n-0-1-n-2-n-1-2n-1-p-q-p-q-p-q-

Question Number 215091 by mnjuly1970 last updated on 28/Dec/24

\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \frac{Γ^{2} (n + 1)}{Γ (2 n + 1)} = ?

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β (p, q) = \frac{Γ (p) Γ (q)}{Γ (p + q)}

Answered by MrGaster last updated on 28/Dec/24

Commented by mnjuly1970 last updated on 29/Dec/24

t h a n k s a l o t s i r