Show that $\prod_p1+\frac{1}{p^{4s}}+\frac{1}{p^{5s}}$ converges

Question

I want to show that $$\prod_p1+\frac{1}{p^{4s}}+\frac{1}{p^{5s}}$$ for the Re(s)$\geq\frac{1}{3}$. ( $p$ prime) Should I compare it to the Riemann zeta function, although that only converges for Re(s)>1?

Answer 1

Absolute convergence of the infinite product.
Note that $\sum n^a$ converges for $a<-1$ by the integral test.
Let $p_n$ be the $n$th prime. Let $s>0$ and write $\sigma = \mathrm{Re }(s)$. As $n \to \infty$, \begin{align} p_n &\sim n\log n \\ |p_n^{-4s}| &= p_n^{-4\sigma} \sim n^{-4\sigma}(\log n)^{-4\sigma} \\ |p_n^{-5s}| &\sim n^{-5\sigma}(\log n)^{-5\sigma} \\ |p^{-4s}+p^{-5s}| &\sim n^{-4\sigma}(\log n)^{-4\sigma} \end{align} This means that series $$ \sum_n \left(p_n^{-4s}+p_n^{-5s}\right) $$ converges absolutely if $-4\sigma<-1$, that is $\sigma>1/4$. Thus the infinite product $$ \prod_p (1+p_n^{-4s}+p_n^{-5s}) $$ converges absolutely if $\text{Re}(s)>1/4$.