# Bound on a scaled sum of the Liouville function

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 $n.$

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.
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 $$\sum_{n\le x} \frac{\lambda(n)}{n} \le \frac{x+\sqrt{x}}{x}.$$