Square of Riemann Zeta Function

Theorem
We have the following results for the inverse and square of the Riemann zeta function $\zeta$:


 * $\displaystyle \frac 1 {\zeta(z)} = \sum_{k \mathop = 1}^\infty \frac{\mu(k)}{k^z}$


 * $\displaystyle \zeta^2 (z) = \sum_{k \mathop = 1}^\infty \frac{d(k)}{k^z}$

where $\mu$ is the Moebius function and $d$ is the divisor function.

Proof
Take the inverse part of the theorem first.

We write out the zeta function explicitly:


 * $\displaystyle \frac 1 {\zeta(z)} = \prod_{p \text{ prime}} (1-p^{-z}) = \left({1 - \frac 1 {2^z}}\right) \left({1 - \frac 1 {3^z}}\right) \left({1 - \frac 1 {5^z}}\right) \left({1 - \frac 1 {7^z}}\right) \left({1 - \frac 1 {11^z}}\right) \cdots$

We notice that the expansion of this product will be:


 * $\displaystyle 1 + \sum_{n \text{ prime}} \left({ \frac{-1}{n^z} }\right) + \sum_{n \mathop = p_1 p_2} \left({ \frac{-1}{p_1^z} \frac{-1}{p_2^z} }\right) + \sum_{n \mathop = p_1p_2p_3} \left({ \frac{-1}{p_1^z} \frac{-1}{p_2^z} \frac{-1}{p_3^z} }\right) + \dots$

which of course is precisely:


 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac{\mu(n)}{n^z}$

as desired.

Now we examine the square of the zeta function:

Expanding this product, we get:

We see that each $\dfrac 1 {n^z}$ term in this sum will occur as many times as there are ways represent $n$ as $ab$, counting order.

But this is precisely the number of divisors of $n$, since each way of representing $n = ab$ corresponds to the first term of the product, $a$.

Hence this sum is:


 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac{d(n)}{z^n}$

as desired.