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I'm trying to estimate asymptotically the following sums: $$ S_1(m, n) = \sum_{1\leq i \leq n, (m,i)=1}{\frac{1}{i}} $$

$$ S_2(n) = \sum_{i=1}^n{i\phi(i)}, $$

where $(m,i) = GCD(m,i)$, and $\phi(i)$ is the Euler totient function of $i$.

I would be grateful for any hints.

foveo
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  • Use Möbius inversion/inclusion-exclusion on $$\sum_{1 \leqslant i \leqslant n} \frac{1}{i} = \sum_{d \mid m} \sum_{\substack{1 \leqslant i \leqslant n \ (m,i) = d}} \frac{1}{i}.$$ – Daniel Fischer Jul 27 '17 at 13:15
  • @DanielFisher, I am confused how Möbius inversion could be used here. The left part of your equation depends on $n$ and the first sum of the right part is among $d|m$ (not $d|n$). In addition the sum $$\sum_{1 \leq i \leq n \ (m,i)=d}{\frac{1}{i}}$$ is a function depending on 3 variables: $m, n, d$. – foveo Jul 27 '17 at 15:56
  • View it as an inclusion-exclusion problem. Consider $n$ fixed. Partition the set $A = {1, 2, \dotsc, n}$ into the sets $A_d = { i \in A : (m,i) = d}$. – Daniel Fischer Jul 27 '17 at 16:02
  • @DanielFischer I have posted this argument below. – Marko Riedel Jul 27 '17 at 22:08

2 Answers2

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You can use that $\frac{\phi(n)}{n} = \sum_{d | n} \frac{\mu(d)}{d}$ to obtain $$\sum_{n \le x} \frac{\phi(n)}{n} = \sum_{d \le x} \frac{\mu(d)}{d} \lfloor x/d \rfloor= x \sum_{d \le x} \frac{\mu(d)}{d^2} + \mathcal{O}(\sum_{d \le x} |\mu(d)/d|)= \frac{x}{\zeta(2)}+ \mathcal{O}(\log x)$$ which implies by summation by parts $$\sum_{n \le x} n \phi(n) = x^2\sum_{n \le x} \frac{\phi(n)}{n}+\sum_{n \le x-1} (\sum_{m \le n} \frac{\phi(m)}{m})(n^2-(n+1)^2) = \frac{x^3}{3 \zeta(2)} + \mathcal{O}(x^2 \log x)$$

reuns
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  • Thanks for your answer, it is very helpful for me. I wish I could mark two comments as answers. – foveo Aug 01 '17 at 21:08
  • @foveo what is it useful for ? – reuns Aug 02 '17 at 12:26
  • I am trying to estimate asymptotically the number of different {0, 1}-valued functions over integer grid which can be defined by a system of two linear inequalities. Eventually I came up with several sums involving $\phi(n)$. – foveo Aug 14 '17 at 21:45
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For the first sum, use inclusion-exclusion and consider the Hasse diagram of the divisor poset of $m$, with the weights of the nodes $d|m$ being $\mu(d).$ A node $d$ here represents the set of values $q\le n$ that are multiples of $d.$ A value $q$ appears in all nodes $d|m$ where $d|q$ and hence the total weight of a value $q$ is

$$\sum_{d|(q,m)} \mu(d).$$

This is one when $(q,m)=1$ and zero when $(q,m)\gt 1$ which means that these weights are an exact representation of the problem. Observe that the contribution from node $d$ is

$$\frac{1}{d} H_{\lfloor n/d\rfloor}$$

and hence

$$S_1(m, n) = \sum_{d|m} \mu(d) \frac{1}{d} H_{\lfloor n/d\rfloor}.$$

Using the dominant two terms $H_n \sim \log n + \gamma$ this becomes

$$(\log n + \gamma) \sum_{d|m} \mu(d) \frac{1}{d} - \sum_{d|m} \mu(d) \frac{1}{d} \log d.$$

Now for the asymptotics with respect to $n$ the first is

$$(\log n + \gamma) \prod_{p|m} \left(1-\frac{1}{p}\right)$$

and we find for the leading two terms (logarithm followed by next term, a constant)

$$\bbox[5px,border:2px solid #00A000]{ \frac{\varphi(m)}{m} \log n + \frac{\varphi(m)}{m} \gamma - \sum_{d|m} \mu(d) \frac{1}{d} \log d.}$$

which is precisely as it ought to be (initial term may be conjectured by inspection). Note that additional terms from the expansion of $H_n$ like $H_n \sim \log n + \gamma + \frac{1}{2n} - \frac{1}{12n^2}$ only contribute lower order terms, i.e. terms in inverse powers of $n$ times constants dependent on $m,$ e.g. the next term happens to be zero:

$$\sum_{d|m} \mu(d) \frac{1}{d} \frac{1}{2n/d} = \frac{1}{2n} \sum_{d|m} \mu(d) = 0$$

and the one after that is

$$\sum_{d|m} \mu(d) \frac{1}{d} \frac{1}{12n^2/d^2} = \frac{1}{12n^2} \sum_{d|m} \mu(d) d.$$

Fixing $m$ yields a bona fide asymptotic expansion in $n$.

Marko Riedel
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    Very well explained. A minuscule nitpick, though: the next term is zero only if $m > 1$, which isn't explicitly demanded in the question. But for $m = 1$ we just have $S_1(1,n) = H_n$ anyway, so then it's trivial. – Daniel Fischer Jul 27 '17 at 22:42
  • Thank you for the kind remark. Good point about $m=1.$ – Marko Riedel Jul 27 '17 at 22:45
  • We also use that $\chi(n) = 1_{gcd(n,m)=1}$ is a $m$-periodic Dirichlet character with $L(s,\chi) = \sum_{n=1}^\infty \chi(n) n^{-s} = \zeta(s)\prod_{p| m} (1-p^{-s})$ thus $\sum_{n \le x} \frac{\chi(n)}{n} \sim Res(L(s,\chi),1) \sum_{n \le x} \frac{1}{n} \sim \log x \ \ \frac{\phi(m)}{m}$. – reuns Jul 30 '17 at 00:46