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 $\psi(z)$.

  1. Anyone knows how to prove/disprove this?
  2. 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.

  • $\begingroup$ Neither polygamma nor gamma split particularly nicely into real/imaginary parts (in particular the argument is a rather complicated expression), so I doubt that there's a clean solution for your second problem. $\endgroup$ Jun 12, 2012 at 1:51
  • $\begingroup$ I came across a similar problem recently where the sum was real, but the terms were complex. Fortunately, it was only Arcsin[(-1)^(1/p)] which Maple was able to split (as pointed out by R. Israel) hence one could simplify the expression to contain only real terms. But I'm happy with Zander's elegant answer. $\endgroup$ Jun 12, 2012 at 15:49

1 Answer 1


$$ \begin{align} \sum_{n=1}^\infty\left[\zeta(pn)-1\right] & = \sum_{n=1}^\infty \sum_{k=2}^\infty \frac{1}{k^{pn}} \\ & = \sum_{k=2}^\infty \sum_{n=1}^\infty (k^{-p})^n \\ & = \sum_{k=2}^\infty \frac{1}{k^p-1} \end{align} $$ 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.

  • 2
    $\begingroup$ Zander: Beautiful! I was trying to find a formula for even p as well, but you not only found the general formula for all integer p > 1, it's wonderfully concise, too. Thanks! $\endgroup$ Jun 12, 2012 at 15:39

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