On the zeta sum $\sum_{n=1}^\infty[\zeta(5n)-1]$ and others For p = 2, we have,
$\begin{align}&\sum_{n=1}^\infty[\zeta(pn)-1] = \frac{3}{4}\end{align}$
It seems there is a general form for odd p. For example, for p = 5, define $z_5 = e^{\pi i/5}$.  Then,
$\begin{align} &5 \sum_{n=1}^\infty[\zeta(5n)-1] = 6+\gamma+z_5^{-1}\psi(z_5^{-1})+z_5\psi(z_5)+z_5^{-3}\psi(z_5^{-3})+z_5^{3}\psi(z_5^{3}) = 0.18976\dots \end{align}$
with the Euler-Mascheroni constant $\gamma$ and the digamma function $\psi(z)$. 


*

*Anyone knows how to prove/disprove this? 

*Also, how do we split $\psi(e^{\pi i/p})$ into its real and imaginary parts so as to express the above purely in real terms? 


More details in my blog.
 A: $$
\begin{align}
\sum_{n=1}^\infty\left[\zeta(pn)-1\right] & = \sum_{n=1}^\infty \sum_{k=2}^\infty \frac{1}{k^{pn}} \\
& = \sum_{k=2}^\infty \sum_{n=1}^\infty (k^{-p})^n \\
& = \sum_{k=2}^\infty \frac{1}{k^p-1}
\end{align}
$$
Let $\omega_p = e^{2\pi i/p} = z_p^2$, then we can decompose $1/(k^p-1)$ into partial fractions
$$
\frac{1}{k^p-1} = \frac{1}{p}\sum_{j=0}^{p-1} \frac{\omega_p^j}{k-\omega_p^j}
= \frac{1}{p}\sum_{j=0}^{p-1} \omega_p^j \left[\frac{1}{k-\omega_p^j}-\frac{1}{k}\right]
$$
where we are able to add the term in the last equality because $\sum_{j=0}^{p-1}\omega_p^j = 0$. So
$$
p\sum_{n=1}^\infty\left[\zeta(pn)-1\right] = \sum_{j=0}^{p-1}\omega_p^j\sum_{k=2}^{\infty}\left[\frac{1}{k-\omega_p^j}-\frac{1}{k}\right]
$$
Using the identities
$$
\psi(1+z) = -\gamma-\sum_{k=1}^\infty\left[\frac{1}{k+z}-\frac{1}{k}\right]
= -\gamma+1-\frac{1}{1+z}-\sum_{k=2}^\infty\left[\frac{1}{k+z}-\frac{1}{k}\right]\\
\psi(1+z) = \psi(z)+\frac{1}{z}
$$
for $z$ not a negative integer, and
$$
\sum_{k=2}^\infty\left[\frac{1}{k-1}-\frac{1}{k}\right]=1
$$
by telescoping, so finally
$$
\begin{align}
p\sum_{n=1}^\infty\left[\zeta(pn)-1\right]
& = 1+\sum_{j=1}^{p-1}\omega_p^j\left[1-\gamma-\frac{1}{1-\omega_p^j}-\psi(1-\omega_p^j)\right] \\
& = \gamma-\sum_{j=1}^{p-1}\omega_p^j\psi(2-\omega_p^j)
\end{align}
$$
So far this applies for all $p>1$. Your identities will follow by considering that when $p$ is odd $\omega_p^j = -z_p^{2j+p}$, so
$$
\begin{align}
p\sum_{n=1}^\infty\left[\zeta(pn)-1\right]
& = \gamma+\sum_{j=1}^{p-1}z_p^{2j+p}\psi(2+z_p^{2j+p})\\
& = \gamma+\sum_{j=1}^{p-1}z_p^{2j+p}\left[\frac{1}{1+z_p^{2j+p}}+\frac{1}{z_p^{2j+p}}+\psi(z_p^{2j+p})\right] \\
& = \gamma+p-1+S_p+\sum_{j=1}^{p-1}z_p^{2j+p}\psi(z_p^{2j+p})
\end{align}
$$
where
$$
\begin{align}
S_p & = \sum_{j=1}^{p-1}\frac{z_p^{2j+p}}{1+z_p^{2j+p}} \\
& = \sum_{j=1}^{(p-1)/2}\left(\frac{z_p^{2j-1}}{1+z_p^{2j-1}}+\frac{z_p^{1-2j}}{1+z_p^{1-2j}}\right) \\
& = \sum_{j=1}^{(p-1)/2}\frac{2+z_p^{2j-1}+z_p^{1-2j}}{2+z_p^{2j-1}+z_p^{1-2j}} \\
& = \frac{p-1}{2}
\end{align}
$$
which establishes your general form.
I don't have an answer for your second question at this time.
