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.


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.

  • $\begingroup$ Thank you for your answer. I had known (and used) identity (0), so for me the crucial point of your proof is the consideration of the evenness of $F_p(j)$ in $j$. However, using your formulas (2) (with $k$ in place of $p$) and (1), I get that for even $p$ the polynomial $F_p(j)$ in $j$ contains the odd monomial $j^{p-1}/2^p$; this is caused by the term outside the sum in the brackets in (2). Thus, $F_p(j)$ does not not seem even in $j$ for even $p$. I am then mystified how you still get the "right answer", $B_p$ -- because the rest of your solution of Problem 2 seems correct to me. $\endgroup$ – Iosif Pinelis Oct 18 '16 at 17:53
  • $\begingroup$ I'll try to double-check later. Hope you can do the same. $\endgroup$ – T. Amdeberhan Oct 18 '16 at 18:06
  • $\begingroup$ It looks like my mistake was using (2) with $0$ in place of $p$, whereas it holds only for $p\ge1$ . So, after all, $F_p(j)$ does seem even in $j$. I will carefully recheck your entire answer. $\endgroup$ – Iosif Pinelis Oct 18 '16 at 19:19
  • $\begingroup$ Everything looks fine now. $\endgroup$ – Iosif Pinelis Oct 19 '16 at 0:18

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