I am trying to prove the following for a very, very long time: $$ 2k=\sum_{j=1}^k \binom{2k+1}{2j}2^{2j}B_{2j} $$ Here, $B_{2j}$ are Bernoulli numbers.
I would be extremely happy if somebody could help me with this!
$\begingroup$
$\endgroup$
1
-
$\begingroup$ The following MSE link points to a computation that can easily be adapted to prove the result being asked for here. $\endgroup$– Marko RiedelAug 6, 2014 at 2:23
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
Let $B_n(X) := \sum_{i=0}^n \binom{n}{i}B_i X^{n-i}$ denote the $n$-th Bernoulli polynomial. It satisfies the symmetry relation $B_n(1-X) = (-1)^n B_n(X)$ which implies $B_n(1/2) = 0$ for odd numbers $n$. Hence $$0 = 2^{2k+1} B_{2k+1}(1/2) = \sum_{i=0}^{2k+1} \binom{2k+1}{i} B_i 2^i = 1 + (2k+1)2B_1 + \sum_{i=2}^{2k+1} \binom{2k+1}{i} B_i 2^i.$$ All summands with odd $i$ vanish since the odd-numbered Bernoulli numbers are zero, the only exception being $B_1 = -1/2$. It follows that $$0 = -2k + \sum_{j=1}^k \binom{2k+1}{2j}B_{2j}2^{2j}$$ which proves your proposition.
