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What are the coefficients of the series for: $$\frac x{e^x+1}$$ It looks similar to the Bernoulli generating function, but the $+$ sign is throwing me off.

I already found the series for its inverse. If $x=\frac y{e^y+1}$, then: $$y=\sum_{n=1}^\infty\left(\sum_{k=1}^n\binom nkk^{n-1}\right)\frac{x^n}{n!}$$ —that is, that finite series in the middle is the $n$th coefficient of the exponential generating function. (Wolfram Alpha can't reduce it to a simpler form.) (If you're bored later today, by the way, it's a nice challenge to derive that.)

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They are Genocchi numbers, when you multiply by 2. In particular even Genocchi numbers are related to Bernoulli numbers:

$$G_{2k}=2(1-2^{2k})B_{2k}.$$

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  • $\begingroup$ Thank you! That means that$$\frac x{e^x+1}=\frac x{e^x-1}-\frac{2x}{e^{2x}-1}$$(which is easily checked). That's a very interesting formula. $\endgroup$ – Akiva Weinberger Jul 22 '15 at 19:55
  • $\begingroup$ @columbus8myhw: Neat observation. You can probably get a more holistic explanation from the fact that $\eta(s)=(1-2^{1-s})\zeta(s)$, the alternating zeta function vs the Riemann zeta function. Indeed, $\zeta(s)\Gamma(s) = \int_0^\infty \frac{u^{s-1}}{e^u-1}du$ and $\eta(s)\Gamma(s)=\int_0^\infty \frac{u^{s-1}}{e^u+1}du$, but the proof for the identity is much simpler from the series representation of $\zeta(s)$ and $\eta(s)$. $\endgroup$ – Alex R. Jul 22 '15 at 20:05
  • $\begingroup$ Ah, that's a nice connection! I wonder if that has any relation to the inverse function series I found. $\endgroup$ – Akiva Weinberger Jul 22 '15 at 20:10

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