Upper bound on Trigamma function $\psi^{(1)}(x)$ for $x\in\Bbb N$

Question

The Trigamma function is defined as $\psi^{(1)}(x):=\frac{\mathrm d}{\mathrm dx^2} \ln (\Gamma(x))$, where $\Gamma(x)$ is the Gamma function. I am trying to find an upper bound for this function where $x\in \mathbb{Z}, x>0$.

I'm looking at Wiki on Digamma function and see that $\psi^{(0)}(x)=\frac{d}{dx} \ln (\Gamma(x))= \frac{\Gamma'(x)}{\Gamma(x)}\sim \ln x-\frac{1}{2x} $. If I go with the approximation of Digamma function, then $\psi^{(1)}(x)=\frac{d}{dx^2} \ln (\Gamma(x)) \sim \frac{1}{x}+\frac{1}{2x^2}<\frac{1}{x}+\frac{1}{x^2}$. But I'm wondering if there is a known upper bound or if we can show that $\psi^{(1)}(x)<\frac{1}{x}+\frac{1}{x^2}$ for all $x\in \mathbb{Z},x>0$?

Answer 1

Indeed, you are looking on the wrong wiki page @manifolded. Moving over the the wiki page for the polygamma function, $\psi^{(m)}(x):=\partial_x^{m+1}\log\Gamma(x)$, we find the following inequality:

For all $m\geq 1$ and $x > 0$,

$$ \frac{(m-1)!}{x^m}+\frac{m!}{2x^{m+1}}\leq(-1)^{m+1}\psi^{(m)}(x)\leq\frac{(m-1)!}{x^m}+\frac{m!}{x^{m+1}}. $$

You can find on this page details about the proof of this inequality too. For the specific case of the trigamma function, $m=1$; hence, $$ \frac{1}{x}+\frac{1}{2x^2}\leq\psi^{(1)}(x)\leq\frac{1}{x}+\frac{1}{x^2}, $$ which is precisely the bounds you derived! Note that these bounds hold for all $x>0$ so that the additional requirement $x\in\Bbb Z$ is automatically satisfied.