I was curious if there were any papers, Wikipedia articles, or YouTube videos that cover videos similar to the sum in the title.
I’ve been looking for a while now, and the closest thing I got to was Michael Penn’s YouTube channel where he evaluated:
$$\sum_{k=1}^{\infty} (\zeta(4k) - 1)$$
Bonus.. It would help to have some direction to solve:
$$\sum_{k=1}^{\infty} (-1)^{k-1} (\zeta(3k) - 1)$$
Is it something to do with the roots of unity theorem filter?
For $\Re a>1$, we substitute $\zeta(ak)-1=\sum_{n=2}^\infty n^{-ak}$ and sum over $k$ first (absolute convergence!): $$\sum_{k=1}^\infty(-1)^k\big(\zeta(ak)-1\big)=\sum_{n=2}^\infty\sum_{k=1}^\infty(-1/n^a)^k=-\sum_{n=2}^\infty\frac{1}{n^a+1}.$$ When $a$ is an integer, this is evaluated in terms of the digamma function. We do partial fractions for $1/(z^a+1)$: let $z_k=\exp\big((2k-1)\pi\mathrm{i}/a\big)$ be the roots of $z^a+1=0$ (for $1\leqslant k\leqslant a$), then $$\frac{1}{z^a+1}=\sum_{k=1}^{a}\frac{1}{az_k^{a-1}(z-z_k)}=-\frac1a\sum_{k=1}^{a}\frac{z_k}{z-z_k};$$ to apply the result from the first link above, we replace $n$ by $n+1$ (and sum from $n=1$). This gives $$\sum_{n=2}^\infty\frac{1}{n^a+1}=\frac1a\sum_{k=1}^{a}z_k\psi(1-z_k).$$