I am trying to express the function $ f(z) = \arccos(-z), \lvert z \lvert \leqslant 1 $ with a Maclaurin expansion using only non-negative coefficients in a compact form of $ f(z) = \sum_{n=0}^\infty a_nz^n, a_n\geqslant0 $ (1)
From trigonometric identities and standard expansion of arccos/arcsin in infinite series with binomial coefficients it seems easy as $$ \arccos(x) = \frac{\pi}{2}-\arcsin(x) = \frac{\pi}{2} - \sum_{k=0}^\infty\frac{(2k)!}{2^{2k}(k!)^2(2k+1)}x^{2k+1}$$ and substituting $x$ with $-z,$ yields $$f(z)=\frac{\pi}{2} + \sum_{k=0}^\infty\frac{(2k)!}{2^{2k}(k!)^2(2k+1)}z^{2k+1}$$ However, even if this expansion is using only non-negative coefficients is not exactly as in (1). Any hints?
As suggested i used the transformation $n=2k+1 \Rightarrow $ $$ f(z)=\frac{\pi}{2} + \sum_{n=1}^\infty\frac{(n-1)!}{2^{n-1}(\frac{(n-1)!}{2!})^2n}z^{n} = \frac{\pi}{2} + \sum_{n=1}^\infty\frac{2^3}{2^{n}(n-1)!n}z^{n} = \frac{\pi}{2} + \sum_{n=1}^\infty\frac{2^{3-n}}{n!}z^{n} $$ which is closer but not the same