# An Equation Involving the Trigamma Function

Let $$N$$ be a positive integer, and consider the equation

$$$$\frac{1}{2} \sum_{n=1}^{N} \psi^{(1)} \left( \frac{x+1-n}{2} \right) = \frac{N}{x},$$$$ in the real unkown $$x > N - 1$$, where $$\psi^{(1)}(x)$$ is the trigamma function.

I have conjectured that this equation has no solution. Does someone have an idea of a possible proof?

Any help is welcome.

NB From the series representation of the trigamma function we know that the map $$(0,\infty) \ni x \mapsto \psi^{(1)}(x)$$ is strictly decreasing. So our statement is proved if we can prove the inequality $$$$\psi^{(1)}(x) > \frac{1}{x} \quad (x > 0),$$$$ but I do not know how to prove it for now.

• The last inequality is just $$\sum_{k = 0}^\infty \frac 1 {(k + x)^2} > \int_0^\infty \frac {dk} {(k + x)^2}.$$ – Maxim May 28 at 13:04
• @Maxim Wonderful one-life proof!!!! Thank you very very very ... much, Maxim for your help. I would never be able to find it by myself! – Maurizio Barbato May 28 at 15:25

$$$$\psi^{(1)}(x) > \frac{1}{x} + \frac{1}{2x^2} \quad (x > 0).$$$$ For a proof, see e.g. Elbert and Laforgia On Some Properties of the Gamma Function, Section 2; Gordon, A Stochastic Approach to the Gamma Function, Theorem 4; Alzer, On Some Inequalities for the Gamma Function, Theorem 9.