Compute $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}$ and $\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}$

Question

How to prove

$$S_1=\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}=1+\frac{35}{128}\pi\zeta(3)+\frac{1}{48}\zeta(4)-\frac1{384}\psi^{(3)}\left(\frac14\right)$$ $$S_2=\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}=1+\frac18G\zeta(2)-\frac{35}{64}\pi\zeta(3)-\frac{15}{16}\zeta(4)+\frac1{768}\psi^{(3)}\left(\frac14\right)$$ where $H_n=\sum_{n=1}^\infty\frac1n$ is the $n$th harmonic number, $G$ denotes the Catalan's constant, $\zeta$ denotes the Riemann Zeta function and $\psi^{(n)}$ designates the polygamma function.

These two sums were proposed by Cornel and can be found here and here . My solution to $S_1$ can be found in the first link but its long, so can we find a better way to find $S_1$ and $S_2$ ?

Thanks.

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Note: Using the generating function of $\ \sum_{n=1}^\infty x^n\frac{H_n}{n^3}$ to evaluate $S_1$ is not allowed.

Answer 1

The first sum $S_1$:

From this solution we have

$$I=\int_0^1\frac{\operatorname{Li}_3(x)}{1+x^2}\ dx=\frac1{384}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4-9\pi\zeta(3)-64\pi^2G\right)\tag{1}$$

But \begin{align} I&=\sum_{n=0}^\infty(-1)^{n}\int_0^1 x^{2n}\operatorname{Li}_3(x)\ dx, \quad \text{apply integration by parts}\\ &=\sum_{n=0}^\infty(-1)^{n}\left(\frac{\zeta(3)}{2n+1}-\frac{\zeta(2)}{(2n+1)^2}+\frac{H_{2n+1}}{(2n+1)^3}\right)\\ &=\frac{\pi}4\zeta(3)-G\zeta(2)+\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}\tag{2} \end{align}

From (1) and (2) we get

$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac1{384}\psi^{(3)}\left(\frac14\right)-\frac{1}{48}\pi^4-\frac{35}{128}\pi\zeta(3)$$

Or

$$S_1=\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}}{(2n+1)^3}=1+\frac{35}{128}\pi\zeta(3)+\frac{1}{48}\zeta(4)-\frac1{384}\psi^{(3)}\left(\frac14\right)$$

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The solution for the second sum $S_2$ is by Cornel Valean and as follows:

By Cauchy product we have

$$\operatorname{Li}^2_2(x)=\sum_{n=1}^\infty x^n\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}\right)-6\operatorname{Li}_4(x)$$

set $x=i$ and take the imaginary parts of both sides we have

$$\Im\sum_{n=1}^\infty(i)^n\frac{H_n^{(2)}}{n^2}=\frac12\Im\operatorname{Li}^2_2(i)+3\Im\operatorname{Li}_4(i)-2\Im\sum_{n=1}^\infty(i)^n\frac{H_n}{n^3}$$

Using the fact that

$$\Im\sum_{n=1}^\infty (i)^n a_n=\sum_{n=0}^\infty (-1)^n a_{2n+1}$$

we have

$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}^{(2)}}{(2n+1)^2}=\frac12\Im\operatorname{Li}^2_2(i)+3\Im\operatorname{Li}_4(i)-2\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}$$

substitute $\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac1{384}\psi^{(3)}\left(\frac14\right)-\frac{1}{48}\pi^4-\frac{35}{128}\pi\zeta(3)$ along with $\Im\operatorname{Li}^2_2(i)=-\frac{\pi^2}{24}G$ and $\Im\operatorname{Li}_4(i)=\frac1{768}\psi^{(3)}\left(\frac14\right)-\frac{\pi^4}{96}$ we get

$$\sum_{n=0}^\infty (-1)^{n}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}=-\frac18G\zeta(2)+\frac{35}{64}\pi\zeta(3)+\frac{15}{16}\zeta(4)-\frac1{768}\psi^{(3)}\left(\frac14\right)$$

Or

$$S_2=\sum_{n=1}^\infty (-1)^{n-1}\frac{H_{2n+1}^{(2)}}{(2n+1)^2}=1+\frac18G\zeta(2)-\frac{35}{64}\pi\zeta(3)-\frac{15}{16}\zeta(4)+\frac1{768}\psi^{(3)}\left(\frac14\right)$$

Answer 2

Different approach to evaluate $S_1$:

From here we have

$$I=\int_0^1 \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\frac{\pi^3}{16}\ln2-\frac{7\pi}{64}\zeta(3)-\frac{\pi^4}{96}+\frac1{768}\psi^{(3)}\left(\frac14\right)\tag1$$

On the other hand

$$I=\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx$$

For the first integral, use $\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1}$ and for the second integral, use the identity $\frac{\arctan x}{1+x^2}=\frac12\sum_{n=0}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}$ we have

$$I=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln^2x\ dx-\frac12\sum_{n=0}^\infty(-1)^n(H_n-2H_{2n})\int_0^1x^{2n}\ln^2x\ dx$$

$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty(-1)^n\frac{H_n-2H_{2n}}{(2n+1)^3}$$

$$=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}-\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^3},\quad H_{2n}=H_{2n+1}-\frac{1}{2n+1}$$

$$=\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}+2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^3}\tag2$$

Combine $(1)$ and $(2)$ and substitute

$$\sum_{n=0}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$

we obtain that

$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac1{384}\psi^{(3)}\left(\frac14\right)-\frac{1}{48}\pi^4-\frac{35}{128}\pi\zeta(3)$$