqrt{1+\frac{(k-1)}{n}}},Put\frac{k-1}{n}=xand\frac{1}{n}=dx$
$\Rightarrow y={lim}_{n\to \infty }{\int }_0^{\frac{n-1}{n}}\frac{dx}{\sqrt{1+x}}={lim}_{n\to \infty }2{\left[\sqrt{1+x}\right]}_0^{\left(\frac{n-1}{n}\right)}$
$\Rightarrow y=2{lim}_{n\to \infty }[\sqrt{\frac{2n-1}{n}}-1]=2{lim}_{n\to \infty }\sqrt{\frac{2n-1}{n}}-2$
$\Rightarrow y=2{lim}_{n\to \infty }\sqrt{2-\frac{1}{n}}-2=2\sqrt{2}-2$