Let $\mu(n)$ the Möbius function, see it definition in this Wikipedia, and we define for $0<x<1$ $$f(x):=\sum_{n=1}^\infty\frac{\mu(n)}{n}x^n.\tag{1}$$
This morning I was drawing with the help of Wolfram Alpha online calculator some graphs for $$f_N(x):=\sum_{n=1}^N\frac{\mu(n)}{n}x^n$$ and their derivatives. Play with these codes for different values of $N=2000,5000,..,10000$:
plot sum mu(n)/n t^n, from n=1 to 10000, for 0<t<1
plot sum mu(n) t^(n-1), from n=1 to 5000, for 0<t<1
plot sum (n-1)mu(n) t^(n-2), from n=1 to 5000, for 0<t<1
From this experiments, and since we know the derivative test, see this Wikipedia I believe that the function $f(x)$ defined in $(1)$ for $0<x<1$ has a maximum around at $x\approx 0.6$.
Question. Is it possible to prove that $$\sum_{n=1}^\infty\frac{\mu(n)}{n}x^n,$$ defined for reals $0<x<1$, has such maximum in this interval? Many thanks.
Appendix. Similar experiments can be done with other arithmetic function, and closely related, the Liouville function
plot sum LiouvilleLambda(n)/n t^n, from n=1 to 1000, for 0<t<1
and the difference between both functions:
plot sum (mu(n)-LiouvilleLambda(n))/n t^n, from n=1 to 5000, for 0<t<1
$$f(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n}x^n, \qquad f_N(x) = \sum_{n=1}^N \frac{\mu(n)}{n}x^n$$ We have the trivial bound $|f(x)-f_N(x)| < \frac{1}{1-x^N}$
And the same for the derivatives, so the behavior on $[0,a],a < 1$ is not complicated, and we can evaluate numerically and find the maximum an minimum.
The problem is to show $\lim_{x \to 1} f(x) =0$ and to find a bound of the form $|f(x)| < \frac{A}{|\ln(1-x)|}$.
To achieve this, use the summation by parts, some bounds for $(1-x)\sum_{n=2}^\infty \frac{x^n}{\ln^k n}$ and the PNT in the form $|\sum_{n= N}^\infty \frac{\mu(n)}{n}| < \frac{C}{\ln^k N}$ (that's the complicated part).
Once this is done, you'll get that the minimum and maximum is in $[0,a]$ for some $ a < 1$ and you'll be able to evaluate it numerically.