On Euler’s totient function series

Let $\phi$ denote Euler’s totient function. Prove that for $s>2$ it holds that:

$\sum_{n=1}^{\infty} \frac{(-1)^n \phi(n)}{n^s} =-\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^s-3}{2^s-1}$

where $\zeta$ stands for the Riemann zeta function.

Solution

Well by Euler’s product we have,

$\sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \prod_{p}\left(1+\frac{\phi(p)}{p^s}+\frac{\phi(p^2)}{p^{2s}}+\frac{\phi(p^3)}{p^{3s}}+\cdots\right)= \prod_{p}\frac{p^s-1}{p^s-p}$

thus,

(1) $\begin{equation*} \sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \prod_p \frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)} \end{equation*}$

and

(2) $\begin{equation*} \sum_{\substack{n\geq 1\\n\text{ odd}}}\frac{\phi(n)}{n^s} = \prod_{p>2} \frac{1-\frac{1}{p^{s}}}{1-\frac{1}{p^{s-1}}}=\frac{\zeta(s-1)}{\zeta(s)}\cdot\frac{2^s-2}{2^s-1} \end{equation*}$

Combining we get the result.

Note: It also holds that

$\sum_{n=1}^\infty \frac{\phi(2n-1)}{(2n-1)^s} = \frac{(1-2^{1-s})\zeta(s-1)}{1-2^{-s} \zeta(s)}$

On Euler’s totient function series

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