Square of Riemann Zeta Function

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Theorem

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

where:

$\zeta$ is the Riemann zeta function
$d$ is the divisor function.


Proof

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


Expanding this product, we get:

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


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:

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

as desired.

$\blacksquare$