Vtyjkyuk

Define $phi:mathbbNto mathbbN$ by $phi(k)=k^m$, where $mgeq 2$ is some fixed number. I want to investigate, if there exists $C_m>0$ such that $$fracphi(k+2)-phi(k+1)phi(k+1)-phi(k)=frac(k+2)^m-(k+1)^m(k+1)^m-k^mleq C_m$$ for all $kgeq 1$. Since
$$
frac(k+2)^m-(k+1)^m(k+1)^m-k^m=fracleft ( frac1+2/k1+1/k right )^m-11-left ( frac11+1/k right )^m,
$$
we define $f:(0,1]tomathbbR$ by
$$
f(x)=fracleft ( frac1+2x1+x right )^m-11-left ( frac11+x right )^m
$$
I suspect that $f(x)$ is increasing, and so $f(1/x)$ must be decreasing (through some calculator graphs), which implies that
$$
frac(k+2)^m-(k+1)^m(k+1)^m-k^mleq frac(1+2)^m-(1+1)^m(1+1)^m-1^m=frac3^m-2^m2^m-1=:C_m
$$
for all $kgeq 1$. 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 the mean value theorem, you can write
$$g_m(k)=frac(k+2)^m-(k+1)^m(k+1)^m-k^m =fracmc_1^m-1mc_2^m-1
$$ where $c_1 in (k+1,k+2)$ and $c_2 in (k,k+1)$. Therefore
$$0le g_m(k) le left(frack+2kright)^m-1le 3^m-1$$ which is a coarse bound.

We can also approach the monotonicity of a bounded approaximation of $f(x)$ as follows.

begineqnarray*
f(x) &=& fracleft(frac1+2x1+xright)^m -11-left(frac11+xright)m \
&=& frac(1+2x)^m-(1+x)^m(1+x)^m-1^m \
&=& fracx^m left(displaystylesum_i=0^m-1(1+x)^i(1+2x)^m-1-i right)x left(displaystylesum_i=0^m-1(1+x)^iright) \
&=& fracx^m-1 left(displaystylesum_i=0^m-1(1+x)^i(1+2x)^m-1-i right) left(displaystylesum_i=0^m-1(1+x)^iright)
endeqnarray*

For $x ge -frac12$,

begineqnarray*
f(x) &=& fracx^m-1 left(displaystylesum_i=0^m-1(1+x)^i(1+2x)^m-1-i right) left(displaystylesum_i=0^m-1(1+x)^iright) \
&ge& fracx^m-1 left(displaystylesum_i=0^m-1(1+x)^i right) left(displaystylesum_i=0^m-1(1+x)^iright) \
&=& x^m-1.
endeqnarray*

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,.

begineqnarray*
f(x) &=& fracx^m-1 left(displaystylesum_i=0^m-1(1+x)^i(1+2x)^m-1-i right) left(displaystylesum_i=0^m-1-i(1+x)^iright) \
&=& fracx^m-1 left(displaystylesum_i=0^m-1(1+x)^m-1left(frac1+2x1+xright)^m-1-i right) left(displaystylesum_i=0^m-1frac(1+x)^m-1(1+x)^m-1 (1+x)^iright) \
&=& fracx^m-1 left(displaystylesum_i=0^m-1left(frac1+2x1+xright)^m-1-i right) left(displaystylesum_i=0^m-1frac1(1+x)^m-1-i right) \
&=& x^m-1 fracdisplaystylesum_i=0^m-1left(1+fracx1+xright)^m-1-i displaystylesum_i=0^m-1left(1-fracx1+xright)^m-1-i \
&=& x^m-1 fracdisplaystylesum_j=0^m-1left(1+fracx1+xright)^j displaystylesum_j=0^m-1left(1-fracx1+xright)^j
endeqnarray*

Since $fracx1+x<1$, the numerator is monotonically increasing and denominator monotonically decreasing. The ratio then is monotonically increasing.

Using $$a^m-b^m=(a-b)left(a^m-1+a^m-2b+a^m-3b^2+...+ab^m-2+b^m-1right)$$
For a fixed $m$ we have:
$$frac(k+2)^m-(k+1)^m(k+1)^m-k^m=
frac(k+2)^m-1+(k+2)^m-2(k+1)+...+(k+2)(k+1)^m-2+(k+1)^m-1(k+1)^m-1+(k+1)^m-2k+...+(k+1)k^m-2+k^m-1=\
left(frack+2k+1right)^m-1frac1+frack+1k+2+...+left(frack+1k+2right)^m-11+frackk+1+...+left(frackk+1right)^m-1 rightarrow 1, krightarrowinfty$$
and any sequence with a finite limit is bounded.

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 $krightarrowinfty$ and thus must be bounded. We may see this explicitly in this case as follows:

Using the identity $a^m-b^m=(a-b)sum_i=0^m-1a^m-1-ib^i$ we see that
$$fracphi(k+2)-phi(k+1)phi(k+1)-phi(k)=fracsum_i=0^m-1(k+2)^m-1-i(k+1)^isum_i=0^m-1(k+1)^m-1-ik^i$$

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
$$fracsum_i=0^m-1(k+2)^m-1-i(k+1)^isum_i=0^m-1(k+1)^m-1-ik^ileqfracm(k+2)^mmk^m=frack^m+sum_i=0^m-1a_ik^ik^m$$
where the $a_i$ depend only on $m$ and not on $k$. Finally, the above sum becomes
$$1+sum_i=0^m-1a_ifrack^ik^mleq 1+sum_ia_i$$ since $frack_ik^mleq1$

Playing naively...

$beginarray\
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\
&=dfrac1+2m/k+O(1/k^2)-(1+m/k+O(1/k^2)(1+m/k+O(1/k^2))-1\
&=dfracm/k+O(1/k^2)m/k+O(1/k^2)\
&=1+O(1/k)\
endarray
$

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)
=dfrac3^m-2^m2^m-1
=dfrac(3/2)^m-11-2^-m
gt (3/2)^m-1
$.

If we can show that
$r_m(k)$
is a decreasing function of $k$,
we are essentially done.

$beginarray\
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)\
&=dfracmx+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)\
&=dfracmx+3m(m-1)x^2+O(x^3)mx+m(m-1)x^2/2+O(x^3)\
&=dfrac1+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)\
endarray
$

Therefore,
for small enough $x$,
$f(1/x)$ is increasing
so $f(k)$
is decreasing.

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 $f'(x)$:

$f'_m(x)=mfrac[(x+2)^m-1-(x+1)^m-1][(x+1)^m-x^m]-[(x+2)^m-(x+1)^m][(x+1)^m-1-x^m-1]((x+1)^m-x^m)^2$

After some algebra one can rearrange this expression into the following:

beginalignf'_m(x)=&m[x(x+1)(x+2)]^m-1((x+1)^m-x^m)^-2times\&Big[frac1(x+1)^m-1-frac1x^m-1-Big(frac1(x+2)^m-1-frac1(x+1)^m-1Big)Big]endalign

so it suffices to check if the auxiliary function $g_m(x)=frac1(x+1)^m-1-frac1x^m-1$ is increasing.

Indeed:

$g'_m(x)=(m-1)(frac1x^m-frac1(x+1)^m)>0 hspace0.3 cmforall x>0, m>1$

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

$f_m(x)leqfrac3^m-2^m2^m-1equiv C_m$ and this concludes the proof for the strictest bound of the required form on $f_m$.