Let's have the following zeta binomial $\sum\limits_{n=1}^\infty (1/n-1/(n+1))^k$, where $k$ a natural number and $k>1$. From the expansion of these binomials we obtain polynomials of $\pi$ where one of the terms is always an integer. Does anyone know how to calculate this integer, for all values of $k$, without expanding the zeta binomial?
Constant term of zeta binomials
2
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zeta-functions
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0@BR: The partial fraction decomposition would just be the binomial expansion of the original $k$th power of a difference. Vassilis: There is a fairly straightforward recursive procedure for computing these polynomials, but it requires the binomial expansion, and in order to find a closed form (if it exists) would require some combinatorics I can't think of. What I'm curious about is why you're asking for this to be done "**without expanding the zeta binomial**" which pretty much ruins all hope you have of getting an answer, in my opinion. – 2012-02-24