How to prove

$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)\stackrel ?=\frac{\pi^3}{32}-2G\ln2,$$ where $G$ is the Catalan's constant.


For the first sum, $$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_{2k}=\Re\left\{\sum_{k=1}^{\infty}\frac{i^k}{(k+1)^2}H_{k}\right\},$$ which can be evaluated by using the formula in this post: $$\sum_{n=1}^\infty \frac{H_n}{n^2}\, x^n=\zeta(3)+\frac{\ln(1-x)^2\ln(x)}{2}+\ln(1-x)\operatorname{Li}_2(1-x)+\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x),$$ but we cannot apply the similar approach to the second sum $$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_k.$$ Then, I tried to write the sum as $$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}\int_0^1\frac{2x^{2k}+x^k-3}{x-1}~\mathrm dx$$ and it become more complicated.


Are we able to evaluate the sum directly (avoid calculating integrals and polylogs as much as possible)? The integral given by @Jack D'Aurizio is a bit complicated (see this post).

  • 1
    $\begingroup$ Please look e.g. at math.stackexchange.com/questions/917154/… , where the second series is discussed as a side effect (constant $K$ , answer of Lucian) . $\endgroup$ – user90369 Nov 20 '18 at 17:45
  • $\begingroup$ @user90369 Thank you for the link! $\endgroup$ – Tianlalu Nov 21 '18 at 7:59
  • $\begingroup$ You are welcome! :) $\endgroup$ – user90369 Nov 21 '18 at 10:56

The series involving $H_k$ and $H_{2k}$ can be studied in a similar way: since $$ \frac{-\log(1-x)}{1-x} = \sum_{n\geq 1} H_n x^{n} $$ we have $ \frac{-\log(1+x^2)}{1+x^2} = \sum_{n\geq 1} H_n(-1)^n x^{2n} $ and $$ \sum_{k\geq 1}\frac{(-1)^k}{(2k+1)^2}H_k = \int_{0}^{1}\frac{\log(1+x^2)\log(x)}{1+x^2}\,dx$$ boils down to $$ \int_{0}^{\pi/4} -2\log(\cos\theta) \log(\tan\theta)\,d\theta $$ which is simple to tackle through well-known Fourier series. It equals

$$ -\frac{\pi^3}{64}-K\log(2)-\frac{\pi}{16}\log^2(2)+2\,\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)\approx -0.07355395672853217. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.