- In line with the previous exercise, we consider the Dirichlet series for \(1 / \zeta\)
- Prove that for \(\operatorname { R e } ( s ) > 1\)
\[ \frac {1}{\zeta (s)} = \sum_ {n = 1} ^ {\infty} \frac {\mu (n)}{n ^ {s}}, \]
where \(\mu ( n )\) is the M¨obius function defined by
\[ \mu (n) = \left\{ \begin{array}{l l} 1 & \text{if } n = 1, \\ (- 1) ^ {k} & \text{if } n = p_{1} \cdots p_{k} \text{, and the } p_{j} \text{ are distinct primes,} \\ 0 & \text{otherwise.} \end{array} \right. \]
Note that \(\mu ( n m ) = \mu ( n ) \mu ( m )\) whenever n and m are relatively prime. [Hint: Use the Euler product formula for \(\zeta ( s ) . ]\)
- Show that
\[ \sum_ {k \mid n} \mu (k) = \left\{ \begin{array}{l l} 1 & \text { if } n = 1, \\ 0 & \text { otherwise }. \end{array} \right. \]
Solution
For \(\Re s>1\), the Euler product converges absolutely: \[ \zeta(s)=\prod_p(1-p^{-s})^{-1}. \] Therefore \[ \frac1{\zeta(s)}=\prod_p(1-p^{-s}). \] Expanding this absolutely convergent product, each squarefree integer \(n=p_1\cdots p_k\) contributes \((-1)^k n^{-s}\), while integers divisible by a square do not occur. Hence \[ \frac1{\zeta(s)}=\sum_{n=1}^\infty\frac{\mu(n)}{n^s}. \]
For part (b), define \[ S(n)=\sum_{d\mid n}\mu(d). \] If \(n=1\), then \(S(1)=\mu(1)=1\). If \[ n=p_1^{e_1}\cdots p_r^{e_r}>1, \] only squarefree divisors contribute, so \[ S(n)=\prod_{j=1}^r(1+\mu(p_j)) =\prod_{j=1}^r(1-1)=0. \] Thus \[ \sum_{d\mid n}\mu(d)= \begin{cases} 1,&n=1,\\ 0,&n>1. \end{cases} \]