<s>[INST] 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.
 [/INST] Playing naively...
$\begin{array}\\
r_m(k)
&=\dfrac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\\
&=\dfrac{(1+2/k)^m-(1+1/k)^m}{(1+1/k)^m-1}\\
&=\dfrac{1+2m/k+O(1/k^2)-(1+m/k+O(1/k^2)}{(1+m/k+O(1/k^2))-1}\\
&=\dfrac{m/k+O(1/k^2)}{m/k+O(1/k^2)}\\
&=1+O(1/k)\\
\end{array}
$
So $1$ is a candidate.
If
$f(x) = x^m$,
then
$f'(x) = mx^{m-1}$
and
$f''(x) = m(m-1)x^{m-2}$
so
$f''(x) \ge 0$
is $m \ge 2$.
Therefore
$\frac12(f(x)+f(x+2))
\ge f(x+1)
$
so
$f(x)+f(x+2)
\ge 2f(x+1)
$
or
$f(x+2)-f(x+1)
\ge f(x+1)-f(x)
$.
Therefore
$r_m(k) \ge 1$.
$r_m(1)
=\dfrac{3^m-2^m}{2^m-1}
=\dfrac{(3/2)^m-1}{1-2^{-m}}
\gt (3/2)^m-1
$.
If we can show that
$r_m(k)$
is a decreasing function of $k$,
we are essentially done.
$\begin{array}\\
r_m(1/x)
&=\dfrac{(1+2x)^m-(1+x)^m}{(1+x)^m-1}\\
&=\dfrac{(1+2mx+m(m-1)4x^2)-(1+mx+m(m-1)x^2/2)+O(x^3)}{(1+mx+m(m-1)x^2/2)-1+O(x^3)}\\
&=\dfrac{mx+m(m-1)4x^2/2)-m(m-1)x^2/2)+O(x^3)}{mx+m(m-1)x^2/2+O(x^3)}\\
&=\dfrac{mx+3m(m-1)x^2+O(x^3)}{mx+m(m-1)x^2/2+O(x^3)}\\
&=\dfrac{1+3(m-1)x+O(x^2)}{1+(m-1)x/2+O(x^2)}\\
&=(1+3(m-1)x+O(x^2))(1-(m-1)x/2+O(x^2))\\
&=1+2(m-1)x+O(x^2)\\
\end{array}
$
Therefore,
for small enough $x$,
$f(1/x)$ is increasing
so $f(k)$
is decreasing.
</s>