I've seen sums like this, and they can get quite amusing, e.g. the Fourier coefficients of $f(x)$ as a periodic function of $\log(x)$ involve values of the Gamma function at complex arguments (see below); but it seems that this is overkill for the question at hand: there are several ranges of $\mu$ for which $f(\mu) > 0$, e.g. $\mu = 1/4$ works, giving $f(1/4) = 0.0892157+ > 0$. Are you sure this is what you meant?
If I computed everything correctly (and gp corroborates numerically), the following sine-Fourier expansion holds: write $\mu = \exp(-\lambda)$ and $x = \mu^t = \exp(-\lambda t)$; then $$ f(x) = \sum_{n=1}^\infty \phantom. c_n \sin \frac{\pi n t}{2} $$ where $$ c_n = \frac1\lambda \mathop{\rm Im} \left( \Gamma\bigl(1 + \frac{\pi i n}{2\lambda}\bigr) \Bigl/ \alpha^{1 + \frac{\pi i n}{2\lambda}} \right). $$ This does not depend on the choice $\alpha = 1 - \mu^2$.
P.S. See this Mathoverflow answer where such a sum (and its Fourier expansion with complex-Gamma coefficients) arises naturally.
