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I have tried to use the techniques from question 822466 and 135844 but I don't get it right.

For all $d\in\mathbb{N}$ prove

$$\sum_{j=0}^{2d+1} \binom{2d+1}{j}\frac{B_{j+1}}{j+1}(4^{j+1}-2^{j+1})=0$$

How can this be done?

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Coolwater
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1 Answers1

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The given sum equals $$ (2d+1)!\sum_{j=0}^{2d+1}\frac{B_{j+1}}{(j+1)!}(4^{j+1}-2^{j+1})\frac{1}{(2d+1-j)!}. \tag{1}$$ We have $$ \frac{1}{e^x-1}-\frac{1}{x} = \sum_{j\geq 0}\frac{B_{j+1}}{(j+1)!} x^j \tag{2}$$ hence $(1)$ can be written as $(2d+1)!$ times the coeffient of $x^{2d+1}$ in $$ e^x\left[\frac{4}{e^{4x}-1}-\frac{4}{4x}-\frac{2}{e^{2x}-1}+\frac{2}{2x}\right] =-\frac{1}{\cosh(x)}\tag{3}$$ Since $2d+1$ is odd and $-\frac{1}{\cosh(x)}$ is an even function, the claim is trivial after $(3)$.
The generalized sum (with $2d$ in place of $2d+1$) is related with Euler numbers.

Jack D'Aurizio
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  • What do you do between (2) and (3)? How come the $e^x$ outside parentheses and where did $(2d+1-j)!$ go? – Coolwater Feb 27 '17 at 20:08
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    @Coolwater: I am considering a convolution. $\frac{1}{(2d+1-j)!}$ is the coefficient of $x^{2d+1-j}$ in $e^x$ and $\frac{B_{j+1}}{(j+1)!}(4^{j+1}-2^{j+1})$ is the coefficient of $x^j$ in the bracketed term. – Jack D'Aurizio Feb 27 '17 at 20:20
  • Marko Riedel also uses generating functions in his answer to http://math.stackexchange.com/questions/822466/sum-involving-bernoulli-numbers-sum-r-1n-binom2n2r-1-fracb-2rr – Jack D'Aurizio Feb 27 '17 at 20:21