# Prove that $\sum_{h=1}^k \zeta(s, \frac{h}{k}) = k^s \zeta(s,1), \, k=1,2,\ldots$

For $0<a \leq 1$ and $s >1$, define $$\zeta(s, a) = \sum_{n=0}^{\infty}{\frac{1}{(n+a)^s}}$$

Prove that $\zeta(s,a)$ converges (done) and

Prove that $$\sum_{h=1}^k{\zeta(s, \frac{h}{k})} = k^s \zeta(s,1)\,,\,\,\, k = 1, 2, 3, \ldots$$

Attempt: I'm trying to prove this by induction on $k$. So,

Let $M = \{k \in \mathbb{N} | \sum_{h=1}^k{\zeta(s, \frac{h}{k})} = k^s \zeta(s,1)\}.$

$$\sum_{h=1}^1{\zeta\big(s, \frac{h}{1}\big)} = \zeta(s,1)=1^s \zeta(s,1) \implies 1\in M.$$

Now, assume $1,2,\ldots, k \in M$. We want to show that $k+1 \in M$.

$$\sum_{h=1}^{k+1}{\zeta\big(s, \frac{h}{k+1}\big)} = \zeta\big(s, \frac{1}{k+1}\big) + \zeta\big(s, \frac{2}{k+1}\big)+ \ldots + \zeta\big(s, \frac{k}{k+1}\big)+ \zeta\big(s, 1\big).$$

So, in this state I cannot apply my induction hypothesis. I need to find a relationship between $\zeta\big(s, \frac{h}{k+1}\big)$ and $\zeta\big(s, \frac{h}{k}\big)$ for each $h=1,2,\ldots,k$.

I've been trying some algebraic manipulations but I haven't got anything useful.

Any ideas?

Any help would be appreciated!

## 1 Answer

For each positive integer $m$, there is a unique non-negative integer $n$ and a unique integer $h\in\{1,\dots,k\}$ such that $m=nk+h$.

Applying this to $m=n+1$, we get $$\zeta(s,1)=\sum_{n=0}^{\infty}(n+1)^{-s}=\sum_{h=1}^k\sum_{n=0}^{\infty}(nk+h)^{-s}=k^{-s}\sum_{h=1}^k\sum_{n=0}^{\infty}\Big(n+\frac{h}{k}\Big)^{-s}$$ $$=k^{-s}\sum_{h=1}^k\zeta\Big(s,\frac{h}{k}\Big)$$ Since the series defining $\zeta(s,1)$ converges absolutely for $s>1$, we can rearrange the series as in the second equality above.

• Thanks! I need to take my time to dissect that second equality. Does absolute convergence comes from the fact that all my terms are positive? – Luis Vera Apr 5 '17 at 16:14
• Yes, that's right. – carmichael561 Apr 5 '17 at 18:16