5
$\begingroup$

The Lambert series for the Möbius function is given by : $$\sum_{n=1}^{\infty}\mu(n)\frac{q^{n}}{1-q^{n}}=\sum_{n=1}^{\infty}\frac{\mu(n)}{q^{-n}-1}=q \;\;\;\;\;\; |q|<1$$ This follows from Möbius inversion of the geometric series : $$\frac{q}{1-q}=\sum_{n=1}^{\infty}q^{n}\;\;\;\;\;\;|q|<1$$ Now, consider the slightly different problem : $$\sum_{n=1}^{\infty}\mu(n)\frac{1}{q^{-n}-p^{-n}}=\sum_{n=1}^{\infty}\mu(n)\frac{q^{n}}{1-\left(\frac{q}{p}\right)^{n}}\;\;\;\;\;\;|q|<|p|$$ We have : $$\frac{q}{1-\left(\frac{q}{p}\right)}=p\sum_{n=1}^{\infty}\left(\frac{q}{p}\right)^{n}$$ Which is not an explicit function in $n$ -Please notice the $p$ factor- Thus, Möbius inversion doesn't work.

My question : is there a way to get a closed form for : $$\sum_{n=1}^{\infty}\frac{\mu(n)}{q^{-n}-p^{-n}}$$

EDIT :

By simple manipulation, We have : $$\sum_{n=1}^{\infty}\mu(n)\frac{1}{q^{-n}-p^{-n}}=\sum_{n=1}^{\infty}\mu(n)\frac{p^{n}\left(\frac{q}{p} \right )^{n}}{1-\left(\frac{q}{p} \right )^{n}}=\sum_{m=1}^{\infty}f_{m}(p)\left(\frac{q}{p} \right )^{m}$$ Where the polynomials $f_{m}(p)$ are given by : $$f_{m}(p)=\sum_{n|m}\mu(n)p^{n}$$ And they satisfy : $$\sum_{n|m}f_{m}(p^{\frac{m}{n}})=p$$ And the question becomes, what are these polynomials, and what is their algebraic meaning ?

$\endgroup$
1
  • $\begingroup$ What do mean by Mobius inversion of the geometric series? I only know about Mobius inversion of a divisor sum. Can you please provide more details about this method? $\endgroup$
    – subhajit
    Commented Mar 5, 2019 at 6:40

1 Answer 1

2
$\begingroup$

By standard arguments with Lambert series, your series is given by $$M(p, q) := \sum_{n \geq 1} \frac{\mu(n) p^n (q/p)^n}{1-(q/p)^n} = \sum_{m \geq 1} \left(\sum_{d|m} \mu(d) p^d\right) (q/p)^m,\ |q/p| < 1.$$ Is this what you had in mind? Not sure about algebraic interpretations of the polynomials in $p$ defined by the coefficients of the first series. What are the applications of this problem?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .