Difference between revisions of "Polygamma series representation"
From specialfunctionswiki
(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\psi^{(m)}(z)=(-1)^{m+1} m! \...") |
|||
Line 1: | Line 1: | ||
− | + | ==Theorem== | |
− | + | The following formula holds: | |
$$\psi^{(m)}(z)=(-1)^{m+1} m! \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{(z+k)^{m+1}},$$ | $$\psi^{(m)}(z)=(-1)^{m+1} m! \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{(z+k)^{m+1}},$$ | ||
where $\psi^{(m)}$ denotes the [[polygamma]] and $m!$ denotes the [[factorial]]. | where $\psi^{(m)}$ denotes the [[polygamma]] and $m!$ denotes the [[factorial]]. | ||
− | + | ||
− | + | ==Proof== | |
− | + | ||
− | + | ==References== | |
+ | |||
+ | [[Category:Theorem]] | ||
+ | [[Category:Unproven]] |
Latest revision as of 06:34, 11 June 2016
Theorem
The following formula holds: $$\psi^{(m)}(z)=(-1)^{m+1} m! \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{(z+k)^{m+1}},$$ where $\psi^{(m)}$ denotes the polygamma and $m!$ denotes the factorial.