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