Find identities for $ζ^\star(2,1,2)$ and $ζ^\star(\bar{2},\bar{1},2)$.

Question

Could someone help find identities for these two?

I started with

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

and

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

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

Thanks!

Answer 1

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.