Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?

Question

Define $f(n)$ to be:

$$ \sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d} $$

But $\sigma_0(d) = 2^{\omega(d)}$ for any $d \mid n\#$ a primorial, so:

$$ f(n) = \prod_{p \text{ prime} \\ p \leq n} \dfrac{1}{\left(1 - \dfrac{2}{p}\right)} $$

Mertens' third theorem has as $1$ instead of a $2$ there.

So what would be an asymptotic estimate of $f(n)$?

Answer 1

We have \begin{align*} \prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ & = \exp \left( {\sum\limits_{2<p \le n} {\frac{2}{p}} - \sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} } \right). \end{align*} Now by Mertens' second theorem $$ \sum\limits_{2<p \le n} {\frac{2}{p}} = 2\log \log n + 2M -1+ \mathcal{O}\!\left( {\frac{1}{{\log n}}} \right), $$ where $$ M = \gamma + \sum\limits_p {\left[ {\frac{1}{p} + \log \left( {1 - \frac{1}{p}} \right)} \right]} $$ is the Meissel–Mertens constant. Also \begin{align*} \sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} - \sum\limits_{p > n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} \\ & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} - \sum\limits_{p > n} {\mathcal{O}\!\left( {\frac{1}{{p^2 }}} \right)} \\ & = \sum\limits_{p>2} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p}} \right)} \right]} + \mathcal{O}\!\left( {\frac{1}{{n\log n}}} \right). \end{align*} Taking exponentials and simplifying $$ \prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } = C(\log n)^2 +\mathcal{O}(\log n), $$ where $$ C = \frac{{e^{2\gamma } }}{4}\prod\limits_{p > 2} {\left( {1 + \frac{1}{{p(p - 2)}}} \right)} = \frac{{e^{2\gamma } }}{4\Pi_2}=1.201303559967362\ldots, $$ with $\Pi_2$ being the twin prime constant.

Answer 2

Asymptotically,

$$g(n) = \prod_{2 < p < n}^{p \text{ prime}} 1-\frac2p \sim \frac{4C_2e^{-2\gamma}}{(\log n)^2} \approx \frac{0.83244}{(\log n)^2}$$

And presumably your expression is simply the reciprocal of that, i.e.,

$$f(n) \sim \frac{e^{2 \gamma}(\log n)^2}{4C_2} \approx 1.2013(\log n)^2$$

where $C_2$ is the twin prime constant:

$$C_2 = \prod_{p>2}^{p \text{ prime}} \frac{p(p-2)}{(p-1)^2} \approx 0.66016$$

Note: or just look at the other answer, which does a better job and shows the full derivation.