Show-that-n-1-n-2-2n-1-3-2pi-3-27-

Question Number 197935 by Frix last updated on 04/Oct/23

Show that

\underset{n = 1}{\sum^{\infty}} \frac{{(n!)}^{2}}{(2 n)!} = \frac{1}{3} + \frac{2 π \sqrt{3}}{27}

Answered by witcher3 last updated on 05/Oct/23

\sum_{n ⩾ 1} \frac{Γ (n + 1) Γ (n) . n}{Γ (2 n + 1)}

= \sum_{n ⩾ 1} n β (n, n + 1) = \sum_{n ⩾ 1} \int_{0}^{1} {nt}^{n} {(1 - t)}^{n - 1} dt

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

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

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

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

f^{'} (a) = \int \frac{- dx}{{(x^{2} + a)}^{2}} \dots

f (a) = \frac{2}{\sqrt{a}} \tan^{- 1} (\frac{1}{2 \sqrt{a}})

f^{'} (a) = - {(a)}^{- \frac{3}{2}} \tan^{- 1} (\frac{1}{2 \sqrt{a}}) + \frac{2}{\sqrt{a}} . (- \frac{1}{4} a^{- \frac{3}{2}} . \frac{4 a}{4 a + 1} .)

A = - \frac{1}{2} . f^{'} (\frac{3}{4})

= \frac{1}{2} (\frac{8}{3 \sqrt{3}} . \frac{π}{6}) + \frac{2}{\sqrt{3}} (\frac{1}{4} . \frac{8}{3 \sqrt{3}} . \frac{3}{4})

= \frac{2 π}{9 \sqrt{3}} + \frac{1}{3} = \frac{2 π \sqrt{3}}{27} + \frac{1}{3}

\Leftrightarrow \sum_{n ⩾ 1} \frac{{(n!)}^{2}}{2 n} = \frac{2 π \sqrt{3}}{27} + \frac{1}{3}

Commented by witcher3 last updated on 05/Oct/23

y^{'} re Welcom

Commented by Frix last updated on 05/Oct/23

Thank you!

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