Upper bound on Trigamma function $\psi^{(1)}(x)$ for $x\in\Bbb N$ 
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 $x\in \mathbb{Z},x>0$?
 A: 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.
A: It follows from Theorem 1.3 of the paper https://royalsocietypublishing.org/doi/10.1098/rspa.2017.0363 that for any $N\geq 1$ and $x>0$,
$$
\psi ^{(1)} (x) = \zeta (2,x) = \frac{1}{x} + \frac{1}{{2x^2 }} + \sum\limits_{n = 1}^{N - 1} {\frac{{B_{2n} }}{{x^{2n + 1} }}}  + \theta _N (x)\frac{{B_{2N} }}{{x^{2N + 1} }}
$$
where $B_n$ denotes the Bernoulli numbers and $0<\theta _N (x)<1$ is a suitable number depending on $N$ and $x$. (Empy sums are interpreted as zero.) In particular,
$$
\frac{1}{x} + \frac{1}{{2x^2 }} < \psi ^{(1)} (x) < \frac{1}{x} + \frac{1}{{2x^2 }} + \frac{1}{{6x^3 }}$$
for any $x>0$.
A: If you use the series expansion
$$\psi ^{(1)}(x)=\frac{1}{x}+\frac{1}{2 x^2}+\frac{1}{6 x^3}-\frac{1}{30 x^5}+\frac{1}{42
   x^7}-\frac{1}{30 x^9}+\frac{5}{66
   x^{11}}+O\left(\frac{1}{x^{13}}\right)$$ you have simple bounds such
$$\frac{1}{x}+\frac{1}{2 x^2}+\frac{1}{6 x^3}-\frac{1}{30 x^5}<\psi ^{(1)}(x)<\frac{1}{x}+\frac{1}{2 x^2}+\frac{1}{6 x^3}$$
