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It seems that the $n$th cumulant of the uniform distribution on the interval $[-1,0]$ is $B_n/n$, where $B_n$ is the $n$th Bernoulli number.

And also $-\zeta(1-n) = B_n/n$, where $\zeta$ is Riemann's $\zeta$ function.

Is there some reason why one should expect these to be the same, as opposed to proofs that convince you that they are?

  • 0
    Since we're looking at the analytic continuation of $\zeta$, the usual $p$-series concept is out the window - except through formal regularization.2011-09-14
  • 0
    Also posted to MathOverflow.2011-09-16

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