find-the-sum-0-2-1-1-2-3-2-2-5-

Question Number 143846 by mathmax by abdo last updated on 18/Jun/21

find the sum \frac{{(0!)}^{2}}{1!} + \frac{{(1!)}^{2}}{3!} + \frac{{(2!)}^{2}}{5!} + \dots . .

Answered by Ar Brandon last updated on 18/Jun/21

\frac{(n!) (n!)}{(n + n + 1)!} = \frac{Γ (n + 1) Γ (n + 1)}{Γ (2 n + 2)} = β (n + 1, n + 1)

S = \frac{{(0!)}^{2}}{1!} + \frac{{(1!)}^{2}}{3!} + \frac{{(2!)}^{2}}{5!} + \cdot \cdot \cdot = \underset{n = 0}{\sum^{\infty}} β (n + 1, n + 1)

= \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} x^{n} {(1 - x)}^{n} = \int_{0}^{1} \frac{1}{1 - (x - x^{2})} dx

= \int_{0}^{1} \frac{dx}{x^{2} - x + 1} = \int_{0}^{1} \frac{dx}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}

= {[\frac{2}{\sqrt{3}} \arctan (\frac{2 x - 1}{\sqrt{3}})]}_{0}^{1} = \frac{2}{\sqrt{3}} (\frac{π}{6} + \frac{π}{6}) = \frac{2 \sqrt{3} π}{9}

Commented by mathmax by abdo last updated on 19/Jun/21

thank you sir

Commented by Ar Brandon last updated on 19/Jun/21

Your welcome, Sir .