Difference between revisions of "Digamma at z+n"
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(Created page with "==Theorem== The following formula holds for $n=1,2,3,\ldots$: $$\psi(z+n)=\dfrac{1}{z} + \dfrac{1}{z+1} + \ldots + \dfrac{1}{z+n-1} + \psi(z),$$ where $\psi$ denotes the dig...") |
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Revision as of 15:56, 23 June 2016
Theorem
The following formula holds for $n=1,2,3,\ldots$: $$\psi(z+n)=\dfrac{1}{z} + \dfrac{1}{z+1} + \ldots + \dfrac{1}{z+n-1} + \psi(z),$$ where $\psi$ denotes the digamma function.
Proof
References
- 1953: Harry Bateman: Higher Transcendental Functions Volume I ... (previous): $\S 1.7 (10)$
