In search of π π

Calculus Level 4

1 + 1 2 ( 1 2 ⋅ 1 ! ) 2 + 1 3 ( 1 ⋅ 3 2 2 ⋅ 2 ! ) 2 + 1 4 ( 1 ⋅ 3 ⋅ 5 2 3 ⋅ 3 ! ) 2 + ⋯ = a b π 1+\frac{1}{2} \left(\frac{1}{2 \cdot 1!}\right)^2 +\frac{1}{3} \left(\frac{1 \cdot 3}{2^2 \cdot 2!}\right)^2 + \frac{1}{4} \left(\frac{1 \cdot 3 \cdot 5}{2^3 \cdot 3!}\right)^2 +\cdots=\frac{a}{bπ}

The equation above holds for positive coprime integers a a and b b . Find a − 4 b a-4b .

The problem is original

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The answer is 0.

Dwaipayan Shikari by Dwaipayan Shikari , 17, India

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1 solution

The series is ∑ n ≥ 0 ∞ ( 1 2 ) n 2 ( 2 ) n n ! = 2 F 1 ( 1 2 , 1 2 ; 2 ; 1 ) \large \sum_{n≥0}^∞ \frac{(\frac{1}{2})^2_n }{(2)_n n!} = _2F_1 (\frac{1}{2},\frac{1}{2};2;1)

Where ( a ) n = ∏ k ≥ 0 n − 1 ( a + k ) (a)_n = \prod_{k≥0}^{n-1} (a+k)

Now 2 F 1 ( a , b ; c ; 1 ) = Γ ( c ) Γ ( c − a − b ) Γ ( c − a ) Γ ( c − b ) _2F_1 (a,b;c;1) = \frac{\Gamma{(c)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}} In this case it is 2 F 1 ( 1 2 , 1 2 ; 2 ; 1 ) = 1 Γ 2 ( 3 2 ) = 4 π _2F_1 (\frac{1}{2},\frac{1}{2};2;1)= \frac{1}{\Gamma^2 {(\frac{3}{2})}}= \frac{4}{π} Answer is a − 4 b = 0 \color{#20A900}{\boxed{a-4b=0}}

2 F 1 ( a , b ; c ; z ) = ∑ n ≥ 0 ∞ ( a ) n ( b ) n ( c ) n n ! z n _2F_1 (a ,b;c;z) = \sum_{n≥0}^∞ \frac{(a)_n (b)_n}{(c)_n n!}z^n Gaussian hypergeometric function

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