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Calculus Level 5

∑ n = 1 ∞ Γ ( n + 1 2 ) n 2 Γ ( n ) = A π B ln ⁡ C \sum _{ n=1 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) }{ { n }^{ 2 }\Gamma (n) } } =A\sqrt [ B ]{ \pi } \ln { C }

If the above equation holds true for positive integers A A , B B and C C , where C C is square-free, then find A B C ABC .

Notation : Γ ( ⋅ ) \Gamma(\cdot) denotes the Gamma function .


The answer is 8.

Hamza A by Hamza A , 19, USA

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1 solution

First Last
Jul 6, 2017

∑ n = 1 ∞ Γ ( n + 1 2 ) n 2 Γ ( n ) = π ∑ n = 1 ∞ ( 2 n − 1 ) ! ! n ( 2 n ) ! ! \displaystyle\sum_{n=1}^\infty\frac{\Gamma(n+\frac1{2})}{n^2\Gamma(n)}=\sqrt{\pi}\sum_{n=1}^\infty\frac{(2n-1)!!}{n(2n)!!}

Using the taylor series for arcsin ⁡ ( x ) = 1 + 1 2 x 3 3 + 1 × 3 2 × 4 x 5 5 + . . . \displaystyle\arcsin(x) = 1+\frac1{2}\frac{x^3}{3}+\frac{1\times3}{2\times4}\frac{x^5}{5}+...

1 1 − x = ∑ n = 1 ∞ ( 2 n − 1 ) ! ! ( 2 n ) ! ! x n + 1 \displaystyle\frac1{\sqrt{1-x}}=\sum_{n=1}^\infty\frac{(2n-1)!!}{(2n)!!}x^n+1

1 x 1 − x − 1 x = ∑ n = 1 ∞ ( 2 n − 1 ) ! ! ( 2 n ) ! ! x n − 1 \displaystyle\frac1{x\sqrt{1-x}}-\frac1{x}=\sum_{n=1}^\infty\frac{(2n-1)!!}{(2n)!!}x^{n-1}

π ∑ n = 1 ∞ ( 2 n − 1 ) ! ! n ( 2 n ) ! ! 1 n = π ∫ 0 1 ∑ n = 1 ∞ ( 2 n − 1 ) ! ! ( 2 n ) ! ! x n − 1 d x = π ∫ 0 1 1 x 1 − x − 1 x d x = 2 π ln ⁡ 2 \displaystyle\sqrt{\pi}\sum_{n=1}^\infty\frac{(2n-1)!!}{n(2n)!!}1^n=\sqrt{\pi}\int_0^1\sum_{n=1}^\infty\frac{(2n-1)!!}{(2n)!!}x^{n-1}dx=\sqrt{\pi}\int_0^1\frac1{x\sqrt{1-x}}-\frac1{x}dx=2\sqrt{\pi}\ln{2}

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