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Does $$\frac{1}{N^2}\sum _{d=1}^N \log d \sum _{n=1}^{N/d} \frac{\phi(n)}{\log (dn)},$$

converges or not when $N$ goes to infinity?

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up vote 8 down vote accepted

The expression converges to $0$, even when $\phi(n)$ is replaced by the larger $n$. The contribution from $n\le\sqrt N/d$ can be given by ignoring the logarithm in the denominator: \begin{align*} \frac1{N^2} \sum_{d=1}^N \log d \sum_{n=1}^{\sqrt N/d} \frac{\phi(n)}{\log dn} &\le \frac1{N^2} \sum_{d=1}^N \log d \sum_{n=1}^{\sqrt N/d} n \\\ &\lesssim \frac1{N^2} \sum_{d=1}^N \log d \cdot \frac12 \bigg( \frac{\sqrt N}d \bigg)^2 \\\ &= \frac1{2N} \sum_{n=1}^\infty \frac{\log d}{d^2} = \frac{|\zeta'(2)|}{2N}. \end{align*} (The exact value of the sum over $d$ isn't important - the fact that it converges is enough to show that the limit equals $0$ as $N\to\infty$.) The remaining contribution can be given by simplifying the logarithm in the denominator: \begin{align*} \frac1{N^2} \sum_{d=1}^N \log d \sum_{n=\sqrt N/d}^{N/d} \frac{\phi(n)}{\log dn} &\le \frac1{N^2} \sum_{d=1}^N \log d \sum_{n=1}^{N/d} \frac n{\log \sqrt N} \\\ &\lesssim \frac2{N^2\log N} \sum_{d=1}^N \log d \cdot \frac12 \bigg( \frac Nd \bigg)^2 \\\ &= \frac1{\log N} \sum_{n=1}^\infty \frac{\log d}{d^2} = \frac{|\zeta'(2)|}{\log N}, \end{align*} which also tends to $0$ as $N\to\infty$.

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It is also possible to obtain an asymptotic formula with main term $C\frac{1}{\log N}$ for some positive constant $C$. – i707107 May 7 '13 at 2:53
Agreed. I think the constant is $|\zeta'(2)|/2\zeta(2)$, though I might have missed something. – Greg Martin May 7 '13 at 17:50
I think so too. – i707107 May 7 '13 at 19:52
@i707107 Doing some rearrangements, we can rewrite the double sum as a sum with the Von Mangoldt function: $$\frac{1}{N^{2}}\sum_{ab\leq N}\frac{\Lambda(a)b}{\log ab}.$$ From here, we can easily bound the sum over $ab\leq( N/\log(N))^2$, and then show that $$\frac{1}{N^{2}}\sum_{\frac{N}{\left(\log N\right)^{2}}\leq ab\leq N}\frac{\Lambda(a)b}{\log ab}\sim\frac{1}{N^{2}\log N}\sum_{\frac{N}{\left(\log N\right)^{2}}\leq ab\leq N}\Lambda(a)b, $$ and by the prime number theorem, the right hand side above is asymptotic to $$-\frac{\zeta'(2)}{2\zeta(2)}\frac{1}{\log N}.$$ – Eric Naslund Jul 12 '13 at 22:43

For a precise asymptotic, we have that $$\sum_{d=1}^{N}\log d\sum_{n=1}^{N/d}\frac{\phi(n)}{\log(nd)}=-\frac{\zeta'(2)}{\zeta(2)}\text{li}(N^2)+O\left(N\right),$$ where $\text{li}(N)$ is the logarithmic Integral.

Proof: We may rewrite the above sum as $$\sum_{d\leq N}\log d\sum_{n\leq N,\ d|n}\frac{\phi\left(\frac{n}{d}\right)}{\log(n)},$$ and uppon switching the order this is

$$\sum_{n\leq N}\frac{1}{\log(n)}\sum_{d|n}\phi\left(\frac{n}{d}\right)\log d.$$ By using the fact that $\phi*\log=\Lambda*\text{Id},$ the above equals $$\sum_{ab\leq N}\frac{\Lambda(a)b}{\log(ab)}=\sum_{k\leq N}\frac{\left(\Lambda*\text{Id}\right)(k)}{\log k}.$$ Now, we have that

$$\sum_{ab\leq N}\Lambda(a)b=\sum_{a\leq N}\Lambda(a)\sum_{b\leq\frac{N}{a}}b=\frac{1}{2}\sum_{a\leq N}\Lambda(a)\left(\left[\frac{N}{a}\right]^{2}+\left[\frac{N}{a}\right]\right)$$

$$=\frac{N^{2}}{2}\sum_{a\leq N}\frac{\Lambda(a)}{a^{2}}+O\left(N\log N\right)=-\frac{\zeta^{'}(2)}{2\zeta(2)}N^{2}+O(N\log N).$$

Since $$\sum_{k\leq N}\frac{\Lambda*\text{Id}(k)}{\log k}=\int_{2}^{N}\frac{1}{\log x}d\left(\sum_{k\leq x}\Lambda*\text{Id}(k)\right),$$ by applying partial summation, we are able to recover that $$\sum_{ab\leq x}\frac{\Lambda(a)b}{\log(ab)}=-\frac{\zeta'(2)}{\zeta(2)}\text{li}(N^2)+O(N).$$

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