How to obtain an upper bound for $\prod_{p\mid N} (1 + 1/\sqrt{p})$ where $N$ is square free?

Question

I am interested in obtaining an upper bound for $\prod_{p|N} (1 + 1/\sqrt{p})$ when $N$ is squarefree. It's not too hard to show that $$ \prod_{p\mid N} (1 + 1/\sqrt{p}) \ll C^{\omega(N)} \ll N^{\varepsilon} $$ for any $\varepsilon > 0$. I was wondering is it possibly to prove an upper bound $\ll \log N$ or some power of $\log N$? Any comments would be appreciated. Thank you.

Answer 1

The $N$ for which this product is largest will be of the form $N = \prod_{p<y} p$, so that $\log N \sim y$. For this $N$, $$ \log \prod_{p\mid N} \bigg( 1+\frac1{\sqrt p} \bigg) = \sum_{p\mid N} \log \bigg( 1+\frac1{\sqrt p} \bigg) \sim \sum_{p\mid N} \frac1{\sqrt p} = \sum_{p<y} \frac1{\sqrt p} \sim \frac{\sqrt y}{\log y} \sim \frac{\sqrt{\log N}}{\log\log N}, $$ and so $$ \prod_{p\mid N} \bigg( 1+\frac1{\sqrt p} \bigg) = \exp\bigg( (1+o(1)) \frac{\sqrt{\log N}}{\log\log N} \bigg) $$ for these $N$ (and the right-hand side will be an upper bound for all $N$). In particular, an upper bound of $(\log N)^A$ is too ambitious.

Reality check: replacing $1+1/\sqrt p$ by $1+1$ and carrying this computation through will recover the more familiar bound $$ 2^{\omega(N)} \le \exp\bigg( (\log2+o(1)) \frac{\log N}{\log\log N} \bigg). $$

Answer 2

The hint is the condition of $N$ be square free!

Use this condition to introduce Legendre symbol as follows:

Rewrite your series to $\sum{\log(1+(\sum(\frac{y}{p})e^{2\pi{iy/p}})^{-1})}=\sum{\log(1+((\frac{l}{p})\sum(\frac{y}{p})e^{2\pi{iy/p}})^{-1})}$

This fact will change its positive and negative symbols and give an explanation to the case of $<<C^{w(N)}<<N^{\epsilon}$.