19
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician Jun 12 '12 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. $\endgroup$ – Tito Piezas III Jun 12 '12 at 15:49
20
$\begingroup$

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

$\endgroup$
  • 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! $\endgroup$ – Tito Piezas III Jun 12 '12 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.