Even harmonic sums? How do we calculate this?
$ \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{2n}}{2n} $
I am stuck that the integrals isn't converging for harmonic (even) numbers . somebody please help .
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\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,\,{H_{2n} \over 2n}} & =
{H_{2} \over 2} - {H_{4} \over 4} + {H_{6} \over 6} - {H_{8} \over 8} + \cdots =
\sum_{n = 0}^{\infty}{H_{4n + 2} \over 4n + 2} -
\sum_{n = 0}^{\infty}{H_{4n + 4} \over 4n + 4}
\\[5mm] & =
\sum_{n = 0}^{\infty}{H_{n + 2} \over n + 2}
\,{1 + \pars{-1}^{n} + i^{n} + \pars{-\ic}^{n} \over 4} -
\sum_{n = 0}^{\infty}{H_{n + 4} \over n + 4}
\,{1 + \pars{-1}^{n} + i^{n} + \pars{-\ic}^{n} \over 4} =
\\[5mm] & =
\sum_{n = 2}^{\infty}{H_{n} \over n}
\,{1 + \pars{-1}^{n} - i^{n} - \pars{-\ic}^{n} \over 4} -
\sum_{n = 4}^{\infty}{H_{n} \over n}
\,{1 + \pars{-1}^{n} + i^{n} + \pars{-\ic}^{n} \over 4}
\\[5mm] & =
-\,\half\sum_{n = 1}^{\infty}{H_{n} \over n}
\bracks{i^{n} + \pars{-\ic}^{n}} =
-\Re\sum_{n = 1}^{\infty}H_{n}\,\ic^{n}\int_{0}^{1}x^{n - 1}\,\dd x
\\[5mm] & =
-\Re\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n}\,\pars{\ic x}^{n}\,{\dd x \over x} =
\Re\int_{0}^{1}{\ln\pars{1 - \ic x} \over x\pars{1 - \ic x}}\,\dd x
\\[5mm] & =
\color{#f00}{{5 \over 96}\,\pi^{2} - {1 \over 8}\,\ln^{2}\pars{2}} \approx
0.4540
\end{align}
