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.

Using $$a^m-b^m=(a-b)\left(a^{m-1}+a^{m-2}b+a^{m-3}b^2+...+ab^{m-2}+b^{m-1}\right)$$
For a fixed $m$ we have:
$$\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}=
\frac{(k+2)^{m-1}+(k+2)^{m-2}(k+1)+...+(k+2)(k+1)^{m-2}+(k+1)^{m-1}}{(k+1)^{m-1}+(k+1)^{m-2}k+...+(k+1)k^{m-2}+k^{m-1}}=\\
\left(\frac{k+2}{k+1}\right)^{m-1}\frac{1+\frac{k+1}{k+2}+...+\left(\frac{k+1}{k+2}\right)^{m-1}}{1+\frac{k}{k+1}+...+\left(\frac{k}{k+1}\right)^{m-1}} \rightarrow 1, k\rightarrow\infty$$
and any sequence with a finite limit is bounded.
