# Abscissa of absolute convergence for a particular Dirichlet series

For $$n=p_1^{\alpha_1}p_2^{\alpha_2}\cdot\cdot\cdot p_k^{\alpha_k}$$ we denote $$\alpha(n)=\alpha_1\alpha_2\cdot\cdot\cdot\alpha_k$$. Show that $$F(s)=\sum_{n\geq 1}\frac{\alpha(n)}{n^s}$$ is absolutely convergent for $$\sigma>1$$.

My attempt:

Notice that $$\alpha(n)\leq n$$ because $$\begin{equation*} \alpha_i\leq p_i^{\alpha_i},\;\text{ for every p_i prime}\\ \alpha(n)=\alpha_1\cdot\cdot\cdot\alpha_k\leq p_1^{\alpha_1}\cdot\cdot\cdot p_k^{\alpha_k}=n \end{equation*}$$ Thus, $$\sum_{n\geq 1}\frac{\alpha(n)}{n^s}\leq \sum_{n\geq 1}\frac{n}{n^s}=\sum_{n\geq 1}\frac{1}{n^{s-1}}=\zeta(s-1)$$ that is absolutely convergent for $$\sigma>2$$.

But I don't find the way to ensure convergence for $$\sigma>1$$.

Thanks for any suggestion.

• $F(s) = \prod_p (1+\sum_{k=1}^\infty k p^{-sk})$, $\frac{F(s)}{\zeta(s)} = \prod_p( 1+\sum_{k=2}^\infty p^{-sk})$ which converges for $\Re(s) > 1/2$. – reuns Nov 25 '18 at 21:51
• Doesn't that first equality hold only if the series converge absolutely? – Kale36 Nov 26 '18 at 1:14
• @reuns How did you get the expresion for $\frac{F(s)}{\zeta(s)}$? – Kale36 Nov 26 '18 at 17:40
• $F(s) =\sum_{n\geq 1}\frac{\alpha(n)}{n^s}= \prod_p (1+\sum_{k=1}^\infty k p^{-sk})$ converges for $\Re(s) > 1$ and $\frac{F(s)}{\zeta(s)}=\prod_p (1+\sum_{k=1}^\infty k p^{-sk})(1-p^{-s})= \prod_p( 1+\sum_{k=2}^\infty p^{-sk})$ converges for $\Re(s) > 1/2$ – reuns Nov 26 '18 at 18:10