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.
 A: By way of making up for my earlier error (note to self: don't math late at night!), here is another attempt. We have
$$S_{n} \,=\, \sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}} \,=\, \underbrace{\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + \cdots + \frac{1}{\sqrt{n^2+n}}}_{n \text{  terms} }.$$ 
From here it should be clear that
$$\frac{n}{\sqrt{n^2+n}} \,=\, \frac{n}{\sqrt{n(n+1)}} \,<\, S_{n} \,<\, \frac{n}{\sqrt{n^2+1}} \qquad \forall\,n\in\mathbb{N}.$$
You may already have this bound as you mention you have shown $S_{n}\rightarrow 1$ as $n\rightarrow \infty$; this follows from above and the Squeeze theorem. Using this, we also have
$$S_{n+1}\,>\,\frac{n+1}{\sqrt{(n+1)(n+2)}}. $$
Since all the terms in the inequalities are positive, we have:
$$\begin{align}
\frac{S_{n+1}}{S_{n}} &\,>\, \frac{n+1}{\sqrt{(n+1)(n+2)}}\cdot\frac{\sqrt{n(n+1)}}{n} =  \sqrt{\frac{n(n+1)^{3}}{n^{2}(n+1)(n+2)}\,} =  \sqrt{\frac{n(n+1)^{2}}{n^{2}(n+2)}\,} \\[0.2cm]
&\,=\, \sqrt{\frac{n^{3}+2n^{2}+n}{n^{3}+2n^{2}}\,} = \sqrt{1+\frac{n}{n^{3}+2n^{2}}\,} \,>\, 1.
\end{align}$$
Since $S_{n}>0$ for all $n\in\mathbb{N}$, we have (after multiplication by $S_{n}$) that $S_{n+1}>S_{n}$ for all $n\in\mathbb{N}$ and hence $(S_{n})$ is monotonically increasing. 
A: 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}
$
