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What is a good upper bound for $$\sum_{d\leq z} \mu(d)\frac{\tau(d)}{d}$$ where $\tau(d)$ is the divisor function?

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By "the" divisor function, do you mean the one that counts the number of divisors? –  Barry Cipra May 18 '12 at 21:27
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A naive argument leads to the bound O((log z)^2), a slightly more refined one gets you down to O(log z). Are you hoping for more than that? –  Alan Haynes May 18 '12 at 21:50
    
yes, the divisor function is the function that counts the divisors –  Alex Botros May 18 '12 at 22:46
    
well, if we defined the sum over all $d\vert P_z$ we could bound it above nicely by $1/\log(z)^2$ using Euler products...it's the truncation of the sum that kills it. I was hoping that some work had been done on it. –  Alex Botros May 18 '12 at 22:48
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I guess this is a variation on the prime number theorem since the generating function for the sum is $\zeta(1+s)^{-2}$ times an Euler product that is absolutely convergent up to $\text{Re}(s) > -1/2$. So the bound should be $O( (\log{z})^{-A})$ for $A$ arbitrarily large. –  Matt Young May 19 '12 at 0:03
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3 Answers

Let me elaborate on Matt Young's comment and show, for a sufficiently small absolute constant $c>0$, $$ \sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} \ll \exp(-c\sqrt{\log z}). $$ All references will be from Montgomery-Vaughan: Multiplicative number theory I.

We can assume that $z>2$ is not an integer. The associated Dirichlet series $$ F(s):=\sum_{d=1}^\infty\mu(d)\frac{\tau(d)}{d^{s+1}} =\prod_p\left(1-\frac{2}{p^{s+1}}\right)=\frac{G(s)}{\zeta(s+1)^2} $$ is absolutely convergent in $\Re s>0$, and $G(s)$ is given by an absolutely convergent Euler product in $\Re s>-1/2$. In particular, $G(s)$ is holomorphic, bounded, and bounded away from zero in any half-plane $\Re s>-1/2+\epsilon$. By Perron's formula (Theorem 5.2 and Corollary 5.3), we have $$ \sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} =\frac{1}{2\pi i}\int_{\sigma_0-iT}^{\sigma_0+iT}F(s)\frac{z^s}{s}dz+R, $$ where $T>0$ is arbitrary, $\sigma_0:=\frac{1}{\log z}$, and $$ R \ll \sum_{ z/2 < d < 2z } \frac{\tau(d)}{d} \min\left(1,\frac{z}{T|z-d|}\right) +\frac{1}{T}\sum_d\frac{\tau(d)}{d^{1+\sigma_0}}. $$ It is straightforward to estimate the right hand side to yield $$ R \ll z^{-\frac{1}{2}+\epsilon}+\frac{\log^2 z}{T}. $$ To see this, we estimate the first term as $$ \sum_{ z/2 < d < 2z } \frac{\tau(d)}{d} \min\left(1,\frac{z}{T|z-d|}\right) \ll z^{-\frac{1}{2}+\epsilon}+\frac{1}{T}\sum_{{z/2 < d < 2z}\atop{|z-d|>\sqrt{2z}}}\frac{\tau(d)}{|z-d|}$$ $$ \ll z^{-\frac{1}{2}+\epsilon}+\frac{1}{T}\sum_{ \ell < \sqrt{2z}}\frac{1}{\ell}\sum_{z/(2\ell) < m < 2z/\ell} \min\left(1,\left|\frac{z}{\ell}-m\right|^{-1}\right), $$ and for the inner sum we apply the argument on page 180 in the book. We shall use $T:=\exp(\sqrt{c\log z})$ for some small absolute constant $c>0$, so that $$ \sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} =\frac{1}{2\pi i}\int_{\sigma_0-iT}^{\sigma_0+iT}F(s)\frac{z^s}{s}dz +O\left(\frac{\log^2 z}{T}\right).$$ Using Theorem 6.7 we can see that the integrand is holomorphic in the rectangle with vertices $\sigma_0\pm iT$ and $\sigma_1\pm iT$, for $\sigma_1:=-\frac{c}{\log T}$ and $c>0$ a small absolute constant. Moreover, we can estimate the integrand on the sides of the rectangle. Hence applying Cauchy's theorem and estimates as on page 181 of the book, we obtain $$ \sum_{d\leq z} \mu(d)\frac{\tau(d)}{d} \ll z^{-\sigma_1}(\log z)^3+\frac{\log^2 z}{T}.$$ The right hand side is $\ll\exp(-(c/2)\sqrt{\log z})$, proving the original claim.

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It seems to me that the accepted answer to this question can be adapted, step by step, to give the estimate you need.

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This is also directly related to the size of $\sum_{n \leq X} \mu(n)$ as follows :

Assume that the estimate $\sum_{n \leq X} \frac{\mu(n)}{n} = O(\varepsilon(X))$ holds for some non increasing function $\varepsilon$ (with $ \varepsilon(X) = O(\log X ) $ ). From Dirichlet hyperbola principle we have

$$\sum_{n \leq X} \frac{\mu \star \mu(n)}{n} = 2\sum_{n \leq \sqrt{X}} \frac{\mu(n)}{n} \sum_{dn \leq X} \frac{\mu(d)}{d} - \left( \sum_{n \leq \sqrt{X}} \frac{\mu(n)}{n} \right)^2 $$ and thus

$$\sum_{n \leq X} \frac{\mu \star \mu(n)}{n} = O \left( \varepsilon(\sqrt{X}) \log X +\varepsilon(\sqrt{X})^2 \right) = O \left( \varepsilon(\sqrt{X}) \log X \right) $$

Now set $g = \mu \tau \star 1 \star 1$. As noted above by Matt Young and GH, the series $\sum_{n \geq 1} g(n) n^{- \sigma}$ converges absolutely for $\sigma > \frac{1}{2}$. Observe that

$$\sum_{n \leq X} \frac{\mu(n) \tau (n)}{n} = \sum_{n \leq X} \frac{g(n)}{n} \sum_{dn \leq X} \frac{\mu \star \mu(d)}{d} = O \left( \sum_{n \leq X} \frac{|g(n)|}{n} \varepsilon(\sqrt{\frac{X}{n}}) \log \frac{X}{n} \right) $$

Since $\sum_{n \geq X^{\frac{1}{2}}} |g(n)| n^{- 1} = O \left( X^{-\frac{1}{6}} \sum_{n} |g(n)| n^{- \frac{2}{3}} \right) = O \left( X^{-\frac{1}{6}} \right) $, we get

$$\sum_{n \leq X} \frac{\mu(n) \tau (n)}{n} = O \left( \varepsilon(X^{\frac{1}{4}}) \log X + X^{-\frac{1}{7}} \right) $$

For example, an estimate $\sum_{n \leq X} \frac{\mu(n)}{n} = o(\frac{1}{\log X})$ ensures that $\sum_{n \leq X} \frac{\mu(n) \tau (n)}{n} = o \left( 1 \right) $

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