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.