Terence Tao has shown [see his blog post](https://terrytao.wordpress.com/2009/08/30/an-elementary-inequality-involving-the-mobius-function/) 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 $n.$



Bound on a scaled sum of the Liouville function
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.


