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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $(\ast)$.