Prove $ \sum_{i=0}^{2a+1} {2a+1 \choose i} B_{2a+1-i} [ (n+1)^i+(-n)^i ] =0 $ for Bernoulli numbers $B_{n}$

Question

Prove for the Bernoulli numbers $B_n$, that for all $a \in \mathbb{N}$, that $ \sum_{i=0}^{2a+1} {2a+1 \choose i} B_{2a+1-i} [ (n+1)^i+(-n)^i ] =0 $. As much as this is a neat identity, it's a crucial step in another result concerning powersums. I have no idea how to prove it, aside from noting that it cannot be directly evaluated by powersums. Any ideas for a proof?

Answer 1

Dividing your expression by $(2a+1)!$ and using the definition of binomial coefficients, we see that you would like to prove that $$ \sum_{i=0}^{2a+1} \frac{1}{(2a+1-i)!i!}B_{2a+1-i}((n+1)^i+(-n)^i)=0 $$ or, in other words, $$ \sum_{b+c=2a+1}\frac{B_b}{b!}\frac{(n+1)^c+(-n)^c}{c!}=0. $$ Replacing $2a+1$ by arbitrary $m$, multiplying by $x^m$ and evaluating the sum, we see that your desired identity is the statement that the function $$ f(x)=\sum_{m=0}^{+\infty}x^m\sum_{b+c=m}\frac{B_b}{b!}\frac{(n+1)^c+(-n)^c}{c!} $$ is even, i.e. $f(x)=f(-x)$. Next, $f(x)$ can easily be expressed as a product of series: $$ f(x)=g(x)h(x)=\sum_{b=0}^{+\infty}\frac{B_b}{b!}x^b\cdot \sum_{c=0}^{+\infty}\frac{(n+1)^c+(-n)^c}{c!}x^c. $$ By one of the definitions for $B_n$, we have $$ g(x)=\frac{x}{e^x-1} $$ and the second function is, of course, the sum of exponents $$ h(x)=e^{(n+1)x}+e^{-nx}. $$ Now, $$ g(-x)=\frac{-x}{e^{-x}-1}=\frac{x}{1-e^{-x}}=\frac{xe^x}{e^x-1}=e^xg(x) $$ and $$ h(-x)=e^{-(n+1)x}+e^{nx}=e^{-x}(e^{-nx}+e^{(n+1)x})=h(x)e^{-x}. $$ Therefore, $$ f(-x)=g(-x)h(-x)=g(x)e^xh(x)e^{-x}=f(x), $$ as needed.