Bernoulli numbers identity:$\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}=0$,for all $n\ge1$,$m\ge0$

Question

For all $n\geq 1$ and $m\geq0$, I'm trying to prove that

$\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}=0$

where $B_n$ are the Bernoulli numbers with $B_{1}=-\frac{1}{2}$.

I made a couple of attempts using the recursive relationship of Bernoulli numbers and induction, but unsuccessful. Now I'm wondering if this is a viable proof strategy. Any comments are appreciated!

Edit: This is a conjecture, I evaluated it numerically for $n,m\leq 20$.

Answer 1

The instructive derivation of @RenéGy reduces OPs problem to show the nice symmetric Bernoulli number identity: \begin{align*} \color{blue}{(-1)^n \sum_{g=0}^m}&\color{blue}{ \binom{m}{g}\frac{B_{n+g+1}}{n+g+1} +(-1)^m \sum_{g=0}^n \binom{n}{g}\frac{B_{m+g+1}}{m+g+1}}\\ &\color{blue}{= - \frac{1}{(n+m+1){\binom{n+m}{m}}}}\tag{1} \end{align*}

Here we show the identity (1) follows from Theorem 2 in Bernoulli Numbers and a New Binomial Transform Identity by H.W. Gould.

Given a sequence $(a_n)_{n\geq 0}$ we consider the sequence of binomial transforms $(b_n)_{n\geq 0}$. We so have the binomial inverse pair \begin{align*} b_n=\sum_{k=0}^n\binom{n}{k}a_k\qquad\mathrm{and}\qquad a_n=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}b_k\tag{2} \end{align*} Setting $a_k=B_k$, the $k$-th Bernoulli number we have thanks to a well known recursion formula of the Bernoulli numbers the binomial inverse pair \begin{align*} a_k=B_k\quad\mathrm{and}\quad b_n=\sum_{k=0}^n\binom{n}{k}a_k=\sum_{k=0}^n\binom{n}{k}B_k=(-1)^nB_n\qquad\qquad n\geq 0 \end{align*} Theorem 2 in the paper states the following identity for binomial inverse pairs: \begin{align*} \sum_{k=0}^m\frac{\binom{m}{k}}{\binom{n+k+s}{s}}a_{n+k+s} &=\sum_{k=0}^n(-1)^{n-k}\frac{\binom{n}{k}}{\binom{m+k+s}{s}}b_{m+k+s}\\ &\quad+\sum_{j=0}^{s-1}\sum_{i=0}^{s-1-j}\binom{s-1-j}{i}\binom{s-1}{j}\frac{(-1)^{n+1+i}sa_j}{(m+n+1+i)\binom{m+n+i}{n}} \end{align*} Setting $s=1$ the identity reduces to \begin{align*} \sum_{k=0}^m\binom{m}{k}\frac{a_{n+k+1}}{n+k+1}&=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\frac{b_{m+k+1}}{m+k+1}\\ &\qquad+\frac{(-1)^{n+1}a_0}{(m+n+1)\binom{m+n}{n}}\tag{3} \end{align*}

Putting the Bernoulli numbers (2) in (3) we obtain with $a_0=B_0=1$: \begin{align*} \sum_{k=0}^m\binom{m}{k}\frac{B_{n+k+1}}{n+k+1} &=\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}\frac{(-1)^{m+k+1}B_{m+k+1}}{m+k+1}\\ &\qquad+\frac{(-1)^{n+1}}{(m+n+1)\binom{m+n}{n}}\\ \color{blue}{(-1)^n\sum_{k=0}^m\binom{m}{k}\frac{B_{n+k+1}}{n+k+1}} &\color{blue}{=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^{m+1}B_{m+k+1}}{m+k+1}}\\ &\qquad\color{blue}{-\frac{1}{(m+n+1)\binom{m+n}{n}}} \end{align*} and the claim (1) follows.

Answer 2

This is not an answer but the derivation of equivalent statements involving single sums, which are symetric with respect to $n$ and $m$ and might be useful.

Equivalent statements which need to be proven, for $n \ge 1$ and $m \ge1$ : $$ \sum_{h=1}^\frac{n+m}{2} \Big((-1)^n\binom{2h-1}{n} + (-1)^{m}\binom{2h-1}{m}\Big)\binom{n+m+1}{2h} B_{2h}=-1 \tag 1$$ or $$ (-1)^n \sum_{g= 0}^m \frac{B_{n+g+1}}{n+g+1}{m \choose g}+(-1)^m \sum_{g= 0}^n \frac{B_{m+g+1}}{m+g+1}{n \choose g} = - \frac{1}{(n+m+1){n+m\choose m}}\tag 2$$

Let $S[m,n]:=\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}$

$ S[m,n]=\frac{n!m!}{(n+m+1)!}\sum_{k=0}^n\sum_{l=0}^m\frac{(n+m+1)!(k+l)!}{(n+m-k-l+1)!k!l!(k+l)!}(-1)^l B_{k+l}$

$S[m,n]=\frac{n!m!}{(n+m+1)!}\sum_{k=0}^n\sum_{l=0}^m\binom{n+m+1}{k+l}\binom{k+l}{l}(-1)^l B_{k+l}$

Let $T[m,n]:=\sum_{k=0}^n\sum_{l=0}^m\binom{n+m+1}{k+l}\binom{k+l}{l}(-1)^l B_{k+l}.$

Since the odd-index Bernoulli numbers, but $B_1$, are zero, we have

$T[m,n]=(n+m+1) \Big(\frac{1}{2} - \frac{1}{2}\Big)+ \sum_{ k \ even \le n}\sum_{ l \ even \le m }\binom{n+m+1}{k+l}\binom{k+l}{l} B_{k+l}-\sum_{ k \ odd \le n}\sum_{ l \ odd \le m }\binom{n+m+1}{k+l}\binom{k+l}{l} B_{k+l} $

$T[m,n]= \sum_{ k \ even \le n}\sum_{ l \ even \le m }\binom{n+m+1}{k+l}\binom{k+l}{l} B_{k+l}-\sum_{ k \ odd \le n}\sum_{ l \ odd \le m }\binom{n+m+1}{k+l}\binom{k+l}{l} B_{k+l} $

$T[m,n]= \sum_{ h \ even \le m+n}\binom{n+m+1}{h} B_{h} \Big( \sum_{h-m \le 2k \le n}\binom{h}{2k} -\sum_{h-m \le 2k+1 \le n}\binom{h}{2k+1} \Big) $

$T[m,n]= \sum_{ h \ even \le m+n}\binom{n+m+1}{h} B_{h} \sum_{h-m \le k \le n}(-1)^k\binom{h}{k} $

$T[m,n]= 1+ \sum_{2\le h \ even \le m+n}\binom{n+m+1}{h} B_{h} \sum_{h-m \le k \le n}(-1)^k\binom{h}{k} $

$T[m,n]=1+ \sum_{2\le h \ even \le m+n}\binom{n+m+1}{h} B_{h} \Big(\sum_{0 \le k \le n}(-1)^k\binom{h}{k} - \sum_{0 \le k \le h-m-1}(-1)^k\binom{h}{k}\Big)$

$T[m,n]=1+ \sum_{2 \le h \ even \le m+n}\binom{n+m+1}{h} B_{h} \Big((-1)^n\binom{h-1}{n} - (-1)^{h-m-1}\binom{h-1}{h-m-1}\Big) \square $

Equation (2) is then obtained from Equation (1) thanks to standard manipulations of the binomial coefficients and the indices.

Answer 3

With the work by @RenéGy and @epi163sqrt we are trying to prove the statement

$$(-1)^n \sum_{g=0}^m \frac{B_{n+g+1}}{n+g+1} {m\choose g} + (-1)^m \sum_{g=0}^n \frac{B_{m+g+1}}{m+g+1} {n\choose g} = - \frac{1}{n+m+1} {n+m\choose m}^{-1}$$

We prove this for $n\ge m$, it then follows by symmetry for $m\ge n.$ Using

$$B_n = (-1)^n \sum_{k=0}^n {n\choose k} B_k$$

we get for the first piece

$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g} (-1)^{g+1} \sum_{k=0}^{n+g+1} {n+g+1\choose k} B_k.$$

Extracting $k=0$ we get

$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g} (-1)^{g+1} = \sum_{g=0}^m \frac{1}{n+m-g+1} {m\choose g} (-1)^{m-g+1} \\ = [z^{n+m+1}] \log\frac{1}{1-z} \sum_{g=0}^m z^g {m\choose g} (-1)^{m-g+1} \\ = (-1)^{m+1} [z^{n+m+1}] \log\frac{1}{1-z} (1-z)^m = - [z^{n+m+1}] \log\frac{1}{1-z} (z-1)^m.$$

Recall from MSE 4316307 that with $1\le k\le n$

$${n\choose k}^{-1} = k [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$

We put $n := n+m+1$ and $k := n+1$ to get

$$-\frac{1}{n+1} {n+m+1\choose n+1}^{-1} = -\frac{1}{n+m+1} {n+m\choose n}^{-1}.$$

This also could have been obtained by summing residues of a suitable function. Good, we have the RHS. Now we get for the remainder

$$\sum_{g=0}^m \frac{1}{n+g+1} {m\choose g} (-1)^{g+1} \sum_{k=1}^{n+g+1} {n+g+1\choose k} B_k \\ = \sum_{g=0}^m {m\choose g} (-1)^{g+1} \sum_{k=1}^{n+g+1} {n+g\choose k-1} \frac{B_k}{k} \\ = \sum_{g=0}^m {m\choose g} (-1)^{g+1} \sum_{k=0}^{n+g} {n+g\choose k} \frac{B_{k+1}}{k+1}.$$

We get two components, the first is,

$$\sum_{g=0}^m {m\choose g} (-1)^{g+1} \sum_{k=0}^{m-1} {n+g\choose k} \frac{B_{k+1}}{k+1} \\ = \sum_{k=0}^{m-1} \frac{B_{k+1}}{k+1} \sum_{g=0}^m {m\choose g} (-1)^{g+1} {n+g\choose k}.$$

The inner sum is

$$[z^k] (1+z)^n \sum_{g=0}^m {m\choose g} (-1)^{g+1} (1+z)^g = - [z^k] (1+z)^n (-1)^m z^m = 0$$

since $k\lt m$. That leaves

$$\sum_{g=0}^m {m\choose g} (-1)^{g+1} \sum_{k=m}^{n+g} {n+g\choose k} \frac{B_{k+1}}{k+1} \\ = \sum_{g=0}^m {m\choose g} (-1)^{g+1} \sum_{k=0}^{n-m+g} {n+g\choose m+k} \frac{B_{m+k+1}}{m+k+1} \\ = \sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1} \sum_{g=k+m-n}^m {m\choose g} (-1)^{g+1} {n+g\choose m+k}.$$

Now when $n+g\lt m+k$ or $g\lt m-n+k$ the second binomial coefficient is zero, so we may lower $g$ to zero (observe that the first binomial coefficient is zero when $g\lt 0$ so we also may raise to zero when $k+m-n \lt 0$):

$$\sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1} \sum_{g=0}^m {m\choose g} (-1)^{g+1} {n+g\choose m+k}.$$

The inner sum is

$$[z^{m+k}] (1+z)^n \sum_{g=0}^m {m\choose g} (-1)^{g+1} (1+z)^g = - [z^{m+k}] (1+z)^n (-1)^m z^m \\ = - (-1)^m [z^k] (1+z)^n = - (-1)^m {n\choose k}.$$

We have obtained

$$- (-1)^m \sum_{k=0}^n \frac{B_{m+k+1}}{m+k+1} {n\choose k},$$

which is minus the second piece and concludes the argument.

Addendum. Obviously what we have proved here is with $b_n= (-1)^n \sum_{k=0}^n {n\choose k} a_k$ then

$$(-1)^n \sum_{g=0}^m \frac{b_{n+g+1}}{n+g+1} {m\choose g} + (-1)^m \sum_{g=0}^n \frac{a_{m+g+1}}{m+g+1} {n\choose g} = - \frac{a_0}{n+m+1} {n+m\choose m}^{-1}.$$

We get for an ordinary binomial transform $b_n = \sum_{k=0}^n {n\choose k} a_k$ the relation

$$\sum_{g=0}^m (-1)^{g+1} \frac{b_{n+g+1}}{n+g+1} {m\choose g} + (-1)^m \sum_{g=0}^n \frac{a_{m+g+1}}{m+g+1} {n\choose g} = - \frac{a_0}{n+m+1} {n+m\choose m}^{-1}.$$

Answer 4

Now, I propose an autonomous proof of $$ (-1)^n \sum_{g= 0}^m \frac{B_{n+g+1}}{n+g+1}{m \choose g}+(-1)^m \sum_{g= 0}^n \frac{B_{m+g+1}}{m+g+1}{n \choose g} = - \frac{1}{(n+m+1){n+m\choose m}}.\tag 2$$

I make use of well-known properties of generating functions: let $f(x)$ be a formal serie, such that $ f(x) = \sum_{j\ge 0}f_j\frac{x^j}{j!}$, let $[D^kf]$ be the $k$th derivative of $f$, then

$$f_{g+m}= [\frac{x^g}{g!}][D^mf](x) =[\frac{x^m}{m!}][D^gf](x), \tag a$$

if $g(x)= f(-x)$ then $[D^m f] (-x) = (-1)^m[D^m g] (x). \tag b$

Moreover $[\frac{x^m}{m!}] (\exp \cdot [D^mf]) (x) = \sum _ {g = 0}^m {m \choose g} [\frac{x^g}{g!}][ D^m f] (x) \tag c$

and $[D^m (f\cdot \exp)] (x) = \exp x \sum _ {g = 0}^m {m \choose g}[ D^g f] (x). \tag d$

Let

$ h (t) := \frac {1}{\exp t - 1} - \frac{1}{t}$

$f (t) := \exp t \cdot h(t) $

$g (t) := f (-t),$

we have $ g (t) = -h (t) + j (t) $ with $ j (t) = \sum_{i\ge 0}\frac{(-1)^{i + 1}}{i + 1} \frac{ t^i}{i!}.$

Also, since $ \frac{B_{h + 1}}{h + 1} = [\frac{x^h}{h!}] h(x)$, we have

\begin{align} \sum_{g=0}^m {m \choose g}\frac{B_{g+n+1}}{g+n+1} &= \sum_{g=0}^m {m \choose g}[\frac{x^n}{n!}][D^gh](x)\\ &=[\frac{x^n}{n!}] \sum_{g=0}^m {m \choose g}[D^gh](x)\\ &=[\frac{x^n}{n!}] \exp(-x)[D^mf](x)\\ (-1)^n\sum_{g=0}^m {m \choose g}\frac{B_{g+n+1}}{g+n+1} &=[\frac{(-x)^n}{n!}] \exp(-x)[D^mf](x)\\ &= [\frac{x^n}{n!}]\exp(x)[D^mf](-x)\\ &= (-1)^m [\frac{x^n}{n!}]\exp(x)[D^mg](x)\\ &= -(-1)^m [\frac{x^n}{n!}]\exp(x)[D^mh](x)+ (-1)^m [\frac{x^n}{n!}]\exp(x)[D^mj](x). \end{align} On the other hand \begin{align} (-1)^m\sum_{g=0}^n {n \choose g}\frac{B_{g+m+1}}{g+m+1} &= (-1)^m\sum_{g=0}^n {n \choose g}[\frac{x^g}{g!}][D^mh](x)\\ &=(-1)^m [\frac{x^n}{n!}]\exp(x)[D^mh](x). \end{align}

The lhs of Equation (2) is then \begin{align}(-1)^m\sum_{g=0}^n {n \choose g}\frac{B_{g+m+1}}{g+m+1}+(-1)^n\sum_{g=0}^m {m \choose g}\frac{B_{g+n+1}}{g+n+1}&= (-1)^m [\frac{x^n}{n!}]\exp(x)[D^mj](x)\\ &= (-1)^m \sum_{i = 0}^{n}{n \choose i}[x^i/i!][D^m j] (x) \\ &=- \sum_{i = 0}^{n}{n \choose i}\frac{(-1)^{i}}{i+m+1} . \end{align} The last step is to show that $$\sum_{i = 0}^{n}{n \choose i}\frac{(-1)^{i}}{i+m+1}= \frac{1}{(n+m+1){n+m \choose m}}$$ which can be done by induction on $n$. For $n=0$, it is obviously true for all $m$ and then, the basic recursion for binomial coefficients can be used.