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.

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.