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Suppose $x > \frac{1}{2}$. Let $\psi^{(0)}$ and $\psi^{(1)}$ denote the digamma and trigamma functions, respectively. Does the following inequality hold for any such $x$?

$\psi^{(0)}(x) - \psi^{(0)}\left(x + \frac{1}{2} \right) + x \left( \psi^{(1)}(x) - \psi^{(1)}\left(x + \frac{1}{2}\right) \right) > 0$

Graphing it suggests that it does, but I don't know how to prove it. Thanks in advance!

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  • $\begingroup$ I find that $f_n(x) = \frac{1}{x+n}-\frac{1}{x+1/2+n}+x(\frac{-1}{(x+n)^2}-\frac{-1}{(x+1/2+n)^2}) = \frac{1/2(x+n)(x+1/2+n)- x/4-x(x+n)}{(x+n)^2(x+1/2+n)^2}$ and your LHS is $-\sum_{n=0}^\infty f_n(x)$ $\endgroup$ – reuns Oct 14 '16 at 8:43
  • $\begingroup$ "Functional analysis" is a completely different thing (operators on normed vector spaces). And do you know the partial fraction decomposition of $\psi(z)$ ? $\endgroup$ – reuns Oct 14 '16 at 10:22
  • $\begingroup$ My apologies! And I didn’t (thank you!). I can understand how you arrived at the above. It transforms the problem to one that requires $\sum_{n=0}^{\infty} \frac{n(2n+1)-2x^2}{(x+n)^2(2x+2n+1)^2} > 0$ (I believe). I will have a go at working with this form. $\endgroup$ – KMBR Oct 14 '16 at 10:46
  • $\begingroup$ For showing $\frac{\Gamma'(s)}{\Gamma(s)} = -\gamma+\sum_{n \ge 0} \frac{1}{n+1}-\frac{1}{s+n}$ you need to say that the poles of $\Gamma(s)$ are of order $1$ at $s = -n$, then show it has no zeros, so that $\frac{\Gamma'(s)}{\Gamma(s)}-\sum_{n \ge 0} \frac{1}{n+1}-\frac{1}{s+n}$ is analytic on the whole complex plane (entire), and then prove that it is bounded, so by Liouville's theorem it is constant $\endgroup$ – reuns Oct 14 '16 at 11:34

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