Editing Digamma function (section)

==Asymptotic expansion==
The digamma function has the asymptotic expansion
:<math>\psi(z) \sim \ln z + \sum_{n=1}^\infty \frac{\zeta(1-n)}{z^n} = \ln z - \sum_{n=1}^\infty \frac{B_n}{nz^n},</math>
where {{mvar|''B''<sub>''k''</sub>}} is the {{mvar|''k''}}th [[Bernoulli number]] and {{mvar|ζ}} is the [[Riemann zeta function]].  The first few terms of this expansion are:
:<math>\psi(z) \sim \ln z - \frac{1}{2z} - \frac{1}{12z^2} + \frac{1}{120z^4} - \frac{1}{252z^6} + \frac{1}{240z^8} - \frac{1}{132z^{10}} + \frac{691}{32760z^{12}} - \frac{1}{12z^{14}} + \cdots.</math>
Although the infinite sum does not converge for any {{mvar|''z''}}, any finite partial sum becomes increasingly accurate as {{mvar|''z''}} increases.

The expansion can be found by applying the [[Euler–Maclaurin formula]] to the sum<ref>{{cite journal| url=http://www.uv.es/~bernardo/1976AppStatist.pdf|first1=José M.|last1= Bernardo|title= Algorithm AS 103 psi(digamma function) computation|year=1976|journal=Applied Statistics|volume=25|pages=315–317|doi=10.2307/2347257|jstor=2347257}}</ref>
:<math>\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{z + n}\right)</math>
The expansion can also be derived from the integral representation coming from Binet's second integral formula for the gamma function.  Expanding <math>t / (t^2 + z^2)</math> as a [[geometric series]] and substituting an integral representation of the Bernoulli numbers leads to the same asymptotic series as above.  Furthermore, expanding only finitely many terms of the series gives a formula with an explicit error term:
:<math>\psi(z) = \ln z - \frac{1}{2z} - \sum_{n=1}^N \frac{B_{2n}}{2nz^{2n}} + (-1)^{N+1}\frac{2}{z^{2N}} \int_0^\infty \frac{t^{2N+1}\,dt}{(t^2 + z^2)(e^{2\pi t} - 1)}.</math>