# An identity involving the Möbius function

$$\sum_{n=1}^{\infty}\frac{1}{n^s}\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=1$$ for $s>1$.

How do I prove this identity?

## 2 Answers

The idea needed here is Dirichlet convolution.

The convolution identity states that if

$$\left(\sum_{j=1}^\infty \frac{a_j}{j^s}\right)\left(\sum_{k=1}^\infty \frac{b_k}{k^s}\right)=\sum_{\ell=1}^\infty \frac{c_\ell}{\ell^s}$$

then

$$\sum_{k\,\mid \,n} a_k b_{n/k} = c_n$$

where the sum ranges over all positive divisors of $n$.

More specifically, your original identity is equivalent to the convolution identity

$$\sum_{k\,\mid \,n} \mu(k) = [n=1]$$

where the two functions being convolved are the constant function $1$ and the Möbius function $\mu(k)$. ($[p]$ here is the Iverson bracket; it evaluates to $1$ if $p$ is true, and $0$ if $p$ is false.)

For $n=1$, the identity is easily verified. To see what happens when $n > 1$, take the prime factorization

$$n=\prod_{k=1}^r p_k^{m_k}$$

It is known that the summatory function of a multiplicative function is itself multiplicative; thus, letting $\xi(n)=\sum\limits_{k\,\mid \,n} \mu(k)$, we have $\xi(qr)=\xi(q)\xi(r)$. We can thus restrict to the case

$$\xi(p^m)=\mu(1)+\mu(p)+\cdots+\mu(p^m)=1-1+0+\cdots+0=0$$

and since $\xi(n)=\xi(p_1^{m_1})\xi(p_2^{m_2})\cdots \xi(p_r^{m_r})$, $\xi(n)=0$ for $n > 1$.

• I was wondering how to show that "The Dirichlet series that generates the Möbius function is the (multiplicative) inverse of the Riemann zeta function." – Kou Nov 20 '11 at 2:45
• Right; remember that the Dirichlet coefficients of the product of two Dirichlet series is the Dirichlet convolution of the Dirichlet coefficients of the individual series. – J. M. is a poor mathematician Nov 20 '11 at 3:03

Let $D_f(s)$ denote the Dirichlet generating series:

$$D_f(s) = \sum_{n =1}^{\infty} \frac{f(n)}{n^s}.$$

When the series convergences absolutely, we can write $D_f(s)$ as an Euler product:

$$D_f(s) = \prod_p \left(1+f(p)p^{−s} +f(p^2)p^{−2s} \dots \right)$$ where the product ranges over all the primes $p$. Looking at the Euler product for $\zeta(s)$ (which corresponds to $f(n) = 1$) we see that

$$\zeta(s) = \prod_p \left(1+p^{-(s-1)} +p^{-2(s-1)}+ \dots \right) = \prod_p \left(1 - p^{-s} \right)^{-1}.$$

Hence,

\begin{align} \frac{1}{\zeta(s)} = \prod_{p} \left(1 - p^{-s} \right) = \sum_{n=1}^{\infty} \frac{g(n)}{n^s}, \end{align}

where $g$ is the multiplicative function such that $g(p) = -1$, and $g(p^n) = 0$ when $n \geq 2$. Note that this is precisely the definition of $\mu$ (and this is probably how the formula for $\mu$ was originally discovered). Thus, we have shown that:

$$\sum_{n=1}^{\infty} \frac{1}{n^s} \cdot \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} = \zeta(s) \cdot \frac{1}{\zeta(s)} = 1$$

which is valid when $\zeta(s)$ converges absolutely (so this is valid for $s = \sigma + it \in \mathbb{C}$ such that $\sigma > 0$).