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 $n.$

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 $n.$

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 mobius 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 $n.$

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 $n.$