Question

Evaluate $\underset{n\to \infty }{lim}[\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+.........+\frac{1}{\sqrt{n^2+2n}}]$

Answer 1

Given:

$F\left(n\right)=\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots +\frac{1}{\sqrt{n^2+2n}}$

Now,

Let $f\left(n\right)=[\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+...+\frac{1}{\sqrt{n^2}}]=\frac{2n+1}{\sqrt{n^2}}$
$...(2n+1)$ terms
$\underset{n\to \infty }{lim}f\left(n\right)=2$
$g\left(x\right)=[\frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+2n}}+....+\frac{1}{\sqrt{n^2+2n}}]$
$...(2n+1)$ terms
$\underset{n\to \infty }{lim}g\left(n\right)=2$

Now,

$g\left(n\right)<F\left(n\right)<f\left(n\right)$

Taking limits on all, we get

$\underset{n\to \infty }{lim}g\left(n\right)<\underset{n\to \infty }{lim}F\left(n\right)<\underset{n\to \infty }{lim}f\left(n\right)$

$\Rightarrow 2<\underset{n\to \infty }{lim}F\left(n\right)<2$

$\therefore \underset{n\to \infty }{lim}[\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+....+\frac{1}{\sqrt{n^2+2n}}]=2$