Show that if $|q| < 1$, then $\displaystyle{\sum_{n=1}^{\infty}} \frac{\mu(n)*q^n}{1-q^n} = q$.

I have a feeling that $\displaystyle{\sum_{n=0}^{\infty}}q^n = \frac{1}{1-q}$ (for $|q|<1$) is going to make an appearance, but from there I am not sure.

As $n$ gets larger, the denominator will approach $1$ and the nominator will approach $0$. But I am not sure if that helps, since this is a sum, not a limit. I can't see how that would help me get $q$.

  • $\begingroup$ Can i sub in $\frac{1}{1-q}$ and then pull them outside the sum since they are not dependent on $n$? It would look something like $\frac{q}{1-q}\displaystyle{\sum_{n=1}^{\infty}}\mu(n)$ $\endgroup$ Feb 24, 2014 at 20:55

1 Answer 1


The answer is pretty simple: $$\sum_{n=1}^{\infty} \frac{\mu(n)q^n}{1-q^n} = \sum_{n=1}^{\infty}\sum_{t=0}^{\infty}\mu(n)q^{nt} = \sum_{n=1}^{\infty}\sum_{d|n}\mu(d)q^{n}=q$$ Caused by the simple lemma:

For $n>1$ the $$\sum_{d|n}\mu(d)=0$$ and for $n=1$ $$\sum_{d|1}\mu(d)=1$$

In general this is the particular case of Lambert series.



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