Editing Polygamma function (section)

==Recurrence relation==
It satisfies the [[recurrence relation]]
:<math>\psi^{(m)}(z+1)= \psi^{(m)}(z) + \frac{(-1)^m\,m!}{z^{m+1}}</math>

which – considered for positive integer argument – leads to a presentation of the sum of reciprocals of the powers of the natural numbers:

:<math>\frac{\psi^{(m)}(n)}{(-1)^{m+1}\,m!} = \zeta(1+m) - \sum_{k=1}^{n-1} \frac{1}{k^{m+1}} = \sum_{k=n}^\infty \frac{1}{k^{m+1}} \qquad m \ge 1</math>

and

:<math>\psi^{(0)}(n) = -\gamma\ + \sum_{k=1}^{n-1}\frac{1}{k}</math>

for all <math>n \in \mathbb{N}</math>, where <math>\gamma</math> is the [[Euler–Mascheroni constant]]. Like the log-gamma function, the polygamma functions can be generalized from the domain {{math|[[Natural number|<math>\mathbb{N}</math>]]}} [[unique (mathematics)|unique]]ly to positive real numbers only due to their recurrence relation and one given function-value, say {{math|''ψ''<sup>(''m'')</sup>(1)}}, except in the case {{math|''m'' {{=}} 0}} where the additional condition of strict [[Monotonic function|monotonicity]] on <math>\mathbb{R}^{+}</math> is still needed. This is a trivial consequence of the [[Bohr–Mollerup theorem]] for the gamma function where strictly logarithmic convexity on <math>\mathbb{R}^{+}</math> is demanded additionally. The case {{math|''m'' {{=}} 0}} must be treated differently because {{math|''ψ''<sup>(0)</sup>}} is not normalizable at infinity (the sum of the reciprocals doesn't converge).