A very strange statement from Wolfram Alpha....

[chisigma]

The so called 'Moebious Function' is a discrete function defined as...

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

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 $\chi$ $\sigma$

 

[alyafey22]

Well, the sum seems convergent to me as well . I ran it on Mathemaica on my PC and it still can not determine convergence .

I tried the following \(\displaystyle \sum_{k\geq 1} \frac{\mu(k)}{2^k } \) still no response !

 

[chisigma]

Today I have found the following 'explicit expression'...

$$\sum_{n=1}^{\infty} \mu(n)\ x^{n} = x - \sum_{a=2}^{\infty} x^{a} + \sum_{b=2}^{\infty} \sum_{a=2}^{\infty} x^ {a\ b} - \sum_{c=2}^{\infty} \sum_{b=2}^{\infty} \sum_{a=2}^{\infty} x^ {a\ b\ c} + \sum_{d=2}^{\infty} \sum_{c=2}^{\infty} \sum_{b=2}^{\infty} \sum_{a=2}^{\infty} x^ {a\ b\ c\ d} - ...\ (1)$$

No surprise about the fact that the amount of computation required overflows also the capability of 'Monster Wolfram' (Sweating) ...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

Today I decided to try a 'nice 'experiment'... I recovered from the cellar a very 'artifact', one of them first Pentium PC [improved after it was discovered that first examples failed in doing multiplications (Sadface)...] that I used about twenty year ago and conserve as 'souvenir' and used it to numerically compute some values of the function...

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

The sum was interrupted after 4000 iteration and the range was $-.99 < x < .99$. The result is reported in the diagram...

http://www.123homepage.it/u/i72335019._szw380h285_.jpg.jfif

Clearly in this range we have 'good convergence' but an obvious question is: what does it happen in the range $-1 < x < -.99$ and $.99 < x < 1$?... The answer is not so easy and You can realize that considering that for x=1 we have...

$$\mu(1) = \lim_{n \rightarrow \infty} M (n)\ (2)$$

... where...

$$M(n) = \sum_{k=1}^{n} \mu(k)\ (3)$$

... is the so called 'Marten's function' , the behaviour of which for 'large' values of n is highly 'oscillatory' so that may be that the $\mu(x)$ in the range $.99 < x < 1$ crosses the zero infinite times (Sweating)... in any case I intend to do more 'investigations'...

Kind regards

$\chi$ $\sigma$

 

[CellCoree]

I would say that the statement from Wolfram Alpha regarding the convergence of the series (2) is not strange, but rather a valid mathematical observation. The ratio test is a commonly used method for determining the convergence of a series, and in this case, it is inconclusive. We may need to use other methods, such as the root test or the comparison test, to determine the convergence of the series. Furthermore, it is important to note that the convergence of a series does not depend on its value at a particular point, but rather on the behavior of the series as a whole. Therefore, the fact that the series (2) is convergent for $0 \le x < 1$ is still a valid statement. As for the "Moebious Function," it is a well-defined discrete function with interesting properties, and its relationship to the series (2) is also a valid mathematical concept. Overall, I would suggest exploring more mathematical concepts and methods to understand the convergence of series and the properties of functions.