## Expansion of the Dirichlet series for the mobius function

Summing to infinity:

$\sum \limits_{n=1}^{\infty} \frac{\mu(n)}{n^{s}} = 1 - \sum \limits_{a=2}^{\infty} \frac{1}{a^{s}} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \frac{1}{(a \cdot b)^{s}} - \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \frac{1}{(a \cdot b \cdot c)^{s}} + \sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} \frac{1}{(a \cdot b \cdot c \cdot d)^{s}} -...$

Mertens function, partial sums of the Mobius function:

$\sum \limits_{k=1}^{n} \mu(k) = 1 - \sum \limits_{a=2}^{a \leq n} 1 + \sum \limits_{a=2}^{a \leq \frac{n}{b}} \sum \limits_{b=2}^{b \leq \frac{n}{a}} 1 - \sum \limits_{a=2}^{a \leq \frac{n}{b \cdot c}} \sum \limits_{b=2}^{b \leq \frac{n}{a \cdot c}} \sum \limits_{c=2}^{c \leq \frac{n}{a \cdot b}} 1 + \sum \limits_{a=2}^{a \leq \frac{n}{b \cdot c \cdot d}} \sum \limits_{b=2}^{b \leq \frac{n}{a \cdot c \cdot d}} \sum \limits_{c=2}^{c \leq \frac{n}{a \cdot b \cdot d}} \sum \limits_{d=2}^{d \leq \frac{n}{a \cdot b \cdot c}} 1 -...$

For the first partial sum:
$a \leq n$

The partial double sum:
$a \cdot b \leq n$

The partial triple sum:
$a \cdot b \cdot c \leq n$

The partial quadruple sum:
$a \cdot b \cdot c \cdot d \leq n$

This notation is not entirely correct since the minimum value for indices in for example the quadruple sum is 2*2*2*2=16 which is too large if n is less than that number

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