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.
 A: Using the mean value theorem, you can write 
$$g_m(k)=\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m} =\frac{mc_1^{m-1}}{mc_2^{m-1}}
$$ where $c_1 \in (k+1,k+2)$ and $c_2 \in (k,k+1)$. Therefore 
$$0\le g_m(k) \le \left(\frac{k+2}{k}\right)^{m-1}\le 3^{m-1}$$ which is a coarse bound.
A: 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. 
A: The numerator and denominator are both polynomials in $k$ of order $m-1$. The ratio of two polynomials of the same degree always has a limit as $k\rightarrow\infty$ and thus must be bounded. We may see this explicitly in this case as follows:
Using the identity $a^m-b^m=(a-b)\sum_{i=0}^{m-1}a^{m-1-i}b^i$ we see that
$$\frac{\phi(k+2)-\phi(k+1)}{\phi(k+1)-\phi(k)}=\frac{\sum_{i=0}^{m-1}(k+2)^{m-1-i}(k+1)^i}{\sum_{i=0}^{m-1}(k+1)^{m-1-i}k^i}$$
Each term in the sum in the numerator is less than or equal to $(k+2)^m$, and every term in the sum in the denominator is greater than or equal to $k^m$ so that
$$\frac{\sum_{i=0}^{m-1}(k+2)^{m-1-i}(k+1)^i}{\sum_{i=0}^{m-1}(k+1)^{m-1-i}k^i}\leq\frac{m(k+2)^m}{mk^m}=\frac{k^m+\sum_{i=0}^{m-1}a_ik^i}{k^m}$$
where the $a_i$ depend only on $m$ and not on $k$. Finally, the above sum becomes
$$1+\sum_{i=0}^{m-1}a_i\frac{k^i}{k^m}\leq 1+\sum_ia_i$$ since $\frac{k_i}{k^m}\leq1$
A: 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.
A: 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.
A: 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 $f_m$. 
