Consider the Bernoulli Polynomials $B_n\in\mathbb{R}$ given as the coefficients of the series:

$$\frac{t}{e^t-1}=\sum\limits_{n=0}^{\infty}B_n\frac{t^n}{n!}$$ and the Bernoulli polynomials gven by the following expansion:

$$\frac{t e^{xt}}{e^t-1}=\sum\limits_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}$$

It seems to be that the following relation holds, but I do not know why:

$$-B_{2k+2}\left(\frac{1}{2}\right)=B_{2k+2}\cdot \left( 1-\frac{1}{2^{2k+1}} \right), \forall k\in\mathbb{N}$$

I calculated a couple of examples and suprisingly the equation is satisfied by the first Bernoulli numbers and polynomials. Does anyone know how to prove it?

Best regards

  • 1
    $\begingroup$ Have you tried to apply the identity $$\frac{t e^{t/2}}{e^t-1}+\frac{t}{e^t-1}=\frac{t}{e^{t/2}-1}$$ ? $\endgroup$ – Jack D'Aurizio Jan 23 '15 at 21:16

The identity just follows from: $$\frac{t e^{t/2}}{e^t-1}+\frac{t}{e^t-1}=\frac{t}{e^{t/2}-1}=2\cdot\frac{t/2}{e^{t/2}-1}.$$

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