Asymptotic expansion of $\sum_{n \leq x} \frac{\mu^2(n)}{\phi(n)}$

Question

How can we derive the asymptotic expansion
$$ \sum_{n \leq x} \frac{\mu^2(n)}{\phi(n)} = \log x + C_0 + \sum_p \frac{\log p}{p(p-1)} + O(x^{-1/2} \log x)\,? $$

This question has been previously asked and answered here, but the provided sketch skips over the key difficulty. The error term appears to have several obstructions, and most of the standard methods that I used failed to work. Below, I outline where I'm stuck.

We follow Eric's approach on that answer but we slightly modify the function that we use to "approximate" $\mu^{2}(n)/\phi(n)$. We use instead the function $1/n$. This may simplify some estimates, as the values of the convolution at prime powers will only differ by sign. Then, we define $g(n)$ as the multiplicative function satisfying
$$ G(s) := \sum_n g(n) n^{-s} = \prod_p \left(1 + \frac{1}{p-1} p^{-s}\right)\left(1 - \frac{1}{p} p^{-s}\right) = \prod_p \left(1 + \frac{1}{p(p-1)} p^{-s} - \frac{1}{p(p-1)} p^{-2s}\right). $$ It is immediate to verify that this converges absolutely for $\sigma > -1/2$. Also, we see that, for $s = 0$, the product is simply $1$ and, by taking the $\log$ derivative above at $s = 0$, we obtain $G'(0) = \sum_{p}\log p/p(p-1)$. After some manipulations, we have $$ \sum_{n \leq x} \frac{\mu^2(n)}{\phi(n)} = (\log x + C_0)\sum_{d \leq x} g(d) - \sum_{d \leq x} g(d) \log d + O\left(\sum_{d \leq x} |g(d)| d\right). $$ Next, write $\sum_{d \leq x} = \sum_d - \sum_{d > x}$. Then, we only need to show that $$ \sum_{d > x} g(d) \ll x^{-1/2}. $$ All of the other terms are relatively manageable and/or follow from this estimate. However, the challenge lies in the fact that $G(s)$ does not converge absolutely at $\sigma = -1/2$. So, to obtain the above estimate we most likely have to bound the partial sums of $g(d) d^{1/2}$ directly. How can this be done?

Answer 1

I believe I saw precisely this tricky part you mentioned, and decided to only write the sketch because of it. Let me add some details to clarify.

I'll use the same $g$ as defined in that sketch:

Let $\frac{f(n)}{n}=\frac{\mu^2(n)}{\phi(n)}$. Set $g(n)=(\mu * f)(n)$ so that $f=(1*g)$.

Then, as mentioned in the sketch, $$g\left(p^{k}\right)=\begin{cases} \frac{1}{p-1} & \text{ if }k=1\\ \frac{-p}{p-1} & \text{ if }k=2\\ 0 & \text{ if }k\geq3 \end{cases} .$$

Let $G(s)=\sum_{n=1}^\infty g(n)n^{-s}$. This $G$ is also slightly different from that defined in your question.

Bounding the Error Term

Our goal is to prove a bound on the following sum:

Goal: Bound $$\sum_{d>x}\frac{g(d)}{d}.$$

Let me begin with a lemma.

Lemma 1: We have that $$\sum_{j>y}\frac{1}{j\phi(j)}=O\left(\frac{1}{y}\right).$$ Proof: We can bound this by using a Riemann-Stieltjes integral and integration by parts. Define $S(t)=\sum_{n\leq t}\frac{n}{\phi(n)}.$ Then $$\sum_{j>y}\frac{1}{j\phi(j)}=\int_{y}^{\infty}\frac{1}{t^{2}}dS(t)=\frac{S(t)}{t^{2}}\biggr|_{t=y}^{t=\infty}+\int_{y}^{\infty}\frac{S(t)}{t^{3}}dt.$$ Since $S(t)=\frac{315}{2\pi^{4}}\zeta(3)t+O(\log t)$, as shown in On The Mean Value of a Multiplicative Function, the lemma follows.

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In the prime-power break-down for $g$, note that $g(p^r)$ is very small for $r=1$, close to $1$ when $r=2$, and zero otherwise. This means we need to split out the squares.

For every integer $d$, let us split $d$ into its power-free part $k$, defined as $k=\prod_{p|d,p^{2}\nmid d}1$, and its powerful part $l$, defined as $d/k$. Then we have $d=k\cdot l$, and $g(d)=g(k)g(l)$. Thus we are trying to bound the sum $$\sum_{\begin{array}{c} lk>x\\ k\text{ pf-free}\\ l\text{ powerful} \end{array}}\frac{|g(kl)|}{kl}.$$ Since $g(p^{r})=0$ for any $r\geq3$, $l$ must be a square. Also, we know that $|g(j^{2})|\leq\frac{j}{\phi(j)}$and $|g(k)|\leq\frac{1}{\phi(k)}$. Thus, we can drop the condition on $k$ to upper bound the sum, and we have $$\sum_{d>x}\frac{|g(d)|}{d}\leq\sum_{j^{2}k>x}\frac{1}{k\phi(k)}\frac{1}{j\phi(j)}$$ $$\leq\sum_{k=1}^{\infty}\frac{1}{k\phi(k)}\sum_{j>\sqrt{\frac{x}{k}}}\frac{1}{j\phi(j)}.$$

Applying Lemma 1, we have that $$\sum_{d>x}\frac{|g(d)|}{d}=O\left(\frac{1}{\sqrt{x}}\sum_{k=1}^{\infty}\frac{1}{k^{\frac{1}{2}}\phi(k)}\right)=O\left(\frac{1}{\sqrt{x}}\right)$$

The other sum with $\log d$ follows similarly.

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Alternative: We could instead apply the Riemann-Stieltjes integral and integration by parts directly to $g$. Let $S_g(x)=\sum_{d\leq x} g(d)$. Then $$\sum_{x>d}\frac{g(d)}{d}=\int_{x}^{\infty}\frac{1}{t^{2}}dS_{g}(x)=-\frac{S_{g}(x)}{x^{2}}+\int_{x}^{\infty}\frac{S_{g}(x)}{t^{3}}dt.$$ The upper bound we desire would then follows from $$\sum_{d\leq x} g(d) = O(\sqrt{x}).$$ As before, splitting into "powerful-free" and "powerful" would directly prove this final bound.

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Clarifying the Sketch

In this section, I'll add some more details to the sketch in the other answer. We are trying to prove:

Theorem 1: We have that $$\sum_{n\leq x }\frac{\mu^2(n)}{\phi(n)}=\log x+\gamma+\sum_{p}\frac{\log p}{p(p-1)}+O\left(x^{-1/2}\log x\right).$$

First we'll prove the following proposition:

Proposition 1: We have that $$\sum_{n\leq x}\frac{\mu^{2}(n)}{\phi(n)}=G(1)\log x+G'(1)+\gamma G(1)+E$$ where $$E=O\left(\log x\sum_{d>x}\frac{|g(d)|}{d}+\sum_{d>x}\frac{g(d)\log d}{d}+\sum_{d\leq x}|g(d)\right).$$

From this proposition, using the arguments above about the error terms, we arrive at the theorem.

Proof of Proposition 1: Observe that $$\sum_{n\leq x}\frac{\mu^{2}(n)}{\phi(n)}=\sum_{n\leq x}\frac{(1*g)(n)}{n} = \sum_{n\leq x}\frac{1}{n}\sum_{d|n}g(d)$$

Rearranging the order of summation, this becomes:

$$\sum_{d\leq x}g(d)\sum_{\begin{array}{c} n\leq x\\ d|n \end{array}}\frac{1}{n}=\sum_{d\leq x}\frac{g(d)}{d}\sum_{k\leq\frac{x}{d}}\frac{1}{k}.$$

At this point, we may use the fact that $$\sum_{k\leq\frac{x}{d}}\frac{1}{k}=\log\frac{x}{d}+\gamma+O\left(\frac{d}{x}\right)$$ where $\gamma$ is the Euler-Mascheroni constant. This yields:

$$\sum_{d\leq x}\frac{g(d)}{d}\log\frac{x}{d}+\gamma\sum_{d\leq x}\frac{g(d)}{d}+O\left(\sum_{d\leq x}|g(d)\right)$$

Expanding $\log(x/d)=\log(x)-\log(d)$, we arrive at

$$\sum_{n\leq x}\frac{\mu^{2}(n)}{\phi(n)}=\log x\sum_{d\leq x}\frac{g(d)}{d}+\sum_{d\leq x}\frac{g(d)\log d}{d}+\gamma\sum_{d\leq x}\frac{g(d)}{d}+O\left(\sum_{d\leq x}|g(d)\right).$$

That proves Proposition 1.

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To prove the Theorem from Proposition 1, one needs to bound the sums in $E$ as was done in the first section above.

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See also: These two answers which use very similar techniques:

• On the mean value of a multiplicative function.
• The Mean Value of a Multiplicative Function close to $n$

Answer 2

We aim to analyze the sum $\sum_{y > n \geq x} g(n)$. Note that every integer $n$ can be uniquely expressed in the form $n = m^2 k$, $k \in \mathcal{Q}_2$, where $\mathcal{Q}_2$ is the set of square-free positive integers. Additionally, we can assume that $m$ is coprime to $k$. If $m$ and $k$ were not coprime, the primary factorization of $n$ would include a prime with an exponent of at least $3$, leading to $g(n) = 0$. Furthermore, we can assume $m \in \mathcal{Q}_2$. If $m \notin \mathcal{Q}_2$, then $m^2$ would have a prime factor with an exponent of at least $4$, which would again imply $g(n) = 0$. Based on these observations, we have $$ \sum_{y > n \geq x} g(n) = \sum_{\substack{y > m^2 k \geq x \\ (m, k) = 1 \\ m, k \in \mathcal{Q}_2}} g(m^2) g(k). $$ For the values of $m$ considered above, we see that $g(m^2) = \mu(m)/m \varphi(m)$, since $g(p^2) = -1/p(p-1)$. Moreover, for those values of $k$, $g(k) \geq 0$. Then, $$ \sum_{\substack{y > m^2 k \geq x\\(m,k)=1\\k \in \mathcal{Q}_2}} g(k) \frac{\mu(m)}{m \phi(m)} \ll \sum_{\substack{k \geq 1\\k\in \mathcal{Q}_2}} g(k) \sum_{(y/k)^{1/2}\geq m > (x/k)^{1/2}}\frac{1}{m \phi(m)} \ll x^{-1/2}\sum_{\substack{k \geq 1\\k\in \mathcal{Q}_2}} g(k) k^{1/2}, $$ where we used Eric's Lemma 1 above on the inner sum over $m$. Note that $\sum_{k} |g(k)| k^{1/2}$ does not converge when summed over all $k$. However, the restriction to square-free integers guarantees convergence, which can be readily verified from the Euler product of $g$. Thus, we conclude that $$ \sum_{y > n \geq x} g(n) \ll x^{-1/2}. $$