Editing Polygamma function (section)

==Inequalities==
The [[hyperbolic cotangent]] satisfies the inequality
:<math>\frac{t}{2}\operatorname{coth}\frac{t}{2} \ge 1,</math>
and this implies that the function
:<math>\frac{t^m}{1 - e^{-t}} - \left(t^{m-1} + \frac{t^m}{2}\right)</math>
is non-negative for all {{math|''m'' ≥ 1}} and {{math|''t'' ≥ 0}}.  It follows that the Laplace transform of this function is completely monotone.  By the integral representation above, we conclude that
:<math>(-1)^{m+1}\psi^{(m)}(x) - \left(\frac{(m-1)!}{x^m} + \frac{m!}{2x^{m+1}}\right)</math>
is completely monotone.  The convexity inequality {{math|''e<sup>t</sup>'' ≥ 1 + ''t''}} implies that
:<math>\left(t^{m-1} + t^m\right) - \frac{t^m}{1 - e^{-t}}</math>
is non-negative for all {{math|''m'' ≥ 1}} and {{math|''t'' ≥ 0}}, so a similar Laplace transformation argument yields the complete monotonicity of
:<math>\left(\frac{(m-1)!}{x^m} + \frac{m!}{x^{m+1}}\right) - (-1)^{m+1}\psi^{(m)}(x).</math>
Therefore, for all {{math|''m'' ≥ 1}} and {{math|''x'' > 0}},
:<math>\frac{(m-1)!}{x^m} + \frac{m!}{2x^{m+1}} \le (-1)^{m+1}\psi^{(m)}(x) \le \frac{(m-1)!}{x^m} + \frac{m!}{x^{m+1}}.</math>
Since both bounds are ''strictly'' positive for <math>x>0</math>, we have:
* <math>\ln\Gamma(x)</math> is strictly [[convex function|convex]].
* For <math>m=0</math>, the digamma function, <math>\psi(x)=\psi^{(0)}(x)</math>, is strictly monotonic increasing and strictly [[concave function|concave]].
* For <math>m</math> odd, the polygamma functions, <math>\psi^{(1)},\psi^{(3)},\psi^{(5)},\ldots</math>, are strictly positive, strictly monotonic decreasing and strictly convex.
* For <math>m</math> even the polygamma functions, <math>\psi^{(2)},\psi^{(4)},\psi^{(6)},\ldots</math>, are strictly negative, strictly monotonic increasing and strictly concave. 
This can be seen in the first plot above.

===Trigamma bounds and asymptote===
For the case of the [[trigamma function]] (<math>m=1</math>) the final inequality formula above for <math>x>0</math>, can be rewritten as:
:<math>
\frac{x+\frac12}{x^2} \le \psi^{(1)}(x)\le \frac{x+1}{x^2}
</math>
so that for <math>x\gg1</math>: <math>\psi^{(1)}(x)\approx\frac1x</math>.