# $\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s} = \frac{L(\chi,s-1)}{L(\chi,s)}$

Let $$\chi$$ be a Dirichlet character mod 4. Show $$\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s} = \frac{L(\chi,s-1)}{L(\chi,s)}$$ and $$\sum_{n=1}^\infty \chi(n)d(n)n^{-s}=L(\chi,s)^2$$. ($$\phi$$ is the Euler totient function and $$d(n)$$ is the number of divisors of $$n$$.)

First, is this true just for characters mod 4 and not true in general? I'm not sure what specific properties about characters mod 4 I should use besides that $$\chi(n)=0$$ for $$n$$ even.

I took the log of both sides and tried to use the following: $$L(\chi,s)=\prod_{p \text{ prime}}\frac{1}{1-\frac{\chi(p)}{p^s}}$$

$$\log L(\chi,s)=\sum_{p \text{ prime}}\sum_{n=1}^\infty \frac{\chi(p)^n}{np^{ns}}$$

$$\chi$$ and $$\phi$$ are multiplicative, so we can express $$\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s}$$ as the Euler product $$\prod_p(1+\frac{\chi(p)\phi(p)}{p^s}+\frac{\chi(p^2)\phi(p^2)}{p^{2s}}+\cdots).$$

Manipulating things are not quite working. Any help would be appreciated.

• If $\chi$ is completely multiplicative and $a_\chi(n) = a(n) \chi(n)$ and $a \ast b = c$ then $a_\chi \ast b_\chi = c_\chi$. It works with analytic functions too : the twist of the logarithm is the logarithm of the twist. – reuns May 4 at 23:41

It is easier to show $$L(\chi, s)\sum_{ n \geq 1} \frac{\chi(n) \phi(n)}{n^{s}} = L(\chi, s-1).$$ It is enough to show for $$\Re (s) >> 0$$ (by analytic continuation), and actually this is true for any Dirichlet character. If you compute LHS directly (assume that the serieses converges absolutely), we get $$\left( \sum_{n\geq 1} \frac{\chi(n)\phi(n)}{n^{s}} \right)\left( \sum_{m\geq 1} \frac{\chi(m)}{m^{s}}\right) = \sum_{n, m\geq 1} \frac{\chi(nm)\phi(n)}{(nm)^{s}} = \sum_{k\geq 1} \frac{\chi(k)}{k^{s}} \left( \sum_{d|k} \phi(d)\right) = \sum_{k\geq 1} \frac{\chi(k)}{k^{s-1}} = L(\chi, s-1)$$ where the last equality follows from $$\sum_{d|k}\phi(d) = k$$.
Looking at $$\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s} L(\chi,s)= \sum_{n=1}^\infty \frac{(χφ * \chi)(n)}{n^s}$$, * being Dirichlet Convolution.
Now looking at $$(χφ * φ)(n)=\sum_{d|n} \chi(d)\phi(d) \chi(\frac{n}{d})= \sum_{d|n} \chi(n)\phi(d)=\chi(n) \sum_{d|n}φ(d)=n\chi(n)$$
(maybe it's worth mentioning that $$\sum_{d|n} \phi(d)=n$$.)