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

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.

The $$N$$ for which this product is largest will be of the form $$N = \prod_{p, so that $$\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 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 $$2^{\omega(N)} \le \exp\bigg( (\log2+o(1)) \frac{\log N}{\log\log N} \bigg).$$

• Thank you very much!! – Johnny T. Feb 19 at 8:47

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 $$<.

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