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Jul
29
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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
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Jun
24
answered Integer sum as binomial coefficient
Jun
17
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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?
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