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In a textbook I'm reading, the author states without proof that $$ \zeta(s,mz) = \frac{1}{m^{s}} \sum_{k=0}^{m-1} \zeta \left(s,z+\frac{k}{m} \right), \tag{1}$$ where $\zeta(s,z) $ is the Hurwitz zeta function

Supposedly, this isn't hard to prove. But is it possible to prove $(1)$ using simply the series definition of the Hurwitz zeta function, that is, $ \displaystyle\zeta(s,z) = \sum_{n=0}^{\infty} \frac{1}{(z+n)^{s}}$?

It might be interesting to note that the polygamma functions (excluding the digamma function) can be expressed in terms of the Hurwitz zeta function.

So from $(1)$ we can derive the multiplication formula $$\psi_{n}(mz) = \frac{1}{m^{n+1}} \sum_{k=0}^{m-1}\psi_{n} \left(z+ \frac{k}{m} \right) , \quad n \in \mathbb{Z}_{>0}.$$

Random Variable
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1 Answers1

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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.

davidlowryduda
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