The Dirichlet Series for the von-Mangoldt Function

Theorem 1: $\displaystyle{\sum_{n=1}^{\infty} \frac{\Lambda (n)}{n^s} = \frac{-\zeta '(s)}{\zeta(s)}}$.

• Proof: Consider the Riemann-zeta function which is the Dirichlet series of the arithmetic function $1$:

(1)

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

• Noting that the derivative of $\frac{1}{n^s}$ with respect to $s$ is $\frac{-\ln (n)}{n^s}$, and since $\zeta (s)$ is absolutely convergent for each $s > 1$, we can different the series term by term (with respect to $s$ to get:

(2)

\begin{align} \quad -\zeta'(s) = \sum_{n=1}^{\infty} \frac{\ln n}{n^s} \quad (*) \end{align}

• We have already proved that:

(3)

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

• From $(*)$ and $(**)$ we see that $\displaystyle{ \frac{-\zeta'(s)}{\zeta (s)} }$ is the Dirichlet series of $\ln * \mu$. But $\ln * \mu = \Lambda$, and thus:

(4)

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