Is the reciprocal of polygamma functions of odd order convex on $ \mathbb{R^+}$, while that of even order above 0 concave? Plotting the functions suggest so, but I've been trying for days to come up with a proof to no avail. The 2nd derivative of the reciprocal is:
$$ \frac{d ^2\psi ^{(m)}(z)^{-1}}{d z^2}=\frac{2 \psi ^{(m+1)}(z)^2-\psi ^{(m)}(z) \psi ^{(m+2)}(z)}{\psi ^{(m)}(z)^3} $$
I can't show the numerator is always positive for $z>0$. Curiously, $\psi ^{(m+1)}(z)^2-\psi ^{(m)}(z) \psi ^{(m+2)}(z)<0$, which I can show. Positivity of the numerator is equivalent to
$$ \frac{(m+2)}{(m+1)}\frac{ \zeta (m+1,z) \zeta (m+3,z)}{ \zeta (m+2,z)^2}<2 $$
so showing this would suffice. Alternatively, since the $z\rightarrow0$ and $z\rightarrow\infty$ limits are not negative, showing the numerator is monotonic also works.