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?

  • $\begingroup$ Why don't you tell us the values for the first few $k$. Maybe there's a pattern to be found. $\endgroup$ – Gerry Myerson Feb 19 '12 at 1:02
  • $\begingroup$ @Gerry Myerson,You can tell us this pattern since you are suggesting that one exists. $\endgroup$ – Vassilis Parassidis Feb 19 '12 at 1:09
  • $\begingroup$ Even with the employment of the binomial expansions the recursion is particularly hard to wade through, or at least so I predict. Figuring this out without such expansion I have even less of an idea of how to go about. $\endgroup$ – anon Feb 19 '12 at 1:20
  • $\begingroup$ The first six terms are 1, -3, 10, -35, 126, -462. $\endgroup$ – anon Feb 19 '12 at 1:32
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    $\begingroup$ Vassilis, write $(1/n-1/(n+1))^k$ as $1/(n(n+1))^k$ and use partial fraction decomposition, then summing everything up, you'll see that terms with odd powers will form telescoping series (thus easy to deal with) and the terms with even powers with contribute powers of $\pi$ via zeta values. Actually calculating the individual numbers in the partial fraction decomposition for general $k$ seems like a chore, though it is easy for any particular $k$. $\endgroup$ – B R Feb 19 '12 at 2:35

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