The value of $\lim _{n \rightarrow+\infty}\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\ldots .

Question

The value of $\lim _{n \rightarrow+\infty}\left(\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\ldots .+\frac{1}{\sqrt{n^2+2 n}}\right)$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

Let $f(x) = \left( \frac{1}{\sqrt{n^2}} + \frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \dots + \frac{1}{\sqrt{n^2+2n}} ight)$

As, $\frac{1}{\sqrt{n^2+2n}} < \frac{1}{\sqrt{n^2}} = \frac{1}{\sqrt{n^2}}$

$\frac{1}{\sqrt{n^2+2n}} < \frac{1}{\sqrt{n^2+1}} < \frac{1}{\sqrt{n^2}}$

On adding, we get $\frac{2n+1}{\sqrt{n^2+2n}} < \frac{1}{\sqrt{n^2}} + \frac{1}{\sqrt{n^2+1}} + \dots + \frac{1}{\sqrt{n^2+2n}} < \frac{2n+1}{\sqrt{n^2}}$

$2 < \lim_{nightarrow\infty} \left( \frac{1}{\sqrt{n^2}} + \frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \dots + \frac{1}{\sqrt{n^2+2n}} ight) < 2$

$	herefore \lim_{nightarrow\infty} \left( \frac{1}{\sqrt{n^2}} + \frac{1}{\sqrt{n^2+1}} + \dots + \frac{1}{\sqrt{n^2+2n}} ight) = 2$