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
 A: If
$s_n
=\sum_{k=0}^n \dfrac1{\sqrt{n+k}}
$
then
$s_n
\ge \sum_{k=0}^n \dfrac1{\sqrt{2n}}
=(n+1)\dfrac1{\sqrt{2n}}
\gt\dfrac{n}{\sqrt{2n}}
=\dfrac{\sqrt{n}}{\sqrt{2}}
\to \infty
$
Note that,
for any $c > 0$,
if
$s_n(c)
=\sum_{k=0}^{cn} \dfrac1{\sqrt{n+k}}
$
then
$\begin{array}\\
s_n(c)
&\ge \sum_{k=0}^{cn} \dfrac1{\sqrt{n+cn}}\\
&\ge cn\dfrac1{\sqrt{(1+c)n}}\\
&=\dfrac{c\sqrt{n}}{\sqrt{1+c}}\\
&\to \infty\\
\end{array}
$
This also works
for any $c(n)$ such that
$c(n)\sqrt{n} \to \infty$.
A: Hint:
For each $k=0,1,\dots,n$, one has $\;\frac 1{\sqrt{n+k}}\ge\frac1{\sqrt{2n}}$. How many such terms do you have? Deduce a lower bound for the sum of these terms.
A: 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}
$.
A: You could have quite good approximations using generalized harmonic numbers
$$S_n=\sum_{i=0}^n \frac 1 {\sqrt{n+i}}=H_{2 n}^{\left(\frac{1}{2}\right)}-H_{n-1}^{\left(\frac{1}{2}\right)}$$
Using the asymptotics
$$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{p}\right)^{1/2}-\frac{1}{24}
   \left(\frac{1}{p}\right)^{3/2}+O\left(\frac{1}{p^{7/2}}\right)$$ Using it twice and continuing with Taylor series
$$S_n=2 \left(\sqrt{2}-1\right) \sqrt{n}+\frac{\left(2+\sqrt{2}\right)}{4 \sqrt n} 
   +O\left(\frac{1}{n^{3/2}}\right)$$ Just trying for $n=100$, the value should be $\approx 8.369654$ while the above truncated series gives $\frac{3}{40} \left(267 \sqrt{2}-266\right) \approx 8.369627$
A: $$\frac1{\sqrt n}\sum_{k=0}^n\frac1{\sqrt{n+k}}=\frac1n\sum_{k=0}^n\frac1{\sqrt{1+\dfrac kn}}\to\int_0^1\frac{dx}{\sqrt{1+x}}$$ is a Cauchy sum which converges to $$2(\sqrt2-1).$$
Your sum is $\sqrt n$ times larger.
