An identity involving binomial coefficients and Bernoulli numbers.

Question

While solving a problem I came across the following identity, which holds by numerical evidence: $$ \sum_{k=1}^i\frac1k\binom{i}{k-1}\binom kj{B_{k-j}}=\delta_{ij}. $$ where $B$ are the Bernoulli numbers.

I have no experience with Bernoulli numbers, so any hint for proving the equality will be appreciated.

Answer 1

Allow me to change the notation as to reserve (as far as possible) to $i$ it's common meaning, and use $k,j,l$ as indices.

First let's simplify the sum working on the properties of the binomial coefficients $$ \eqalign{ & S(n,m) = \sum\limits_{1 \le \,k\, \le \,n} {{1 \over k}\left( \matrix{ n \cr k - 1 \cr} \right)\left( \matrix{ k \cr m \cr} \right)B_{\,k - m} } = \cr & = {1 \over {n + 1}}\sum\limits_{1 \le \,k\, \le \,n} {\left( \matrix{ n + 1 \cr k \cr} \right)\left( \matrix{ k \cr m \cr} \right)B_{\,k - m} } = \cr & = {1 \over {n + 1}}\left( \matrix{ n + 1 \cr m \cr} \right)\sum\limits_{1 \le \,k\, \le \,n} {\left( \matrix{ n + 1 - m \cr k - m \cr} \right)B_{\,k - m} } = \cr & = {1 \over {n + 1}}\left( \matrix{ n + 1 \cr m \cr} \right)\sum\limits_{\max \left( {1 - m,0} \right) \le \,j\, \le \,n - m} {\left( \matrix{ n + 1 - m \cr j \cr} \right)B_{\,j} } \cr} $$

where:
- in the first step, we use the "absorption" indentity;
- in the second step, we use the "trinomial revision" indentity;
- in the third step, we changed the summation index.

Now, assuming $1 \le m$, we can use the fundamental reursive identity of Bernoulli numbers (the "standard" definition $B_{\,j} ^ -$) $$ \sum\limits_{0 \le \,j\, \le \,n - m} {\left( \matrix{ n + 1 - m \cr j \cr} \right)B_{\,j} } = \delta _{\,n - m,\,0} $$

to get $$ S(n,m) = \sum\limits_{1 \le \,k\, \le \,n} {{1 \over k}\binom{n}{k-1} \binom{k}{m} B_{\,k - m}^{\, - } } = \delta _{\,n,\;m} \quad \left| {\;1 \le n,m} \right. $$

Answer 2

We have the following claim where $n\ge j$ (the sum is zero when $n\lt j$ and the claim holds by inspection)

$$\sum_{k=j}^n \frac{1}{k} {n\choose k-1} {k\choose j} B _{k-j} = \delta_{nj}.$$

This is

$$\sum_{k=j}^n {n+1\choose k} {k\choose j} B _{k-j} = \delta_{nj} \times (n+1).$$

Now

$${n+1\choose k} {k\choose j} = \frac{(n+1)!}{(n+1-k)! \times j! \times (k-j)!} = {n+1\choose j} {n+1-j\choose k-j}$$

and we find

$$\sum_{k=j}^n {n+1-j\choose k-j} B_{k-j} = \delta_{nj} \times (n+1) \times {n+1\choose j}^{-1}$$

or

$$\sum_{k=0}^{n-j} {n+1-j\choose k} B_{k} = \delta_{nj} \times (n+1) \times {n+1\choose j}^{-1} \\ = \delta_{nj} \times (n+1) \times {n+1\choose n}^{-1} = \delta_{nj}.$$

To prove this last form we put on the LHS

$$-B_{n+1-j} + \sum_{k=0}^{n+1-j} {n+1-j\choose k} B_{k} \\ = -B_{n+1-j} + (n+1-j)! [z^{n+1-j}] \frac{z}{\exp(z)-1} \exp(z).$$

Observe that

$$\frac{z}{\exp(z)-1} \exp(z) = \frac{z}{\exp(z)-1} (\exp(z)-1) + \frac{z}{\exp(z)-1} \\ = z + \frac{z}{\exp(z)-1}$$

so that we get

$$-B_{n+1-j} + (n+1-j)! [z^{n+1-j}] z + (n+1-j)! [z^{n+1-j}] \frac{z}{\exp(z)-1} \\ = - B_{n+1-j} + (n+1-j)! \delta_{nj} + B_{n+1-j} = \delta_{nj},$$

which is the RHS. This concludes the argument.