The Riemann Zeta Function

Recall from the Dirichlet Series page that if $(a_n)$ is a sequence (or arithmetic function) then the corresponding Dirichlet series is the function:

(1)

\begin{align} \quad A(s) = \sum_{n=1}^{\infty} \frac{a_n}{n^s} \end{align}

When we consider the special arithmetic function $1(n) = 1$, we obtain a special function which we define below.

Definition: The Riemann Zeta Function is the function is the Dirichlet series of the arithmetic function $1(n) = 1$ and is denoted by $\displaystyle{\zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^s}}$.

If $s > 1$ then observe that $\zeta (s)$ converges by various series convergence tests (such as the integral test). When $s \leq 1$, $\zeta (s)$ diverges. We can extend $\zeta$ in a useful manner but the extension is complicated and we will avoid that for the time being.

Proposition 1: $\displaystyle{\sum_{n=1}^{\infty} \frac{\mu (n)}{n^s} = \frac{1}{\zeta (s)}}$.

• On the The Dirichlet Convolution μ * 1 = ꙇ we proved that $\mu * 1 = \iota$. Therefore:

(2)

\begin{align} \quad \left [ \sum_{n=1}^{\infty} \frac{\mu (n)}{n^s} \right ] \left [ \sum_{n=1}^{\infty} \frac{1}{n^s} \right ] &= \sum_{n=1}^{\infty} \frac{\iota (n)}{n^s} \\ \quad \left [ \sum_{n=1}^{\infty} \frac{\mu (n)}{n^s} \right ] \zeta (s) &= 1 \\ \end{align}

Therefore:

(3)

\begin{align} \sum_{n=1}^{\infty} \frac{\mu (n)}{n^s} = \frac{1}{\zeta (s)} \quad \blacksquare \end{align}