The Riemann zeta function (denoted by $\zeta(s)$) is defined

$\zeta(s) =\sum_{n=1}^\infty \frac{1}{n^s}$ where $s \in \mathbb{R}$ and $s>1$.

This function can be extended to all $s \in \mathbb{C}, s \neq 1$ which leads to much more such as the Riemann hypothesis, but I'm just going to focus on $s \in \mathbb{R}$. This is my favorite function and I've seen the formula for its inverse but never seen a derivation. But I think I was able to prove it myself thanks to our homework. So here it is:

One cool identity is called the Euler product formula:

$\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}$

-Proof*:

$\zeta(s) = 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+ \cdots$

$\frac{1}{2^s}\zeta(s) = \frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+\frac{1}{10^s}+ \cdots$

Subtracting the second from the first we remove all elements that have a factor of 2:

$\left(1-\frac{1}{2^s}\right)\zeta(s) = 1+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{9^s}+\frac{1}{11^s}+\frac{1}{13^s}+ \cdots$

Repeating for the next term:

$\frac{1}{3^s}\left(1-\frac{1}{2^s}\right)\zeta(s) = \frac{1}{3^s}+\frac{1}{9^s}+\frac{1}{15^s}+\frac{1}{21^s}+\frac{1}{27^s}+\frac{1}{33^s}+ \cdots$

Subtracting again we get:

$\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s) = 1+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{11^s}+\frac{1}{13^s}+\frac{1}{17^s}+ \cdots$

where all elements having a factor of 3 or 2 (or both) are removed.

It can be seen that the right side is being sieved. Repeating infinitely we get:

$\cdots \left(1-\frac{1}{11^s}\right)\left(1-\frac{1}{7^s}\right)\left(1-\frac{1}{5^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s) = 1$

Dividing both sides by everything but the $\zeta(s)$ we obtain:

$\zeta(s) = \frac{1}{\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\left(1-\frac{1}{7^s}\right)\left(1-\frac{1}{11^s}\right) \cdots }$

This can be written more concisely as an infinite product over all primes ''p'':

$\zeta(s)\;=\;\prod_{p} (1-p^{-s})^{-1}$

*Taken from Wikipedia

Now, we proved in our homework if $f(n)$ is multiplicative and $n = p_{1}^{a_{1}}... \;p_{k}^{a_{k}}$, then $\sum_{d \mid n} \mu(d)f(d)& = \prod_{i=1}^k (1-f(p_{i}))$ where $\mu$ is the Möbius function.

Now let $n = k!$ and $f(n) = n^{-s}$ (which is multiplicative).

So we obtain;

$\sum_{d \mid k!} \mu(d)d^{-s}& = \prod_{i=1}^r (1-p_{i}^{-s})$, but since all $d \leq k!$ divide $k!$

$\sum_{j=1}^{k!} \mu(j)j^{-s}& = \prod_{i=1}^r (1-p_{i}^{-s})$

$\lim_{k \to \infty}\sum_{j=1}^{k!} \mu(j)j^{-s}& = \lim_{r \to \infty}\prod_{i=1}^r (1-p_{i}^{-s})$, so

$\sum_{j=1}^{\infty} \mu(j)j^{-s}& = \prod_{p} (1-p^{-s}) = \frac{1}{\zeta(s)}$

Now we must test $\sum_{j=1}^{\infty} \mu(j)j^{-s}$ for convergence, but this is easy to see since

$\mu(j)j^{-s} \leq j^{-s} \;\forall\; j \in \mathbb{N}$

By the direct comparison test, $\sum_{j=1}^{\infty} \mu(j)j^{-s}$ converges if $\zeta(s)$ does as well.

SO…

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