0
$\begingroup$

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

$\endgroup$
  • 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
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.