Mike Spivey
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 Jul 29 awarded Nice Answer Jul 24 comment Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$ That is interesting. Thanks for the observation. Jul 22 comment How can I find the median of this frequency distribution @Shahab: It depends on whether $n$ is even or odd. If $n$ is even, use $x = (n+1)/2$, as in the OP's example. If $n$ is odd, use $x = n/2$. Jul 2 awarded Curious Jun 24 answered Integer sum as binomial coefficient Jun 17 awarded Good Answer Jun 9 comment Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$ Got it; thanks! And +1. Jun 9 comment Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$ How do you go from $\int_0^1 \frac{1-t^k}{1-t} \, dt$ to $\int_0^1 \ln(1-t) (-kt^{k-1}) \, dt$ in the second step? May 29 awarded Nice Answer May 18 awarded Nice Question Apr 17 awarded Announcer Mar 24 awarded Announcer Mar 20 awarded Enlightened Mar 20 awarded Nice Answer Mar 5 awarded Announcer Feb 24 awarded Notable Question Feb 22 awarded Announcer Feb 20 awarded Nice Answer Feb 16 awarded Enlightened Feb 16 awarded Nice Answer