# Infinite sum of mobius function

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 $q$.

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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 $\frac{q}{1-q}\displaystyle{\sum_{n=1}^{\infty}}\mu(n)$ – Skuttle_Butt Feb 24 '14 at 20:55