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.

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

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