Square of Riemann Zeta Function

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Theorem

$\ds \map {\zeta^2} z = \sum_{k \mathop = 1}^\infty \frac {\map {\sigma_0} k} {k^z}$

where:

$\zeta$ is the Riemann zeta function
$\sigma_0$ is the divisor count function.


Proof

\(\ds \map {\zeta^2} z\) \(=\) \(\ds \paren {\sum_{n \mathop = 1}^\infty \frac 1 {n^z} } \paren {\sum_{n \mathop = 1}^\infty \frac 1 {n^z} }\)
\(\ds \) \(=\) \(\ds \paren {1 + \frac 1 {2^z} + \frac 1 {3^z} + \frac 1 {4^z} + \frac 1 {5^z} + \frac 1 {6^z} + \cdots} \paren {1 + \frac 1 {2^z} + \frac 1 {3^z} + \frac 1 {4^z} + \frac 1 {5^z} + \frac 1 {6^z} + \cdots}\)


Expanding this product, we get:

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


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

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

Hence this sum is:

$\ds \sum_{n \mathop = 1}^\infty \frac {\map {\sigma_0} n} {z^n}$

as desired.

$\blacksquare$