Inverse and Square of the Riemann Zeta Function

Theorem

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

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

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

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:

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

We notice that the expansion of this product will be:

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

which of course is precisely:

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

as desired.

$\blacksquare$

Now we examine the square of the zeta function:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \zeta^2(z)\)	\(=\)	\(\displaystyle \left({\sum_{n=1}^\infty \frac 1 {n^z} }\right) \left({\sum_{n=1}^\infty \frac 1 {n^z} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({1 + \frac 1 {2^z} + \frac 1 {3^z} + \frac 1 {4^z} + \frac 1 {5^z} + \frac 1 {6^z} + \cdots}\right) \left({1 + \frac 1 {2^z} + \frac 1 {3^z} + \frac 1 {4^z} + \frac 1 {5^z} + \frac 1 {6^z} + \cdots}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Expanding this product, we get:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \frac 1 {2^z} + \frac 1 {3^z} + \frac 1 {4^z} + \frac 1 {5^z} + \frac 1 {6^z} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \frac 1 {2^z} + \frac 1 {4^z} + \frac 1 {6^z} + \frac 1 {8^z} + \frac 1 {10^z} + \frac 1 {12^z} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \frac 1 {3^z} + \frac 1 {6^z} + \frac 1 {9^z} + \frac 1 {12^z} + \frac 1 {15^z} + \frac 1 {18^z} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \frac 1 {4^z} + \frac 1 {8^z} + \frac 1 {12^z} + \frac 1 {16^z} + \frac 1 {20^z} + \frac 1 {24^z} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \frac 1 {5^z} + \frac 1 {10^z} + \frac 1 {15^z} + \frac 1 {20^z} + \frac 1 {25^z} + \frac 1 {30^z} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(+\)	\(\displaystyle \frac 1 {6^z} + \frac 1 {12^z} + \frac 1 {18^z} + \frac 1 {24^z} + \frac 1 {30^z} + \frac 1 {36^z} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\vdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \frac 2 {2^z} + \frac 2 {3^z} + \frac 3 {4^z} + \frac 2 {5^z} + \frac 4 {6^z} + \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

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

as desired.

$\blacksquare$