Möbius function

The Möbius function is one of the most important functions in number theory. It is defined as
$$ \mu(n) = \begin{cases}
1 & \text{if } n = 1, \\
(-1)^k & \text{if $n = p_1, \dots p_k$, where $p_i$ are distinct primes}, \\
0 & \text{otherwise}.
\end{cases} $$ It is the signed characteristic function of squarefree integers. The definition above may seem somewhat out of the blue and unmotivated but once one knows about Dirichlet product and Möbius inversion it is easy to see where the definition comes from (actually it is the inverse of the unit function under Dirichlet product). The Möbius function is intimately connected to the Riemann zeta function, $\zeta(s)$, which is one of the most important functions in analytic number theory. For instance, the estimate
$$
M(x) = \sum_{n \leq x} \mu(n) \ll x^{1/2 + \varepsilon}
$$ is equivalent to Riemann Hypothesis, one of the notoriously difficult problems in analytic number theory (or arguably in all of mathematics), which says that all of the nontrivial (interesting) zeros of $\zeta(s)$ lies on the line $\text{Re}(s) = 1/2$. This equivalence was shown by Littlewood. The Riemann Hypothesis is also equivalent to convergence of the Dirichlet series
$$
\sum_{n = 1}^{\infty} \mu(n) n^{-s}
$$ for every $s$ with $\text{Re}(s) > 1/2$. It is known that the above Dirichlet series converges for every $s$ with $\text{Re}(s) \geq 1$ and that the limit is $0$ when $s = 1$. The convergence at $s = 1$ alone (without regard for the limit value) implies prime number theorem, which is one of the celebrated results in analytic number theory.

Around 1897, Stieltjes conjectured that $|M(x)| < \sqrt{x}$ for every $x > 1$. Strangely, it is known as Merten’s conjecture and it was proven to be false by Odlyzko and te Riele in 1984. Their proof did not produce a counterexample but they showed that
$$
\limsup_{x \to \infty} \frac{M(x)}{\sqrt{x}} > 1.06 \quad \text{and} \quad \liminf_{x \to \infty} \frac{M(x)}{\sqrt{x}} < -1.009.$$ It is not known but seems probable that $\limsup_{x \to \infty} |M(x)|/ \sqrt{x} = \infty$. Although it follows from above but it is also easy to see that one cannot have $M(x) \ll x^{\theta}$ for any $\theta < 1/2$ since this would imply that $\zeta(s)$ has no zeros for $\sigma > \theta$.

Next we evaluate the divisor sum of $\mu$. It is precisely this result that connects $\mu$ to the Riemann zeta function.

Theorem. If $n \geq 1$, then $$ \sum_{d | n} \mu(d) = e(n).$$

Proof. If $n = 1$, then the equality is obvious. Now suppose that $n > 1$ and let $n = p_1^{a_1} \cdots p_k^{a_k}$. Because $\mu(d) = 0$ if $d$ is squarefree we can restrict the sum to squarefree divisor of $n$. Note that
\[\sum_{d | n} \mu(d) = \sum_{I \subset [k]} \mu \left( \prod_{i \in I} p_i \right) = \sum_{I \subset [k]} (-1)^{|I|} = \sum_{i = 0}^{k} \sum_{\scriptstyle I \subset [k] \atop \scriptstyle |I| = i} (-1)^{|I|} = \sum_{i = 0}^{k} (-1)^{k} \sum_{\scriptstyle I \subset [k] \atop \scriptstyle |I| = i} 1.\] It thus follows that
\[
\sum_{d | n} \mu(d) = \sum_{i = 0}^{k} \binom{k}{i} (-1)^{i} = (1 -1)^{k} = 0.
\]

As remarked earlier, it is not so easy to show that the series \[
\sum_{n = 1}^{\infty} \frac{\mu(n)}{n}\] converges. However, using the above result about the divisor sum of $\mu$ we can easily bound the partial sums of this series by $1$.

Corollary. If $n \geq 1$, then
\[
\left| \sum_{n \leq x} \frac{\mu(n)}{n} \right| \leq 1.\]

Proof. Note that
\[ 1 = \sum_{n \leq x} e(n) = \sum_{n \leq x} \sum_{d | n}
\mu(d) = \sum_{qd \leq x} \mu(d) = \sum_{d \leq x}\mu(d) \left \lfloor \frac{x}{d} \right \rfloor. \] Writing $\lfloor x/n \rfloor = x/n- \{x/n\}$ we obtain
\[
1 = x\sum_{n \leq x} \frac{\mu(n)}{n}- \sum_{n \leq x} \mu(n) \left\{ \frac{x}{n} \right\}.\] Observe that
\[ \sum_{n \leq x} \left\{ \frac{x}{n} \right\} = {x} + \sum_{2 \leq n \leq x} \left\{ \frac{x}{n} \right\} \leq {x} + \lfloor x \rfloor -1 = x-1. \] Thus we have
\[
\left|x \sum_{n \leq x} \frac{\mu(n)}{n} \right| \leq 1 + \sum_{n \leq x} \left\{\frac{x}{n} \right\} \leq x. \] This completes the proof.