Bound on a scaled sum of the Liouville function

Question

Terence Tao has shown see his blog post that

$$\left| \sum_{n\leq x} \frac{\mu(n)}{n} \right|\leq 1,$$

for $x$ a positive real number, where $\mu(n)$ is the Möbius function. Let $\lambda(n)$ denote Liouville's lambda function which is defined as $(-1)^{\Omega(n)}$ where $\Omega(n)$ is the number of prime divisors of $n$, counted with multiplicity.

My question is whether it's possible to prove a corresponding bound of the form

$$\left| \sum_{n\leq x} \frac{\lambda(n)}{n} \right|\leq M<\infty,$$

for all $x>0,$ for the scaled partial sums of the Liouville function. I am aware that the function $$\sum_{n\leq x} \frac{\lambda(n)}{n^{\alpha}}$$ converges to $\zeta(2\alpha)/\zeta(\alpha)>0$ for all $\alpha>1$ as well as that the unnormalized sum $$\left| \sum_{k\leq n} \lambda(k) \right| > c \sqrt{n}$$ for some small constant $c \in (0,1)$ for infinitely many positive integers $n.$

Answer 1

Yes, and this can be derived from your first inequality in an elementary way. Indeed, the formal Dirichlet series identity $$ \sum_{n=1}^\infty\frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)} $$ is equivalent to the convolution identity $\lambda=1_{\square}\ast\mu$, i.e. $$ \lambda(n)=\sum_{d^2\mid n}\mu\left(\frac{n}{d^2}\right). $$ Using this identity, $$ \sum_{n\leq x}\frac{\lambda(n)}{n} = \sum_{n\leq x}\frac{1}{n}\sum_{d^2\mid n}\mu\left(\frac{n}{d^2}\right) = \sum_{d\leq\sqrt{x}}\frac{1}{d^2}\sum_{k\leq\frac{x}{d^2}}\frac{\mu(k)}{k}. $$ From here we get, using the triangle inequality and your first inequality, $$ \left|\sum_{n\leq x}\frac{\lambda(n)}{n}\right|\leq \sum_{d\leq\sqrt{x}}\frac{1}{d^2}\left|\sum_{k\leq\frac{x}{d^2}}\frac{\mu(k)}{k}\right|\leq\sum_{d\leq\sqrt{x}}\frac{1}{d^2}<\frac{\pi^2}{6}. $$ Of course, by the prime number theorem both the original sum for $\mu$ and the new sum for $\lambda$ tend to zero with $x$, namely they are both $\ll e^{-c\sqrt{\log x}}$ for some explicit $c>0$. Also, the Riemann Hypothesis is equivalent to either of the sums being $\ll x^{-c}$ for any $c<1/2$. Finally, I remark that there are various Tauberian theorems that try to prove or generalize the limit zero result with as little assumption for the underlying Dirichlet series as possible.

Answer 2

Let $f$ be any multiplicative function with $|f(n)| \le 1$ and such that $\sum_{d|n} f(d)$ is non-negative for all $n$. It is easy to check that $\mu$ and $\lambda$ satisfy this constraint. Then $$ 0\le \sum_{n\le x} \sum_{d|n} f(d) = \sum_{d\le x} f(d) \lfloor \frac{x}{d} \rfloor \le \sum_{d\le x} \Big( x\frac{f(d)}{d} + 1\Big), $$ so that $$ \sum_{d\le x} \frac{f(d)}{d} \ge - \frac{\lfloor x\rfloor}{x} \ge -1. $$

The upper bound for $\mu$ and $\lambda$ follows similarly, here making use of $\sum_{d|n} \mu(d) =1$ if $n=1$ and $0$ otherwise, and $\sum_{d|n} \lambda(d) = 1$ if $n$ is a square and zero otherwise. So for $\lambda$ we obtain $$ \sum_{n\le x} \frac{\lambda(n)}{n} \le \frac{x+\sqrt{x}}{x}. $$

For a more thorough discussion of such partial sums (especially with regard to the general lower bound), see this paper of Granville and Soundararajan.

Answer 3

TL;DR: The bound

$$\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|\leq 1$$

holds uniformly.

To prove this result, we can use the fact that in this paper it is shown that for $x\geq 33$

$$\left|\sum_{n<x}\frac{\mu(n)}{n}\right|<\frac{0.19}{\ln(x)}$$

and so for $x\geq33^2=1089$

\begin{align*} \left|\sum_{n<x}\frac{\lambda(n)}{n}\right|&\leq\sum_{d<\sqrt[4]{x}}\frac{1}{d^2}\left|\sum_{k<x/d^2}\frac{\mu(k)}{k}\right|+\sum_{\sqrt[4]{x}<d<\sqrt{x}}\frac{1}{d^2}\left|\sum_{k<x/d^2}\frac{\mu(k)}{k}\right|\\ &\leq\frac{0.625}{\ln^2(x)}+\frac{2}{\sqrt{x}}\\ &\leq\frac{0.837}{\ln(x)}\\ \end{align*}

Thus for $x>1089$ it holds that $\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|<1$. Checking $\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|<1$ for every value between $1$ and $1089$ in Python, we see that $\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|$ only takes on a value $\geq 1$ at $n=1$, and so we can conclude that $\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|\leq1$ uniformly over all real $x$.