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#### chisigma

##### Well-known member

- Feb 13, 2012

- 1,704

$$\mu(n)= \begin{cases} - 1 &\text{if n has an odd number of prime distinct fators} \\ 0 &\text{if n has one or more prime fractors with exponent greater than one}\\ 1 &\text{if n has an even number of prime distinct factors}\end{cases}\ (1)$$

Actually I'm spending some of my time around the [non discrete] function...

$$\mu(x)= \sum_{n=1}^{\infty} \mu (n)\ x^{n}\ (2)$$

In any case is for $0 \le x < 1$ ...

$$|\sum_{n=1}^{N} \mu (n)\ x^{n}| \le \sum_{n=1}^{N} x^{n}\ (3)$$

... I'm sure that for $0 \le x < 1$ the series (2) converges. It seems however not to be so obvious for 'Monster Wolfram'...

sum mu(n) x^n from 1 to infinity - Wolfram|Alpha

... according to that the convergence test fails because 'the ratio test in inconclusive'...

I would be very happy if somebody clarifies mi ideas...

Kind regards

$\chi$ $\sigma$