TITLE: Partial sums of multiplicative functions
QUESTION [28 upvotes]: It is well known that some statements about partial sums of multiplicative functions are extremely hard. For example, the Riemann hypothesis is equivalent to the assertion that $|\mu(1)+\mu(2)+\dots+\mu(n)|$ is bounded above by $n^{1/2+\epsilon}$ for all $\epsilon > 0$. However, I want to know about lower bounds for such partial sums, valid for infinitely many n. For instance, is it known that the sums of the Möbius function must infinitely often be somewhere near $n^{1/2}$ in magnitude? Can one at least prove that they are unbounded? And what about the Liouville function $\lambda$? Are its partial sums unbounded? If so, can anyone give me a good reference or quick proof? And how about general completely multiplicative functions that take values of modulus 1?  Is anything known about those?

REPLY [7 votes]: This paper shows that $L(n) > .061867\sqrt{n}$ for infinitely many $n$.
As for somewhat elementary methods (in the sense of avoiding the Riemann zeta function) to show that $L(n)$ is "usually" of order $\sqrt{n}$, one can use the Lambert series
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)q^n}{1-q^n}} = \sum_{n=1}^{\infty}{q^{n^2}}.$$
As
$$\frac{q^n}{1+q^n} = \frac{q^n}{1-q^n} - 2\frac{q^{2n}}{1-q^{2n}},$$
we have
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{q^{-n}+1}} = \sum_{n=1}^{\infty}{q^{n^2}} - 2\sum_{n=1}^{\infty}{q^{2n^2}}$$
or equivalently, letting $q = e^{-\pi/x}$ and $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}}$, where $x$ is large,
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \psi(1/x) - 2\psi(2/x)$$
Now $\psi(x)$ satisfies the functional equation
$$\frac{1+2\psi(x)}{1+2\psi(1/x)} = \frac{1}{\sqrt{x}},$$
and so we can rewrite this as
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x}.$$
For large $x$, the left-hand side "looks like" $L(x)$, whereas the right-hand side is dominated by the term $\frac{1-\sqrt{2}}{2}\sqrt{x}$. This also explains why $L(n)$ is predominantly negative, as $\frac{1-\sqrt{2}}{2}$ is negative.