# 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 Louisville 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.