I am reading Enumerative Combinatorics by Richard Stanley, and I came across the following expression: $(1-x^n)^{\frac{-\mu(n)}{n}}$, where $\mu(n)$ is the usual Mobius function from number theory.

This expression can also be written as:

$(1-x^n)^{\frac{-\mu(n)}{n}}=\sum_{i \ge 0} \binom{-\mu(n)/n}{i}(-1)^ix^{in}$

This part is clear. But then the author claims that the expansion above shows that $(1-x^n)^{\frac{-\mu(n)}{n}}=1+H(x)$, where $H(x)$ is a formal power series of degree $n$. Why is that?

(Also: the degree of a formal power series is defined here as the smallest integer for which the corresponding coefficient is nonzero.

$\binom{-\mu(n)/n}{i}$ is defined in the usual way, as $\frac{(-\mu(n)-i+1)\dots (-\mu(n))}{i!}$)


As $\mu(n)\in\{0,1,-1\}$ we have one of three cases.

If $\mu(n)=0$ then $$(1-x^n)^{\mu(n)/n}=1$$ so that $H(x)=0$.

If $\mu(n)=1$ then $$(1-x^n)^{\mu(n)/n}=(1-x^n)^{1/n}=1-\frac{x^n}{n}-(n-1)\frac{x^{2n}}{2n^2} -\cdots$$ so that $$H(x)=-\frac{x^n}{n}+\text{higher terms}.$$

If $\mu(n)=-1$ then $$(1-x^n)^{\mu(n)/n}=(1-x^n)^{-1/n}=1+\frac{x^n}{n}+(n+1)\frac{x^{2n}}{2n^2} -\cdots$$ so that $$H(x)=\frac{x^n}{n}+\text{higher terms}.$$

  • $\begingroup$ Thanks. Taylor series, right? $\endgroup$ – odnerpmocon Sep 21 '18 at 17:53

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