Editing Digamma function (section)

==Recurrence formula and characterization==
The digamma function satisfies the [[recurrence relation]]

:<math>\psi(x+1)=\psi(x)+\frac{1}{x}.</math>

Thus, it can be said to "telescope" {{math|{{sfrac|1|''x''}}}}, for one has

:<math>\Delta [\psi](x)=\frac{1}{x}</math>

where {{math|Δ}} is the [[forward difference operator]]. This satisfies the recurrence relation of a partial sum of the [[harmonic series (mathematics)|harmonic series]], thus implying the formula

:<math>\psi(n)=H_{n-1}-\gamma</math>

where {{mvar|γ}} is the [[Euler–Mascheroni constant]].

Actually, {{mvar|ψ}} is the only solution of the functional equation

:<math>F(x+1)=F(x)+\frac{1}{x}</math>

that is [[monotonic function|monotonic]] on {{math|'''[[Real number|R]]'''<sup>+</sup>}} and satisfies {{math|''F''(1) {{=}} −''γ''}}. This fact follows immediately from the uniqueness of the {{math|Γ}} function given its recurrence equation and convexity restriction. This implies the useful difference equation:

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