I wonder if closed expression can be found for sums of harmonic numbers with a squared argument.

Examples are

$$s_{1}=\sum_{k=1}^\infty \frac{ H(k^2)}{k^2} \simeq 3.28709\tag{1}$$
$$s_{2}=\sum_{k=1}^\infty (-1)^{k+1} \frac{ H(k^2)}{k}\simeq 0.456221\tag{2}$$ $$s_{3}(x)=\sum_{k=1}^\infty x^k H(k^2)\tag{3}$$

I tried the representation

$$H(z) = \int_0^1 \frac{1-x^z}{1-z}\,dx\tag{4}$$

which leads to the sums of the type

$$p_1=\sum _{k=1}^{\infty } \frac{y^{k^2}}{k}\tag{5a}$$ $$p_2=\sum _{k=1}^{\infty } \frac{y^{k^2}}{k^2}\tag{5b}$$

related to Jacobi elliptic functions which I could not evaluate. Can you do better?

  • $\begingroup$ Equal to:$\sum _{k=1}^{\infty } \frac{H_{k^2}}{k^2}=-\frac{\pi ^2}{12}+\frac{1}{2} \pi \sum _{j=1}^{\infty } \frac{\coth \left(\sqrt{j} \pi \right)}{j^{3/2}}$, but no hope for closed form.:( $\endgroup$ – Mariusz Iwaniuk Jan 12 at 13:01
  • 1
    $\begingroup$ @Mariusz Iwaniuk Thank you. I find your formula starting from the definition $H_{k^2} = \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+k^2}\right)$. $\endgroup$ – Dr. Wolfgang Hintze Jan 12 at 21:10

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