Show that $ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$ Show that $$ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$$
LHS : $ \lim_{n\to\infty}\frac{1}{n}[\frac{n}{\sqrt n}+\frac{n}{\sqrt {n+1}}+\frac{n}{\sqrt {n+2}}.......\frac{n}{\sqrt {2n}}] = \infty$
Let $a_n=\frac{n}{\sqrt{2n}}$ then $a_n = \frac{1}{\sqrt 2}\sqrt{n}$
hence $\lim_{n\to\infty}a_n = \infty$
I continue the problem therefore using Cauchy's first theorem on limits.
But the way I took $a_n$ is it correct?  
Does it not make $a_1 = \frac{1}{\sqrt{2}}$
If my way of taking $a_n$ is wrong, please suggest me the right way. Thank you

Here is an
elementary proof that
$\dfrac{\sqrt{2}}{3n}
\lt \dfrac{s_n}{\sqrt{n}}-2(\sqrt{2}-1)
\lt \dfrac{2\sqrt{2}}{n}
$.
$\begin{array}\\
\sqrt{m+1}-\sqrt{m}
&=(\sqrt{m+1}-\sqrt{m})\dfrac{\sqrt{m+1}+\sqrt{m}}{\sqrt{m+1}+\sqrt{m}}\\
&=\dfrac{1}{\sqrt{m+1}+\sqrt{m}}\\
&>\dfrac{1}{2\sqrt{m+1}}\\
\text{and}\\
\sqrt{m+1}-\sqrt{m}
&<\dfrac{1}{2\sqrt{m}}\\
\end{array}
$
so
$2(\sqrt{m+1}-\sqrt{m})
<\dfrac{1}{\sqrt{m}}
< 2(\sqrt{m}-\sqrt{m-1})
$.
Therefore
$\begin{array}\\
s_n
&=\sum_{k=0}^n \dfrac1{\sqrt{n+k}}\\
&\gt\sum_{k=0}^n 2(\sqrt{n+k+1}-\sqrt{n+k})\\
&=2(\sqrt{2n+1}-\sqrt{n})\\
&=2\sqrt{n}(\sqrt{2+1/n}-1)\\
&=2\sqrt{n}(\sqrt{2}\sqrt{1+1/(2n)}-1)\\
&\gt2\sqrt{n}(\sqrt{2}(1+1/(6n))-1)
\quad\text{since }\sqrt{1+x} > 1+x/3 
\text{ for } 0 < x < 3\\
&\gt2\sqrt{n}(\sqrt{2}-1)+\dfrac{\sqrt{2}}{3\sqrt{n}}\\
\text{and}\\
s_n
&=\sum_{k=0}^n \dfrac1{\sqrt{n+k}}\\
&\lt\sum_{k=0}^n 2(\sqrt{n+k}-\sqrt{n+k-1})\\
&=2(\sqrt{2n}-\sqrt{n-1})\\
&=2\sqrt{n}(\sqrt{2}-\sqrt{1-1/n})\\
&\lt2\sqrt{n}(\sqrt{2}-(1-1/n))
\quad\text{since }\sqrt{1-x} > 1-x 
\text{ for } 0 < x < 1\\
&=2\sqrt{n}(\sqrt{2}-1)+\dfrac{2\sqrt{2}}{\sqrt{n}}\\
\end{array}
$
Therefore
$\dfrac{\sqrt{2}}{3n}
\lt \dfrac{s_n}{\sqrt{n}}-2(\sqrt{2}-1)
\lt \dfrac{2\sqrt{2}}{n}
$.
