If there are no mistakes using the inequality of the right from the so-called Steffensen's inequality, see this MathWorld or the book by Gradshteyn and Ryzhik, Tables of Integrals, Series, and Products, 6th ed. San Diego, Academic Press (2000) that is referenced, I can to write $$\sum_{n=1}^\infty\frac{\mu(n)}{n}\frac{_1F_2\left(\frac{n}{2}+\frac{1}{2};\frac{1}{2},\frac{n}{2}+\frac{3}{2};-\frac{\pi^2}{16}\right)}{n+1}\leq \frac{2}{\pi}\sin\left(\frac{\pi}{2}\sum_{n=1}^\infty\frac{\mu(n)}{n(n+1)}\right),\tag{1}$$ where $\mu(n)$ is the Möbius function.
Question. Can you prove that $(1)$ is true (without using the mentioned inequality)? Many thanks.
These are some codes that I've computed with Wolfram Alpha online calculator
plot sum mu(n)/n x^n, from n=1 to 1000, for 0<x<1
int cos(pi/2 x)x^n dx, from x=0 to 1
and all issues works thanks to the uniform convergence.
The motivation of this question is try to combine functions defined involving the Möbius function with integral inequalities as previous (or well with the Stolarsky's Inequality) with the purpose to state some interesting facts about the Möbius function or its means, if it is feasible.