The expectation values of the 1D simple random walk $S_n$ can be shown to have the asymptotic behavior of
$$ \lim_{n\to\infty} \frac{a_n}{n^{1/2}} = \sqrt{\frac{2}{\pi}}, \tag{1}\label{1}$$
with $a_n = S_n$.
On the other hand, the Riemann Hypothesis is famously equivalent to the statement that
$$ \lim_{n\to\infty} \frac{a_n}{n^{1/2 +\varepsilon}} = 0, \tag{2}\label{2}$$
for any fixed $\varepsilon>0$ and setting $a_n=L(n)$, the cumulative (or summatory) Liouville function $L(n)=\sum_n \lambda(n)$.
I am interested in about what we know on the difference between the two asymptotic behaviours \eqref{1} and \eqref{2}. Clearly the random walk asymptotics \eqref{1} implies \eqref{2} but not vice versa.
How can it be shown that $a_n=L(n)$ violates \eqref{1}, would its fulfillment have interesting implications? Are there other/simple examples of $a_n$ that fulfill \eqref{2}?
Best Answer
I figured the remarks I gave in the comments deserve to be gathered up into a more coherent form as an answer.
One thing I will start with is that comparing $L(n)$ (or $M(n)$, the Mertens function) to the values $S_n$ is arguably not the right heuristic. Indeed, $S_n$ averages over all random walks, which can have their peaks and troughs at different places, and so they cancel out. Analysing a single random walk gives a different asymptotic: this is governed by the law of the iterated logarithm, which asserts that the random walk will oscillate with extremal values on the order of $\pm\sqrt{2n\log\log n}$.
However it turns out that for subtle arithmetic reasons we do not expect the arithmetic functions of interest to reach the same bounds. Most literature (see e.g. this paper) concerns the Mertens function $M(n)$ so I will discuss this one, but all the relevant heuristics should also hold for $L(n)$. An unpublished conjecture of Gonek is that the maximal order of $|M(n)|$ is not $\sqrt{n\log\log n}$, but rather $\sqrt{n}(\log\log\log n)^{5/4}$, which in particular implies that the ratio $|M(n)|/\sqrt{n}$ is unbounded.
While Gonek's conjecture is still open, as is unboundedness of $|M(n)|/\sqrt{n}$, we do have unconditional results which imply that this sequence does not have a limit. Specifically, we know that $M(n)/\sqrt{n}$ has strictly negative liminf and strictly positive limsup (indeed, they are at least $\pm1.8$ respectively, see Wikipedia for more details and references.) This shows that $|M(n)|/\sqrt{n}$ crosses zero infintiely often, and infinitely often it takes values greater than $1$, so it cannot have a limit. (Same is known for Liouville though with different bounds on liminf and limsup, see Wikipedia.)