Could someone help find identities for these two?

I started with

$ζ^\star(2,1,2)$ = $\sum_{k=1}^{\infty}(\frac{1}{k^2})\sum_{l=1}^{k}(\frac{1}{l})\sum_{m=1}^{l}(\frac{1}{m^2})$


$ζ^\star(\bar{2},\bar{1},2)$ = $\sum_{k=1}^{\infty}(\frac{((-1)^k}{k^2})\sum_{l=1}^{k}(\frac{(-1)^l}{l})\sum_{m=1}^{l}(\frac{1}{m^2})$

but I've only worked on zeta functions with depth 2, so I'm not really sure how to proceed.


  • $\begingroup$ The first is $+\infty$ or undefined, depending on how you take it. $\endgroup$ – Thomas Andrews Sep 9 at 2:42
  • $\begingroup$ See arxiv.org/pdf/math/9910045.pdf $\endgroup$ – reuns Sep 9 at 3:43
  • $\begingroup$ @ThomasAndrews did you perhaps misread the summations? The innermost sum has order $\theta(\ell^{-1})$, so the terms on the second sum are $\theta(\ell^{-2})$, and its sum is thus of order $\theta(k^{-1})$, which means that the terms in the outermost sum are $\theta(k^{-3})$ and it's decidedly convergent... $\endgroup$ – Steven Stadnicki Sep 9 at 17:13
  • $\begingroup$ @StevenStadnicki It is because at first the OP wrote $\sum_{l=1}^\infty$ $\endgroup$ – reuns Sep 9 at 20:57
  • $\begingroup$ @reuns D'oh, I should've checked the edit history first. Thank you, and my apologies! $\endgroup$ – Steven Stadnicki Sep 9 at 21:47

With a bit of patience, summation by parts reduces

$$ \zeta^\star(2,1,2) = \sum_{k\geq 1}\frac{1}{k^2}\sum_{l=1}^{k}\frac{H_l^{(2)}}{l} $$ to $\sum_{k\geq 1}\frac{H_k H_k^{(2)}}{k^2}$, which is computable through the Maclaurin series of $\arcsin^4(x)$, and $$ \sum_{k\geq 1}\frac{1}{k^2}\sum_{l=1}^{k}\frac{H_l}{l^2}= 2\zeta(2)\zeta(3)-\sum_{k\geq 1}\frac{H_{k+1}H_k^{(2)}}{(k+1)^2}, $$ so the determination of $ \zeta^\star(2,1,2)$ can be considered as already solved by nospoon.


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