A good estimation of the inverse $f^{-1}(x)$ of $-\sum_{n=2}^\infty\frac{\mu(n)}{n}x^n$, over the unit inverval if it has mathematical meaning In my home I am trying to get the thesis for a specialization for my example* from a statement published related to functions that satisfy certain  properties. 
I've considered the function*  $$f(x)=-\sum_{n=2}^\infty\frac{\mu(n)}{n}x^n,\tag{1}$$
where $\mu(n)$ denotes the Möbius function and our function $(1)$ is defined as a formal series over the real numbers of the unit interval. Using a CAS I think that my example satisfy (each of those) some of such properties, and with this purpose I wondered next question about one of main properties that I need.

Question. Is it possible to prove that $$f(x)=-\sum_{n=2}^\infty\frac{\mu(n)}{n}x^n$$ is bijective over $[0,1]$? Can you provide a good and explicit estimation (in terms of algebraic functions or elementary transcendental functions, see this Wikipedia $u(x)$ and $l(x)$) of $$l(x)\leq f^{-1}(x)\leq u(x)\tag{2}$$ over the unit interval? Many thanks.

*I know that my example of function $(1)$ was in the literature. If there was  literature about if previous function is bijective and how to estimate its inverse $f^{-1}(x)$ over the unit interval, then answer this question as a reference request for such literature and I try to find and study such statements.
 A: Remarkably $f$ is bijective only over $[0, 1-\delta]$ for a very small
but positive value of $\delta$ (about $10^{-19}$), not over
the entire unit interval $[0,1]$; indeed its derivative
$$
f'(x) = - \sum_{n=2}^\infty \mu(n) x^n,
$$
which one might expect to approach $1 - \frac1{\zeta(0)} = 3$
as $x \to 1$, instead changes sign infinitely often in $(1-\delta, 1)$,
and becomes unbounded as $x \to 1$.  This happens due to an
expansion of $f'(x)$ that contains a term
$$
-\frac{\Gamma(\rho)}{\zeta'(\rho)} (-\log x)^{-\rho}
$$
for each nontrivial zero $\rho$ of the Riemann zeta function;
as $x \to 1-$, the factor $(-\log x)^{-\rho}$ oscillates with
amplitude approaching $\infty$, but the factor $\Gamma(\rho)$
is tiny due to the exponential decay of $\Gamma(s)$ of bounded
real part and large $\left|{\rm Im}(s)\right|$, so $x$ must be
extremely close to $1$ for the oscillation to overwhelm the term
$1 - \frac1{\zeta(0)} = 3$.
The first nontrivial pair of zeros $\frac12 \pm i \gamma_1$ already has
$\gamma_1 > 14.13$ and $\left|\Gamma(\rho)\right| \sim 5.7 \cdot 10^{-10}$.
Combining the Dirichlet series
$$
\frac1{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s}
\quad({\rm Re}(s)>1)
$$
with the contour integral
$$
\frac1{2\pi i} \int_{2-i\infty}^{2+i\infty} \Gamma(s) z^{-s} \, ds = \exp(-z),
$$
we find that
$$
f'(x) = x - \frac1{2\pi i} \int_{2-i\infty}^{2+i\infty} \Gamma(s) z^{-s} \frac{ds}{\zeta(s)}
$$
where $x = \exp(-z)$.
We now move the contour to the left, which can be justified
as in the complex-analytic proof of the Prime Number Theorem.
The expected $1/\zeta(-1) = -2$ appears as the residue at the pole $s=0$;
further poles at $s = -1, -2, -3, -4, \ldots$ (simple at odd $s$,
double at even $s$ because of the trivial zeta zeros) contribute
additional terms that are $O(\left|\log x\right|) = O(1-x)$
as $x \to 1-$.  But there are also the nontrivial zeta zeros
$\rho = \beta + i\gamma$,
which contribute residues that grow as $x \to 1-$, oscillating with
amplitude proportional to $(1-x)^{-\beta}$ (which is $(1-x)^{-1/2}$
under the Riemann hypothesis).  But $x$ must get really close to $1$
for these oscillations to be visible because of the factor $\Gamma(\rho)$.
The location of the first zero of $f'(x)$ depends on the phase of this
oscillation, not just its magnitude; if I computed correctly, this happens
for $x = 1 - \delta$ with $\delta \approx 1.6960 \cdot 10^{-19}$.
