Möbius and Euler’s totient function

The Möbius function, denoted \(\mu\), is defined as
$$
\mu(n) =
\begin{cases}
1 & \text{if } n = 1, \\
(-1)^k & \text{if $n = p_1 \dots p_k$, where $p_1, \dots, p_k$ are distinct primes}, \\
0 & \text{otherwise}.
\end{cases}
$$
It is the signed characteristic function of squarefree positive integers. 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 \(\sum_{n \leq x} \mu(n) \ll x^{1/2 + \epsilon}\) implies Riemann Hypothesis, a notoriously difficult problem in analytic number theory (or arguably in all of mathematics), which says that all of the nontrivial (interesting) zeros of \(\zeta(s)\) lie on the critical line \(\sigma = \frac{1}{2}\). The Möbius function is connected to \(\zeta(s)\) due to the following identity involving Dirichlet series
$$
\sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)},
$$ which holds for \(\sigma = \Re(s) > 1\). It is not known if the Dirichlet series above converges for any \(\sigma < 1\). In fact, the convergence of above Dirichlet series for every \(\sigma > \frac{1}{2}\) implies the above estimate involving summatory function of \(\mu\) due to Kronecker’s lemma. It is however known that the Dirichlet series of \(\mu\) converges for every $ latex s$ with \(\sigma =1\) and the limit is \(0\) when \(s = 1\).

\begin{align} T &= x \\ &=y \end{align}

It is easy to evaluate the divisor sum of \(\mu\). If \(n = 1\), then \(\sum_{d \div n} \mu(d) = \mu(1) = 1\). If \(n > 1\), then $$
\begin{align} \sum_{d \div n} \mu(d) &= \sum_{\substack{d \div n \ d \text{ sq. free}}} \mu(d) = \sum_{\mathcal{P} \subset {p \div n}} \mu \left( \prod_{p \in \mathcal{P}} p \right) = \sum_{\mathcal{P} \subset {p \div n}} (-1)^{|\mathcal{P}|} \ &= \sum_{k = 0}^{\omega(n)} \sum_{\substack{\mathcal{P} \subset {p \div n} \ |\mathcal{P}| = k}} (-1)^{|\mathcal{P}|} = \sum_{k = 0}^{\omega(n)} (-1)^k \sum_{\substack{\mathcal{P} \subset {p \div n} \ |\mathcal{P}| = k}} 1 = \sum_{k = 0}^{\omega(n)} \binom{\omega(n)}{k} (-1)^k \ &= (1-1)^{\omega(n)} = 0. \end{align}$$
Thus we can write the divisor sum of $\mu$ as
[
\sum_{d \div n} \mu(d) = e(n).
]
Although it is not easy to prove that
[
\sum_{n = 1}^{\infty} \frac{\mu(n)}{n} = 0
]
but we can bound the partial sums of above series. For $x \geq 1$ we have
[
1 = \sum_{n \leq x} e(n) = \sum_{n \leq x} \sum_{d \div n} \mu(d) = \sum_{n \leq x} \mu(n) \left \lfloor \frac{x}{n} \right \rfloor = x \sum_{n \leq x} \frac{\mu(n)}{n} – \sum_{n \leq x} \mu(n) \left { \frac{x}{n} \right}.
]
We can bound the right most sum by
[
\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.
]
Combining this with above we get
[
\left| \sum_{n \leq x} \frac{\mu(n)}{n} \right| \leq 1.
]

In one of the very first homeworks of Math 571\footnote{It was taught by Bob Vaughan in spring of 2025 at Penn State.} we were asked to show that for every $k \in \N$ there exist infinitely many $n$ such that
[
\mu(n + 1) = \cdots = \mu(n + k),
]
i.e., there exist infinitely many strings of $k$ consecutive integers which are either all squarefree or nonsquarefree. My solution was cumbersome and was based on induction. A much simpler solution (provided by Bob Vaughan) based on Chinese remainder theorem goes as follows: Let $p_1, \dots, p_k$ be distinct primes. Then by the Chinese remainder theorem there exist infinitely many positive integers $n$ such that $n \equiv -j \mod{p_j^2}$ for every $1 \leq j \leq k$. Thus $p_j^2 \div (n +j)$ for every $1 \leq j \leq k$ and so $n+ j$ is not squarefree, i.e., $\mu(n + j) = 0$ for $1 \leq j \leq k$.