# Asymptotic formula for $\sum_{n \le x} \frac{\varphi(n)}{n^2}$

Here is yet another problem I can't seem to do by myself... I am supposed to prove that $$\sum_{n \le x} \frac{\varphi(n)}{n^2}=\frac{\log x}{\zeta(2)}+\frac{\gamma}{\zeta(2)}-A+O \left(\frac{\log x}{x} \right),$$ where $\gamma$ is the Euler-Mascheroni constant and $A= \sum_{n=1}^\infty \frac{\mu(n) \log n}{n^2}$. I might be close to solving it, but what I end up with doesn't seem quite right. So far I've got: \begin{align*} \sum_{n \le x} \frac{\varphi(n)}{n^2} &= \sum_{n \le x} \frac{1}{n^2} \sum_{d \mid n} \mu(d) \frac{n}{d} \\ &= \sum_{n \le x} \frac{1}{n} \sum_{d \le x/n} \frac{\mu(d)}{d^2} \\ &= \sum_{n \le x} \frac{1}{n} \left( \sum_{d=1}^\infty \frac{\mu(d)}{d^2}- \sum_{d>x/n} \frac{\mu(d)}{d^2} \right) \\ &= \sum_{n \le x} \frac{1}{n} \left( \frac{1}{\zeta(2)}- \sum_{d>x/n} \frac{\mu(d)}{d^2} \right) \\ &= \frac{1}{\zeta(2)} \left(\log x + \gamma +O(x^{-1}) \right) - \sum_{n \le x} \frac{1}{n}\sum_{d>x/n} \frac{\mu(d)}{d^2}. \end{align*} So. I suppose my main problem is the rightmost sum, I have no idea what to do with it! I'm not sure where $A$ comes into the picture either. I tried getting something useful out of \begin{align*} & \sum_{n \le x} \frac{1}{n} \left( \frac{1}{\zeta(2)}- \sum_{d>x/n} \frac{\mu(d)}{d^2} \right) +A-A \\ &= \left( \frac{1}{\zeta(2)} \left(\log x + \gamma +O(x^{-1}) \right) - A \right) - \sum_{n \le x} \frac{1}{n}\sum_{d>x/n} \frac{\mu(d)}{d^2} + A, \end{align*} but I quickly realized that I had no clue what I was doing. Any help would be much appreciated.

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+1 for showing lots of energy and work on your part. – JavaMan Nov 8 '11 at 16:43

I tried this by switching the sums at the beginning so

$\displaystyle\sum_{n\le x}\frac{\phi(n)}{n^2}=\sum_{d\le x}\frac{\mu(d)}{d^2}\sum_{q\le\frac{x}{d}}\frac{1}{q}=\sum_{d\le x}\frac{\mu(d)}{d^2} \left (\log\left(\frac{x}{d}\right)+C+O\left(\frac{d}{x}\right)\right)$ using Thm 3.2(a) of Apostol p.55 (where $C$ is the Euler constant). Then use $\displaystyle\sum_{d\le x}\frac{\mu(d)}{d^2}=\frac{1}{\zeta(2)}+O\left(\frac{1}{x}\right)$ Apostol p.61.

Then use $\displaystyle\sum_{d\le x}\frac{\mu(d)\log d}{d^2}=A-\sum_{d>x}\frac{\mu(d)\log d}{d^2}$.

This last sum is $\displaystyle O\left(\sum_{d>x}\frac{\log d}{d^2}\right)$ and then use:

$0<\displaystyle \sum_{d>x}\frac{\log d}{d^2}=\sum_{d>x}\frac{\log d}{d^\frac{1}{2}}.\frac{1}{d^\frac{3}{2}}<\frac{\log x}{x^\frac{1}{2}}\sum_{d>x}\frac{1}{d^\frac{3}{2}}$ and Thm 3.2(c) p.55 for the error term

$\displaystyle O\left(\frac{\log x}{x}\right)$ and the $A$ in the question.

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Absolutely! The original poster, in the second equality after "So far I've got", seems to be using some lemma that is proved by using this method and then switching the order of summation in $\sum_{d\le x} \frac{\mu(d)}{d^2} \sum_{q\le x/d} \frac1q = \sum_{q\le x} \frac1q \sum_{d\le x/q} \frac{\mu(d)}{d^2}$. But I find apatch's method much more straightforward. – Greg Martin Nov 9 '11 at 5:47
Yes, this is definitely much more straightforward! – Carolus Nov 9 '11 at 6:18

You were indeed almost there. All that's left to do is just switch the order of summation on the last sum. I won't fill in all the details but here's a start:

\begin{align} \sum_{1 \leq n \leq x} ~\sum_{d > x/n} \frac{1}{n} \frac{\mu(d)}{d^2} &= \sum_{d \geq 2} \frac{\mu(d)}{d^2}\sum_{\frac{x}{d}< n \leq x} \frac{1}{n} \end{align}

Noticing that

$$\sum_{\frac{x}{d} < n \leq x}\frac{1}{n} = \sum_{n \leq x}\frac{1}{n} - \sum_{n \leq x/d}\frac{1}{n} = \log d + O\left(\frac{d}{x}\right)$$

should do it. To check that the error terms behave correctly, let's see where we're at. We (okay, You) have shown:

$$\sum_{n \le x} \frac{\varphi(n)}{n^2}=\frac{\log x}{\zeta(2)}+\frac{\gamma}{\zeta(2)}-A + \sum_{d \geq 2} \frac{\mu(d)}{d^2} \cdot O\left( \frac{d}{x} \right).$$

It remains to show that

$$\sum_{d \geq 2} \frac{\mu(d)}{d^2} \cdot O\left( \frac{d}{x} \right) = \sum_{d \geq 2} \frac{\mu(d)}{d} O\left( \frac{1}{x} \right) = O\left(\frac{\log x}{x} \right)\tag{\ast}$$

We can actually do a little better (and unless I'm missing something, I'm not sure where the $\log x$ term comes from - maybe it is just a safety net). First, we use that

$$\sum_{n \geq 1} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}.$$

If you have not seen this before this should be justified. It follows from the formula for $\mu(n)$, when you write the Euler product for the sum. In turn this implies that

$$\sum_{n \geq 1} \frac{\mu(n)}{n} = \lim_{s \to 1^+} \sum_{n \geq 1} \frac{\mu(n)}{n^s} = \lim_{s \to 1^+} \frac{1}{\zeta(s)} = 0.$$

This is nice since we then get that

$$\sum_{d \geq 2} \frac{\mu(d)}{d} = \sum_{d \geq 1} \frac{\mu(d)}{d} - 1 = -1.$$

This proves $(\ast)$.

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Thanks! I've got everything now except the $O(\log x/x)$-part. Does this follow from $\sum_{d \ge 2} \mu(d)/d^2 \cdot O(d/x)$? If it does - how? I'm having a bit of a hard time with asymptotics in general, I don't really know what do do with all those big Ohs... – Carolus Nov 8 '11 at 19:14

Consider Dirichlet series, related to the problem at hand: $$g(s) = \sum_{n=1}^\infty \frac{\varphi(n)}{n^{2+s}} = \frac{\zeta(s+1)}{\zeta(s+2)}$$

We can now recover behavior of $A(x) = \sum_{n \le x} \frac{\varphi(n)}{n^s}$ by employing Perron's formula, using $c > 0$: $$A(x) = \frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} \frac{\zeta(z+1)}{\zeta(z+2)} \frac{x^z}{z} \mathrm{d} z = \mathcal{L}^{-1}_z\left( \frac{\zeta(z+1)}{\zeta(z+2)} \frac{1}{z} \right)(\log x)$$

For large $x$, the main contribution comes from the pole of ratio of zeta functions at $z = 0$. Using $$\frac{\zeta(z+1)}{\zeta(z+2)} \sim \frac{1}{\zeta(2)} \left( \frac{1}{z} + \gamma - \zeta^\prime(2)\right) + O(z)$$ Since $\mathcal{L}^{-1}_z\left( \frac{1}{z^{n+1}} \right)(s) = \frac{s^n}{n!}$ we have $$A(x) = \frac{\log x}{\zeta(2)} + \left( \frac{\gamma}{\zeta(2)} - \frac{\zeta^\prime(2)}{\zeta(2)} \right) + \mathcal{L}^{-1}_z\left( \frac{\zeta(z+1)}{\zeta(z+2)} \frac{1}{z} - \frac{1}{\zeta(2) z^2} - \frac{\gamma - \zeta^\prime(2)}{\zeta(2) z} \right)(\log x)$$ Notice that $\frac{\zeta^\prime(s)}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n) \log(n)}{n^s}$, hence $\frac{\zeta^\prime(2)}{\zeta(2)} = A$.

It remains to be shown that the remainder term is small.

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I'm afraid this is a bit over my head, but thank you anyway. – Carolus Nov 8 '11 at 18:04
@Sasha, There is still a lot left to do, and you should not brush it into a single comment as is done at the end. Showing that the rest of the integral is small is 90% of the work when using Perrons Formula, the residue itself is only 10%. I presume it works out, but in general you should never just think it does. I have made this mistake a few times, a great example is $\sum_{n\leq x}\phi(n)$ and $\sum_{n\leq x}\phi(n)\log\left(\frac{x}{n}\right)$. The optimal error term for both of these is very different for subtle reasons when bounding the integrals... – Eric Naslund Nov 8 '11 at 20:15
@EricNaslund I agree, it might be not the easiest route. By the way, are you aware of a book/article where the bound for the remainder term is discussed ? I only have an intuition about it, but I am not certain how to prove it. – Sasha Nov 8 '11 at 21:12
@Sasha: Have you seen the full proof of the quantitative prime number theorem using Perrons formula, and the zero free region? A good place for this is chapter 6 of Montgomery and Vaughn's Multiplicative number theory. As for bounding integrals in general, you will need bounds on the zeta function and a very good book for this is Titchmarsh's The Riemann Zeta Function. In this particular case we would like to move the contour in close analogously to the proof of the PNT. However that won't work, I encourage you to try it out and learn why! – Eric Naslund Nov 8 '11 at 21:23
When I say it won't work, I mean the bounds on the zeta function don't allow you to pull the contour that far, even if you assume something as powerful as RH. Because of this we have to the proof, and pull the contour close to the zero line to get $O\left(\frac{\log x}{x}\right)$ as or error instead of an error analogous to the prime number theorem: $O\left(\frac{e^{-c\sqrt{\log x}}}{x}\right).$ In fact, this particular case our intuition is incorrect, (or at least mine!) and the error is provably larger then $O(\frac{1}{x})$!! – Eric Naslund Nov 8 '11 at 21:26

Rather belatedly (for the benefit of anyone who stumbles across this page in future), while I like apatch's answer, there is a slight issue with the step to prove $$\left|\sum_{d>x}\frac{\mu(d)\log d}{d^2}\right| \leq \sum_{d>x}\frac{\log d}{d^2} = O\left(\frac{\log x}{x}\right).$$ What follows is now a summary of my post about this question, and the answer contained therein.

Specifically, you can't say $$\sum_{d>x}\frac{\log d}{d^2} = \sum_{d>x}\frac{\log d}{d^\frac{1}{2}}.\frac{1}{d^\frac{3}{2}}<\frac{\log x}{x^\frac{1}{2}}\sum_{d>x}\frac{1}{d^\frac{3}{2}}$$ because $\frac{\log x}{\sqrt{x}}$ only reaches its maximum around $x\approx 7.39$, well above the $x>2$ condition stated in the question.

Instead, you need to approximate the sum by the integral $$\sum_{d>x}\frac{\log d}{d^2} \leq \frac{\log x}{x^2} + \int_x^\infty \frac{\log t}{t^2}dt$$ and then (e.g.) solve the integral using parts.

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