# 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.

## 1 Answer

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

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

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.