Show that, for all $m\geq 2$, there exists $C_m>0$ such that, for all $k\geq 1$, $\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq C_m$ Define $\phi:\mathbb{N}\to \mathbb{N}$ by $\phi(k)=k^m$, where $m\geq 2$ is some fixed number. I want to investigate, if there exists $C_m>0$ such that $$\frac{\phi(k+2)-\phi(k+1)}{\phi(k+1)-\phi(k)}=\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq C_m$$ for all $k\geq 1$. Since
$$
\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}=\frac{\left ( \frac{1+2/k}{1+1/k} \right )^m-1}{1-\left ( \frac{1}{1+1/k} \right )^m},
$$
we define $f:(0,1]\to\mathbb{R}$ by 
$$
f(x)=\frac{\left ( \frac{1+2x}{1+x} \right )^m-1}{1-\left ( \frac{1}{1+x} \right )^m}
$$
I suspect that $f(x)$ is increasing, and so $f(1/x)$ must be decreasing (through some calculator graphs), which implies that 
$$
\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq \frac{(1+2)^m-(1+1)^m}{(1+1)^m-1^m}=\frac{3^m-2^m}{2^m-1}=:C_m
$$
for all $k\geq 1$. The question is, how do I show the monotonicity of $f(x)$ or $f(1/x)$ in a simplest way? Determining $f'$ and then checking if it's greater than $0$ on $(0,1]$ takes a lot time.

We will show that the function $f_m(x)=\frac{(x+2)^m-(x+1)^m}{(x+1)^m-x^m}$ is decreasing for $x>0$ and $m>1$. 
Calculate the derivative $f'(x)$:
$f'_m(x)=m\frac{[(x+2)^{m-1}-(x+1)^{m-1}][(x+1)^{m}-x^m]-[(x+2)^{m}-(x+1)^{m}][(x+1)^{m-1}-x^{m-1}]}{((x+1)^{m}-x^m)^2}$
After some algebra one can rearrange this expression into the following:
\begin{align}f'_m(x)=&m[x(x+1)(x+2)]^{m-1}((x+1)^{m}-x^m)^{-2}\times\\&\Big[\frac{1}{(x+1)^{m-1}}-\frac{1}{x^{m-1}}-\Big(\frac{1}{(x+2)^{m-1}}-\frac{1}{(x+1)^{m-1}}\Big)\Big]\end{align}
so it suffices to check if the auxiliary function $g_m(x)=\frac{1}{(x+1)^{m-1}}-\frac{1}{x^{m-1}}$ is increasing.
Indeed:
$g'_m(x)=(m-1)(\frac{1}{x^m}-\frac{1}{(x+1)^m})>0 \hspace{0.3 cm}\forall x>0, m>1$
from which we conclude that $g_m(x)\lt g_m(x+1)$ for $x>0$ and finally
$f_m'(x)=m[x(x+1)(x+2)]^{m-1}((x+1)^{m}-x^m)^{-2}(g_m(x)-g_m(x+1))<0$ and therefore f is decreasing for $x>0, m>1$  (QED).
Given the restrictions of the problem we conclude that $f_m(x)\leq f_m(1)$ for $x>1$ and thus 
$f_m(x)\leq\frac{3^m-2^m}{2^m-1}\equiv C_m$ and this concludes the proof for the strictest bound of the required form on $f_m$. 
