Difference between revisions of "Value of derivative of trigamma at positive integer plus 1/2"

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==Theorem==
 
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$\psi^{\prime} \left( n + \dfrac{1}{2} \right)=\dfrac{\pi^2}{2} - 4 \displaystyle\sum_{k=1}^n \dfrac{1}{(2k-1)^2},$$
+
$$\psi^{(1)} \left( n + \dfrac{1}{2} \right)=\dfrac{\pi^2}{2} - 4 \displaystyle\sum_{k=1}^n \dfrac{1}{(2k-1)^2},$$
where $\psi^{(m)}$ denotes the [[digamma]] and $\pi$ denotes [[pi]].
+
where $\psi^{(1)}$ denotes the [[trigamma]] and $\pi$ denotes [[pi]].
  
 
==Proof==
 
==Proof==
  
 
==References==
 
==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Value of polygamma at 1/2|next=Polygamma recurrence relation}}: 6.4.5
+
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Value of polygamma at 1/2|next=Polygamma recurrence relation}}: $6.4.5$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 22:45, 17 March 2017

Theorem

The following formula holds: $$\psi^{(1)} \left( n + \dfrac{1}{2} \right)=\dfrac{\pi^2}{2} - 4 \displaystyle\sum_{k=1}^n \dfrac{1}{(2k-1)^2},$$ where $\psi^{(1)}$ denotes the trigamma and $\pi$ denotes pi.

Proof

References