# Thread: Dirichlet series of extended Liouville function

1. ## Dirichlet series of extended Liouville function

I would like to prove that
$\displaystyle \sum_{n = 1}^\infty \frac{\lambda_a}{n^s} = \frac{\zeta(s) \zeta(2s-2a)}{\zeta(s-a)}$
where
$\displaystyle \lambda_a(n) = \sum_{d \vert n} d^a \cdot \lambda(d)$
and
$\displaystyle \lambda(n)$ is Liouville's function, defined as
$\displaystyle \lambda(1) = 1 \mbox{, and if } n=p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \mbox{ then } \lambda(n) = (-1)^{a_1 + a_2 + \cdots + a_k}$

Using multiplication of Dirichlet series, I think I can show that
$\displaystyle \sum_{n = 1}^\infty \frac{\lambda (n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}$
and that
$\displaystyle \sum_{n = 1}^\infty \frac{\sum_{d \vert n} d^a}{n^s} = \zeta(s-a)\zeta(s)$
but I cant see how (or if) this helps.

Any help or advice would be appreciated.

2. Well, notice that since you know : $\displaystyle \sum_{k\geq 1} {\frac{\lambda(k)}{k^s}} = \frac{\zeta(2s)}{\zeta (s)}$

We then have: $\displaystyle \sum_{k\geq 1} {\frac{\lambda(k)}{k^{s-a}}} = \frac{\zeta(2s - 2a)}{\zeta (s-a)}$

But here: $\displaystyle \sum_{k\geq 1} {\frac{\lambda(k)}{k^{s-a}}} = \sum_{k\geq 1} {\frac{k^a \cdot \lambda(k)}{k^s}}$ the generating function of $\displaystyle f(k) = k^a\cdot \lambda ( k)$

Now multiplying this by $\displaystyle \zeta(s)$ we get the Dirichlet-Series-Generating-Function of $\displaystyle \sum_{d|n} {d^a\cdot \lambda( d)}$ which is what you want.

3. Ahh. Thank you. I had not realized that

$\displaystyle \sum_{k \geq 1} \frac{ \lambda (k) }{ k^{s} } = \frac{ \zeta(2s)}{ \zeta(s)}$
would imply
$\displaystyle \sum_{k \geq 1} \frac{ \lambda (k) }{ k^{s-a} } = \frac{ \zeta(2s-2a)}{ \zeta(s-a)}$

Many thanks for the help.