Square of Riemann Zeta Function

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


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


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

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:

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

We notice that the expansion of this product will be

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

which of course is precisely

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

as desired.

Now examine the square of the zeta function:

$$\zeta^2(z) = \left({ \sum_{n=1}^\infty \frac{1}{n^z}  }\right)   \left({ \sum_{n=1}^\infty \frac{1}{n^z}  }\right)

= \left({  1 + \frac{1}{2^z} + \frac{1}{3^z} + \frac{1}{4^z} + \frac{1}{5^z} +  \frac{1}{6^z} + \dots }\right)   \left({   1 + \frac{1}{2^z} + \frac{1}{3^z} + \frac{1}{4^z} + \frac{1}{5^z} +  \frac{1}{6^z} + \dots  }\right)

\ $$

Expanding this product, we receive

$$= 1 + \frac{1}{2^z} + \frac{1}{3^z} + \frac{1}{4^z} + \frac{1}{5^z} + \frac{1}{6^z} + \dots \ $$

$$+ \frac{1}{2^z} + \frac{1}{4^z} + \frac{1}{6^z} + \frac{1}{8^z} + \frac{1}{10^z} + \frac{1}{12^z} + \dots \ $$

$$+ \frac{1}{3^z} + \frac{1}{6^z} + \frac{1}{9^z} + \frac{1}{12^z} + \frac{1}{15^z} + \frac{1}{18^z} + \dots \ $$

$$+ \frac{1}{4^z} + \frac{1}{8^z} + \frac{1}{12^z} + \frac{1}{16^z} + \frac{1}{20^z} + \frac{1}{24^z} + \dots \ $$

$$+ \frac{1}{5^z} + \frac{1}{10^z} + \frac{1}{15^z} + \frac{1}{20^z} + \frac{1}{25^z} + \frac{1}{30^z} + \dots \ $$

$$+ \frac{1}{6^z} + \frac{1}{12^z} + \frac{1}{18^z} + \frac{1}{24^z} + \frac{1}{30^z} + \frac{1}{36^z} + \dots \ $$

$$\vdots \ $$

$$= 1 + \frac{2}{2^z} + \frac{2}{3^z} + \frac{3}{4^z} + \frac{2}{5^z} + \frac{4}{6^z} + \dots \ $$

We see that each $$\frac{1}{n^z} \ $$ 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

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

as desired.