# How to evaluate $\sum_{n=1}^{\infty}\frac{H_{kn}^2-[\gamma+\ln(kn)]^2}{n}?$

From this post @Olivier Oloa gives the closed form for this sum $$(1)$$

$$\sum_{n=1}^{\infty}\frac{H_n^2-(\gamma+\ln n)^2}{n}=\frac{5}{3}\zeta(3)-\frac{2}{3}\gamma^3-2\gamma\gamma_1-\gamma_2\tag1$$

I was trying to generalises $$(2)$$

$$\sum_{n=1}^{\infty}\frac{H_{kn}^2-[\gamma+\ln(kn)]^2}{n}=F(k)\tag2$$ where $$k\ge1$$

but I couldn't only got so far finding out the closed form for $$k=2$$

$$\sum_{n=1}^{\infty}\frac{H_{2n}^2-[\gamma+\ln(2n)]^2}{n}=\frac{11}{12}\zeta(3)-\frac{2}{3}\gamma^3-2\gamma\gamma_1-\gamma_2+\eta(1)[\eta(2)-\gamma^2-2\gamma_1]\tag3$$

where $$\eta(k)$$ is the Dirichlet eta function

Does anyone knows how to find the closed form for $$(2)$$?

• Why not just write $\eta(1)=\ln{(2)}$ and $\eta(2)=\frac{\pi^2}{12}$ – Peter Foreman Apr 17 at 18:35