<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] We can also approach the monotonicity of a bounded approaximation of $f(x)$ as follows.
\begin{eqnarray*}
f(x) &=& \frac{\left(\frac{1+2x}{1+x}\right)^{m} -1}{1-\left(\frac{1}{1+x}\right){m}} \\
&=& \frac{(1+2x)^{m}-(1+x)^{m}}{(1+x)^{m}-1^{m}} \\
&=&  \frac{x^{m} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}(1+2x)^{m-1-i}} \right)}{x \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}}\right)} \\
&=&  \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}(1+2x)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}}\right)} 
\end{eqnarray*}
For $x \ge -\frac{1}{2}$, 
\begin{eqnarray*}
f(x) &=&   \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}(1+2x)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}}\right)} \\
&\ge&  \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i} } \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}}\right)} \\
&=& x^{m-1}.
\end{eqnarray*}
Clearly this bound is monotonically increasing. 
Thinking about it, I guess, we can take it further and prove the monotonicity, without relying on the bound.
Upon re-arranging,.
\begin{eqnarray*}
f(x) &=&  \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}(1+2x)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1-i}{(1+x)^{i}}\right)} \\
&=&  \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{m-1}\left(\frac{1+2x}{1+x}\right)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{\frac{(1+x)^{m-1}}{(1+x)^{m-1}} (1+x)^{i}}\right)} \\
&=&  \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{\left(\frac{1+2x}{1+x}\right)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{\frac{1}{(1+x)^{m-1-i}} }\right)} \\
&=& x^{m-1}  \frac{\displaystyle\sum_{i=0}^{m-1}{\left(1+\frac{x}{1+x}\right)^{m-1-i}} }{ \displaystyle\sum_{i=0}^{m-1}{\left(1-\frac{x}{1+x}\right)^{m-1-i}}} \\
&=& x^{m-1}  \frac{\displaystyle\sum_{j=0}^{m-1}{\left(1+\frac{x}{1+x}\right)^{j}} }{ \displaystyle\sum_{j=0}^{m-1}{\left(1-\frac{x}{1+x}\right)^{j}}} 
\end{eqnarray*}
Since $\frac{x}{1+x}<1$, the numerator is monotonically increasing and denominator monotonically decreasing. The ratio then is monotonically increasing. 
</s>