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I'm having a difficult time showing that $$ \displaystyle \zeta(s,mz) = \frac{1}{m^{s}} \sum_{k=0}^{m-1} \zeta \left(s,z+\frac{k}{m} \right) $$

A couple of authors referred to it as an obvious fact. But I don't see how it simply follows from the series definition of the Hurwitz zeta function (i.e., $ \displaystyle\zeta(s,z) = \sum_{n=0}^{\infty} \frac{1}{(z+n)^{s}})$ .

So I tried to relate it to the multiplication formula for the polylogarithm, that is $$\displaystyle\text{Li}_{s} (z^{m}) = m^{s-1} \sum_{k=0}^{m-1} \text{Li}_{s} (ze^{2 \pi i k/m})$$

But I wasn't successful.

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up vote 4 down vote accepted

We start with $\displaystyle \zeta(s,z) = \sum_{n=0}^{\infty} \frac{1}{(z+n)^{s}}$.

First, substitute in $mz$ for $z$:

$$\begin{align} \zeta(s,mz) &= \sum_{n=0}^{\infty} \frac{1}{(mz+n)^{s}} \\ &= \frac{1}{m^s}\sum_{n=0}^{\infty} \frac{1}{(z+\frac{n}{m})^{s}}. \end{align}$$

Now write $n = n'm + n''$, where now $n'$ will range from $0$ to $\infty $ and $n''$ will range from $0$ to $m-1$. Putting this above gives $$\begin{align} &= \frac{1}{m^s}\sum_{\substack{n'\geq 0 \\ n'' \bmod m}} \frac{1}{(z+\frac{n''}{m} + n')^{s}} \\ &= \frac{1}{m^s} \sum_{k = 0}^{m-1} \sum_{n' \geq 0} \frac{1}{(z + \frac km + n')^s} \\ &= \frac{1}{m^s}\sum_{k =0}^{m-1}\zeta\left(s,z+\frac{k}{m}\right), \end{align}$$

as we wanted to show.

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So you just used the division algorithm? – Random Variable Dec 8 '13 at 11:20
@RandomVariable: Yes, that's how I split $n$ into $n'$ and $n''$. – mixedmath Dec 8 '13 at 12:12

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