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I am wondering if it is possible to solve this problem using Abel summation:

$$\sum_{n \leq x} \frac{\mu (n)}{n} \log^2{\frac{x}{n}}=2\log{x}+O(1)$$

Or maybe I am on the wrong track?

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Does Abel summation refer to this formula: en.wikipedia.org/wiki/Abel's_summation_formula? –  Srivatsan Sep 23 '11 at 1:45
    
Yes, this is what I mean by Abel summation –  Rob Sep 23 '11 at 2:55
    
I'm not sure about Abel summation, but I think you can use the generalized Möbius inversion formula. (See, for example, Theorem 2.23 in Apostol's Introduction to Analytic Number Theory.) –  Mike Spivey Sep 23 '11 at 5:09
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I think the problem with Abel summation is that you're going to need a precise estimate of the dominant term of $\sum_{n=1}^x \frac{\mu(n)}{n}$. I think that will be difficult. –  Mike Spivey Sep 23 '11 at 18:27
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A completely different approach (which may or may not succeed in getting appropriate error terms) is to use the fact that your sum is equal to $\frac{1}{2\pi i} \int \frac1{\zeta(s+1)} \frac{2x^s}{s^3} ds$, where the integral is taken over an infinite vertical line to the right of $s=0$. Moving the contour to the left picks up the residue of the integrand at $s=0$, which is of the form $2\log x + D$ for some constant $D$. The remaining contour integrals would need to be bounded though. See Montgomery and Vaughan, "Multiplicative Number Theory", section 5.1. –  Greg Martin Sep 23 '11 at 18:32
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1 Answer

up vote 11 down vote accepted

Per conversation in the comments, here's how to prove the result using generalized Möbius inversion. In particular, watch how we deal with the error term (which I agree is the tricky part).

Theorem 2.23 in Apostol's Introduction to Analytic Number Theory gives the generalized Möbius inversion formula as the following:

If $\alpha$ is completely multiplicative we have $$G(x) = \sum_{n \leq x} \alpha(n) F\left(\frac{x}{n}\right) \Longleftrightarrow F(x) = \sum_{n \leq x} \mu(n) \alpha(n) G\left(\frac{x}{n}\right).$$

According to the statement we're trying to prove, if $G(x) = \log^2(x)$ then $F(x) = 2 \log (x) + O(1)$ (with the completely multiplicative function $\alpha(n) = 1/n$). So let's try the dominant term in $F(x)$ on the left-hand side and see what happens.

We have $$2\sum_{n \leq x} \frac{\log \left(\frac{x}{n}\right)}{n} = 2\sum_{n \leq x} \left(\frac{\log x}{n} - \frac{\log n}{n}\right) = 2\left(\log x H_x - \sum_{n \leq x}\frac{\log n}{n}\right)$$ $$= 2 \log^2 x + 2\gamma \log x + O\left(\frac{\log x}{x}\right) - \log^2 x - 2A + O\left(\frac{\log x}{x}\right)$$ $$= \log^2 x + 2\gamma \log x - 2A + O\left(\frac{\log x}{x}\right),$$ where $A$ is constant. The second-to-last step uses the asymptotic expansion for the harmonic numbers and the asymptotic order of $\sum_{n \leq x}\frac{\log n}{n}$, the latter of which is Exercise 1 in Chapter 3 of Apostol and can be proved using Euler-Maclaurin summation.

By generalized Möbius inversion, then, $$\sum_{n \leq x} \frac{\mu(n)}{n} \left(\log^2 \frac{x}{n} + R\left(\frac{x}{n}\right)\right) = 2 \log x$$ $$\Longrightarrow \sum_{n \leq x} \frac{\mu(n)}{n} \log^2 \frac{x}{n} = 2 \log x - \sum_{n \leq x} \frac{\mu(n)}{n}R\left(\frac{x}{n}\right),$$ where $R(x)$ is some function satisfying $R(x) = 2\gamma \log x - 2A + O\left(\frac{\log x}{x}\right) $. What remains, then, is to show that $\sum_{n \leq x} \frac{\mu(n)}{n} R\left(\frac{x}{n}\right) = O(1)$.

We know that $\sum_{n \leq x} \frac{1}{n} = \log x + \gamma + O\left(\frac{1}{x}\right)$, so applying generalized Möbius inversion again gives us $$\sum_{n \leq x} \frac{\mu(n)}{n} \left(\log \frac{x}{n} + S\left(\frac{x}{n}\right)\right) = 1,$$ where $S(x)$ is some function satisfying $S(x) = \gamma + O\left(\frac{1}{x}\right)$. We also know that $\sum_{n=1}^{\infty} \frac{\mu(n)}{n} = 0$ (see, for instance, Equation (10) here), so $\sum_{n \leq x} \frac{\mu(n)}{n} = o(1)$. Putting this all together, we have $$\sum_{n \leq x} \frac{\mu(n)}{n} R\left(\frac{x}{n}\right) = \sum_{n \leq x} \frac{\mu(n)}{n} \left(2\gamma \log \frac{x}{n} - 2A + O\left(\frac{n \log \frac{x}{n}}{x}\right)\right)$$ $$=2 \gamma - 2 \gamma \sum_{n \leq x} \frac{\mu(n)}{n} \left(\gamma + O\left(\frac{n}{x}\right)\right) - 2A \sum_{n \leq x} \frac{\mu(n)}{n} + O\left(\frac{1}{x}\right)\sum_{n \leq x} \mu(n) \log \frac{x}{n}$$ $$=2 \gamma - 2 (\gamma^2+A) \sum_{n \leq x} \frac{\mu(n)}{n} + O\left(\sum_{n \leq x} \frac{\mu(n)}{x} \right)+ O\left(\frac{1}{x}\right) \sum_{n \leq x} O\left(\log \frac{x}{n}\right)$$ $$= 2 \gamma + o(1) + O\left(\sum_{n \leq x} \frac{\mu(n)}{n} \right) + O\left(\frac{1}{x}\right) O\left( \sum_{n \leq x} (\log x - \log n)\right)$$ $$= O(1) + o(1) + o(1) + O\left(\frac{1}{x}\right) O\left(x \log x - (x \log x - x + O(\log x))\right)$$ $$ = O(1).$$ (The second-to-last step follows from Stirling's approximation.)

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+1 I've never seen (nor contemplated the existence of) the generalized Mobius inversion. –  JavaMan Sep 23 '11 at 18:26
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