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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
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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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Thanks GH for filling in all the details. My take on it is that if someone knows the usual proof of the prime number theorem then they can adapt it to this particular sum. However, if someone doesn't know the proof of the prime number theorem then they should go study that in any of the standard references. –  Matt Young May 20 '12 at 18:14

3 Answers 3

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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