Let 
\begin{equation}
 \ell\_{m,p}:=\sum\_{j=1}^m\gamma\_{m,j}\sigma\_{p,j},
\end{equation}
where 
\begin{equation}
 \gamma\_{m,j}:=\frac{2 (-1)^{j-1} }{j }\binom{2 m}{m+j}\Big/\binom{2 m}{m},\quad
 \sigma\_{p,j}:=\sum \_{i=0}^{j-1} \left(\frac{j}{2}-i-1\right)^p, 
\end{equation}
$p$ and $m$ are natural numbers, and 
\begin{equation}
m\ge m\_p:=\left\lceil \frac{p+1}{2}\right\rceil. 
\end{equation}



> 
> **Problem 1** (the more important one to me): Show that $\ell\_{m,p}$ does not depend on $m$ (as long as $m\ge m\_p$).
> 
> 
> 




---



> 
> **Problem 2** (a more specific and difficult (?) form of Problem 1): Show that (again for $m\ge m\_p$)  
> 
>  \begin{equation} \ell\_{m,p}=B\_p,
>  \tag{1} \end{equation} where $B\_p$ is the $p$th Bernoulli number --
>  see e.g. \url{<http://mathworld.wolfram.com/BernoulliNumber.html>}
>  and/or \url{<https://en.wikipedia.org/wiki/Bernoulli_number>}. Formula
>  $(1)$ has been verified for $p=1,\dots,20$ and $m=m\_p,\dots,m\_p+5$.
> 
> 
> 


I think I know how to show that the restriction $m\ge m\_p$ cannot be relaxed, for any given natural $p$.


These problems arise in a certain work in approximation theory. 



A representation of the Bernoulli numbers
I'll sketch the idea behind your claims, starting with Problem 1.


**Step 1:** Convince yourself that if we denote
$$F\_p(j):=\frac1j\sum\_{i=0}^{j-1}\left(\frac{j}2-i-1\right)^p$$ 
then $F\_p(j)$ is always an **even** polynomial in $j$; a polynomial in $j^2$. For example, when $p$ is odd, we get $F\_p(j)=-\left(\frac{j}2\right)^{p-1}$. 


**Step 2:** Therefore, from Step 1, it suffices to consider even powers of $j$ instead of $F\_p(j)$. We shall also disregard $\frac{-2}{\binom{2m}m}$ and focus on the sum
$$G\_p(m):=\sum\_{j=1}^m(-1)^j\binom{2m}{m-j}j^{e}$$
where $e\geq0$ is a fixed even integer. Observe that 
$$\sum\_{j=-m}^m(-1)^j\binom{2m}{m-j}j^e
=\begin{cases} 2G\_p(m) \qquad \qquad \text{if $e\neq0$} \\
 2G\_p(m)+1 \qquad \,\,\text{if $e=0$}.\end{cases}$$
**Step 3:** After re-indexing $k=m-j$,
$$\sum\_{j=-m}^m(-1)^j\binom{2m}{m-j}j^e=(-1)^m\sum\_{k=0}^{2m}(-1)^k\binom{2m}k(m-k)^e.$$
It's well-known (actually the crux of the matter here) that if $n>a$ then we've the vanishing of
$$\sum\_{k=0}^n(-1)^k\binom{n}k k^a=0.\tag{0}$$
The reason why you get a persistent value (for $m\geq m\_p$) is caused by the one single term when $e=0$ which happens for $p$ even.


At any rate, we gather that if $e$ is even and $m\geq \frac{e}2$ then
$$G\_p(m)=\begin{cases} \,\,\,\,\,0 \qquad \,\text{when $e\neq0$} \\
-\frac12 \qquad \text{when $e=0$}.
\end{cases}$$


To understand Problem 2, it remains to identify the constant term in $F\_p(j)$ w.r.t. the variable $j$. To this end, the Binomial Theorem furnishes
\begin{align} F\_p(j)
=\frac1j\sum\_{i=1}^j\sum\_{k=0}^p\binom{p}k\left(\frac{j}2\right)^{p-k}(-1)^ki^k
=\sum\_{k=0}^p\binom{p}k\left(\frac{j}2\right)^{p-1-k}(-1)^k\sum\_{i=1}^ji^k
\tag{1}
\end{align}
revealing that the only way to encounter the sought-after constant term is provided that $k=p$. Keeping in mind that $\sum\_{i=1}^ji^k$ is divisible by $j$, we just look at the quantity (using know expressions for Bernoulli numbers)
$$\frac{(-1)^p}j\sum\_{i=1}^ji^p=(-1)^p\left[j^{p-1}+\frac1{p+1}\sum\_{r=0}^p\binom{p+1}rB\_rj^{p-r}\right].\tag{2}$$
Thus, the term we seek is $(-1)^pB\_p=B\_p$ (remember: $p$ is even) leading to
\begin{align}
\frac2{\binom{2m}m}\sum\_{j=1}^m(-1)^{j-1}\binom{2m}{m+j}F\_p(j)
&=\frac{2B\_p}{\binom{2m}m}\sum\_{j=1}^m(-1)^{j-1}\binom{2m}{m+j} \\
&=\frac{2B\_p}{\binom{2m}m}\frac12\binom{2m}m \\
&=B\_p
\end{align}
which is exactly what we want to arrive at. We've already known from above the $B\_p=0$ when $p$ is odd. The proof is complete for both problems.


