Proving that a sequence is monotone Let $ \{s_m\}_{m=1}^\infty$ be the sequence defined:
$s_m = \sum_{k=1}^{m} \frac{1}{\sqrt{m^2 + k}}$ 
I have already proven that $s_m\to 1$ as $m\to \infty $ but i'm having trouble to show that $s_m$ is a monotonic sequence.
Would appreciate any help or advice, thanks in advance.

The power series for
$(1+x)^{-1/2}$
with $0 < x < 1$
is an enveloping series,
in that its sum is between
any two consecutive finite sums.
Since
$(1+x)^{-1/2}
=1-\frac{x}{2}+\frac{3x^2}{8}-\frac{5x^3}{16}+...
$
we have
$1-\frac{x}{2}
\lt (1+x)^{-1/2}
\lt 1-\frac{x}{2}+\frac{3x^2}{8}
$
or
$(1+x)^{-1/2}
=1-\frac{x}{2}+ax^2
$
where
$0 < a < \frac38
$.
This is our basic inequality.
Note:
All the $a$ and $a_m$
in what follows are
values satisfying
$0 < a < \frac38$.
$\begin{array}\\
s_m 
&= \sum_{k=1}^{m} \frac{1}{\sqrt{m^2 + k}}\\
&= \frac1{m}\sum_{k=1}^{m} \frac{1}{\sqrt{1 + k/m^2}}\\
&= \frac1{m}\sum_{k=1}^{m} (1 + k/m^2)^{-1/2}\\
&= 1-\frac1{2m}\sum_{k=1}^{m} 
(k/m^2)+\frac1{m}\sum_{k=1}^{m}a_k(k/m^2)^2\\
&= 1-\frac{m(m+1)}{4m^3}+a_m\frac{m(m+1)(2m+1)}{6m^5}\\
&= 1-\frac{m+1}{4m^2}+a_m\frac{(m+1)(2m+1)}{6m^4}\\
\end{array}
$
In what follows,
"(W)" means that
Wolfy helped.
$\begin{array}\\
s_{m+1}-s_m
&=(1-\frac{m+2}{4(m+1)^2}+a_{m+1}\frac{(m+2)(2m+3)}{6(m+1)^4})-(1-\frac{m+1}{4m^2}+a_m\frac{(m+1)(2m+1)}{6m^4})\\
&=\frac{m+1}{4m^2}-\frac{m+2}{4(m+1)^2}+a_{m+1}\frac{(m+2)(2m+3)}{6(m+1)^4}-a_m\frac{(m+1)(2m+1)}{6m^4}\\
&=\frac{m^2 + 3 m + 1}{4 m^2 (m + 1)^2}+a_{m+1}\frac{(m+2)(2m+3)}{6(m+1)^4}-a_m\frac{(m+1)(2m+1)}{6m^4}
\qquad\text{(W)}\\
&\gt\frac{m^2 + 3 m + 1}{4 m^2 (m + 1)^2}-\frac38\frac{(m+1)(2m+1)}{6m^4}\\
&=\frac{2 m^4 + 5 m^3 - 5 m^2 - 5 m - 1}{16 m^4 (m + 1)^2}
\qquad\text{(W)}\\
&\gt 0\\
\end{array}
$
