Question-189323

Question Number 189323 by mnjuly1970 last updated on 14/Mar/23

Answered by witcher3 last updated on 14/Mar/23

\Leftrightarrow Σ \frac{Γ (n) Γ (n)}{Γ (2 n) . 2^{- n}} = \sum_{n ⩾ 1} \frac{β (n, n)}{2^{- n}}

= \sum_{n ⩾ 1} \int_{0}^{1} \frac{x^{n - 1} {(1 - x)}^{n - 1}}{2^{- n}}

= 2 \int_{0}^{1} \sum_{n ⩾ 0} {(2 x (1 - x))}^{n}

2 x (1 - x) < \frac{1}{2}

= 2 \int_{0}^{1} \frac{1}{1 - 2 x (1 - x)}

= 2 \int_{0}^{1} \frac{dx}{{2 x}^{2} - 2 x + 1}

= 2 \int_{0}^{1} \frac{dx}{2 ((x - \frac{1}{2}) + \frac{1}{4})} = 4 \int_{0}^{1} \frac{dx}{{(2 x - 1)}^{2} + 1}

= {4 \tan}^{- 1} (1) = π

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