Does asymptotic behavior of $\left|\sum_{d|n}f(d)\right|$ imply asymptotic properties of $f(d)$?

Question

The classic example of a function that has a drastic cancelation when summed over divisors is $\mu(n)$, with complete cancellation for every number other than $1$. Another such function is the Liouville function $\lambda(n)$. Both of these functions have have the property that $\sum_{n=1}^{\infty}\frac{f(n)}{n}=0$. Is this a general pattern? I.e do the conditions

\begin{equation} \lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\left|\sum_{d|n}f(d)\right|=0\tag{i} \end{equation}

\begin{equation} f(n)=O(1)\tag{ii} \end{equation}

imply

\begin{equation} \sum_{n=1}^{\infty}\frac{f(n)}{n}=0\tag{iii} \end{equation}

The interesting question here is about the $\mathit{convergence}$ of the sum. If we know that the sum in (iii) converges then we simply note that

$$\lim_{\epsilon\to0^+}\sum_{n=1}^{\infty}\frac{f(n)}{n^{1+\epsilon}}=\lim_{\epsilon\to0^+}\frac{1}{\zeta(1+\epsilon)}\sum_{n=1}^{\infty}\frac{\sum_{d|n}f(d)}{n^{1+\epsilon}}$$

Now since the terms $\sum_{d|n}f(d)$ tend to be small, the sum $\frac{\sum_{d|n}f(d)}{n^{1+\epsilon}}$ will go to infinity slower than $\zeta(1+\epsilon)$ and so

$$\lim_{\epsilon\to0^+}\sum_{n=1}^{\infty}\frac{f(n)}{n^{1+\epsilon}}=0$$

which implies that the value of the sum must be zero. A rigorous proof of the above statement is not much harder than the handwaving done above.

Answer 1

Interestingly enough, it is actually enough to know

\begin{equation} \frac{1}{N}\sum_{n=1}^{N}\left|\sum_{d|n}f(d)\right|=o(1)\tag{1} \end{equation}

to deduce

\begin{equation} \sum_{n=1}^{\infty}\frac{f(n)}{n}=0\tag{2} \end{equation}

to do this, we work instead under the change of variables $g(n):=\sum_{d|n}f(d)$ which shows that the problem $"(1)\implies(2)"$ is equivalent to saying that

\begin{equation} \frac{1}{N}\sum_{n=1}^{N}\left|g(n)\right|=o(1)\tag{3} \end{equation}

implies that

\begin{equation} \sum_{n=1}^{\infty}\frac{\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d)}{n}=0\tag{4} \end{equation}

by Mobius inversion. Working with this sum, we get that

\begin{align*} \sum_{n=1}^{N}\frac{\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d)}{n}&=\sum_{n=1}^{N}\frac{\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d)}{n}\\ &=\sum_{d=1}^{N}\sum_{n=1}^{N/d}\frac{\mu\left(n\right)g(d)}{nd}\\ &=\sum_{d=1}^{N}\frac{g\left(d\right)}{d}M\left(N/d\right) \end{align*}

Where here

$$M(x):=\sum_{n<x}\frac{\mu(n)}{n}$$

is the logarithmically weighted Mertens function. Strong forms of the PNT show that

\begin{equation} |M(x)|=O\left(e^{-C\sqrt{\log(x)}}\right)\tag{5} \end{equation}

for some constant $C$. Thus, for any $L>0$

\begin{align*} \sum_{n=1}^{N}\frac{\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d)}{n}&=\sum_{d=1}^{N}\frac{g(d)}{d}M(N/d)\\ &=\sum_{d=N/L}^{N}\frac{g(d)}{d}M(N/d)+\sum_{d<N/L}\frac{g(d)}{d}M(N/d)\\ &=S_1+S_2 \end{align*}

The second sum $S_2$ can be treated with the inequality (5) to be $o(L)$ as $N\to\infty$. As for $S_1$, we see that

\begin{align*} \sum_{d=N/L}^{N}\frac{g(d)}{d}M(N/d)&=\sum_{j=1}^{L}\sum_{d=N/(j+1)}^{N/j}\frac{g(d)}{d}M(N/d)\\ &=\sum_{j=1}^{L}M(j)\sum_{d=N/(j+1)}^{N/j}\frac{g(d)}{d}\\ \end{align*}

Because of (3), each inner sum $\sum_{d=N/(j+1)}^{N/j}\frac{g(d)}{d}$ tends to $0$ as $N\to\infty$. Thus, first taking $N$ and then $L$ to infinity we yield our result.