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?



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


