6
$\begingroup$

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 $n.$

$\endgroup$
0

3 Answers 3

13
$\begingroup$

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.

$\endgroup$
13
$\begingroup$

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

For a more thorough discussion of such partial sums (especially with regard to the general lower bound), see this paper of Granville and Soundararajan.

$\endgroup$
1
$\begingroup$

TL;DR: The bound

$$\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|\leq 1$$

holds uniformly.

To prove this result, we can use the fact that in this paper it is shown that for $x\geq 33$

$$\left|\sum_{n<x}\frac{\mu(n)}{n}\right|<\frac{0.19}{\ln(x)}$$

and so for $x\geq33^2=1089$

\begin{align*} \left|\sum_{n<x}\frac{\lambda(n)}{n}\right|&\leq\sum_{d<\sqrt[4]{x}}\frac{1}{d^2}\left|\sum_{k<x/d^2}\frac{\mu(k)}{k}\right|+\sum_{\sqrt[4]{x}<d<\sqrt{x}}\frac{1}{d^2}\left|\sum_{k<x/d^2}\frac{\mu(k)}{k}\right|\\ &\leq\frac{0.625}{\ln^2(x)}+\frac{2}{\sqrt{x}}\\ &\leq\frac{0.837}{\ln(x)}\\ \end{align*}

Thus for $x>1089$ it holds that $\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|<1$. Checking $\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|<1$ for every value between $1$ and $1089$ in Python, we see that $\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|$ only takes on a value $\geq 1$ at $n=1$, and so we can conclude that $\left|\sum_{n<x}\frac{\lambda(n)}{n}\right|\leq1$ uniformly over all real $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.