Product Thing

Calculus Level 3

∏ n = 1 ∞ ( 4 n − 3 ) ( 4 π 2 n 2 − 1 ) ( 4 n + 1 ) ( π 2 ( 1 − 2 n ) 2 − 1 ) = a b π tan ⁡ ( c d ) \large \prod _{n=1}^{\infty } \frac{(4 n-3) \left(4 \pi ^2 n^2-1\right)}{(4 n+1) \left(\pi ^2 (1-2 n)^2-1\right)} = \frac{a}{b} \pi \tan \left(\frac{c}{d}\right)

where a a , b b , c c , and d d are positive integers, with gcd ⁡ ( a , b ) = gcd ⁡ ( c , d ) = 1 \gcd(a,b) = \gcd(c,d) = 1 . Submit a + b + c + d a+b+c+d .


The answer is 6.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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2 solutions

Chew-Seong Cheong
Nov 28, 2017

Relevant wiki: Stirling's Formula

P = lim ⁡ m → ∞ ∏ n = 1 m ( 4 n − 3 ) ( 4 π 2 n 2 − 1 ) ( 4 n + 1 ) ( π 2 ( 1 − 2 n ) 2 − 1 ) = lim ⁡ m → ∞ ∏ n = 1 m 4 n − 3 4 n + 1 ∏ n = 1 m 4 π 2 n 2 ( 1 − 1 4 π 2 n 2 ) π 2 ( 2 n − 1 ) 2 ( 1 − 1 π 2 ( 2 n − 1 ) 2 ) = lim ⁡ m → ∞ 1 ⋅ 5 ⋅ 9 ⋯ ( 4 m − 3 ) 5 ⋅ 9 ⋯ ( 4 m − 3 ) ( 4 m + 1 ) ⋅ 4 m ( m ! ) 2 ( ( 2 m − 1 ) ! ! ) 2 ∏ n = 1 m ( 1 − 1 4 n 2 π 2 ) ( 1 − 1 ( 2 n ) 2 π 2 ) ( 1 − 1 ( 2 n − 1 ) 2 π 2 ) ( 1 − 1 ( 2 n ) 2 π 2 ) = lim ⁡ m → ∞ 1 4 m + 1 ⋅ 4 m ( m ! ) 2 ( 2 m m ! ) 2 ( ( 2 m ) ! ) 2 ∏ n = 1 m ( 1 − 1 4 n 2 π 2 ) 2 ( 1 − 1 n 2 π 2 ) Using identity sin ⁡ x x = ∏ k = 1 ∞ ( 1 − x 2 k 2 π 2 ) = lim ⁡ m → ∞ 1 4 m + 1 ⋅ 4 2 m ( m ! ) 4 ( ( 2 m ) ! ) 2 ⋅ ( 2 sin ⁡ 1 2 ) 2 sin ⁡ 1 By Stirling’s formula n ! ∼ 2 π n ( n e ) n = lim ⁡ m → ∞ 1 4 m + 1 ⋅ 4 2 m ( 2 π m m m e − m ) 4 ( 4 π m ( 2 m ) 2 m e − 2 m ) 2 ⋅ 4 sin ⁡ 2 1 2 2 sin ⁡ 1 2 cos ⁡ 1 2 = lim ⁡ m → ∞ 1 4 m + 1 ⋅ 4 2 m 4 π 2 4 2 m m 4 m + 2 e − 4 m 4 π 4 2 m m 4 m + 1 e − 4 m ⋅ 2 sin ⁡ 1 2 cos ⁡ 1 2 = lim ⁡ m → ∞ 1 4 m + 1 ⋅ m π ⋅ 2 tan ⁡ 1 2 = 1 2 π tan ⁡ 1 2 \begin{aligned} P & = \lim_{m \to \infty} \prod_{n=1}^m \frac {(4n-3)(4\pi^2n^2-1)}{(4n+1)(\pi^2(1-2n)^2-1)} \\ & = \lim_{m \to \infty} \prod_{n=1}^m \frac {4n-3}{4n+1} \prod_{n=1}^m \frac {4\pi^2n^2 \left(1-\frac 1{4\pi^2n^2}\right)}{\pi^2(2n-1)^2 \left(1-\frac 1{\pi^2(2n-1)^2}\right)} \\ & = \lim_{m \to \infty} \frac {1\cdot 5 \cdot 9 \cdots (4m-3)}{5 \cdot 9 \cdots (4m-3)(4m+1)} \cdot \frac {4^m(m!)^2}{((2m-1)!!)^2} \prod_{n=1}^m \frac {\left(1-\frac 1{4n^2\pi^2}\right)\left(1-\frac 1{(2n)^2\pi^2}\right)}{\left(1-\frac 1{(2n-1)^2\pi^2}\right)\left(1-\frac 1{(2n)^2\pi^2}\right)} \\ & = \lim_{m \to \infty} \frac 1{4m+1} \cdot \frac {4^m(m!)^2(2^mm!)^2}{((2m)!)^2} \prod_{n=1}^m \color{#3D99F6} \frac {\left(1-\frac 1{4n^2\pi^2}\right)^2}{\left(1-\frac 1{n^2\pi^2}\right)} & \small \color{#3D99F6} \text{Using identity } \frac {\sin x}x = \prod_{k=1}^\infty \left(1-\frac {x^2}{k^2\pi^2}\right) \\ & = \lim_{m \to \infty} \frac 1{4m+1} \cdot \frac {4^{2m}({\color{#D61F06}m!})^4}{({\color{#D61F06}(2m)!})^2} \cdot \color{#3D99F6} \frac {\left(2\sin \frac 12\right)^2}{\sin 1} & \small \color{#D61F06} \text{By Stirling's formula } n! \sim \sqrt{2\pi n}\left(\frac ne\right)^n \\ & = \lim_{m \to \infty} \frac 1{4m+1} \cdot \frac {4^{2m}({\color{#D61F06}\sqrt{2\pi m}m^m e^{-m}})^4}{({\color{#D61F06}\sqrt{4\pi m}(2m)^{2m} e^{-2m}})^2} \cdot \frac {4 \sin^2 \frac 12}{2 \sin \frac 12 \cos \frac 12} \\ & = \lim_{m \to \infty} \frac 1{4m+1} \cdot \frac {4^{2m}4\pi^2 4^{2m}m^{4m+2} e^{-4m}}{4\pi 4^{2m} m^{4m+1} e^{-4m}} \cdot \frac {2 \sin \frac 12}{\cos \frac 12} \\ & = \lim_{m \to \infty} \frac 1{4m+1} \cdot m\pi \cdot 2 \tan \frac 12 \\ & = \frac 12 \pi \tan \frac 12 \end{aligned}

⟹ a + b + c + d = 1 + 2 + 1 + 2 = 6 \implies a+b+c+d = 1+2+1+2 = \boxed{6}

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Here's a hint/partial solution:

∑ n = 1 ∞ ( ( − 1 ) n log ⁡ ( x 2 − π 2 n 2 ) − ( − 1 ) n log ⁡ ( π 2 4 − π 2 n 2 ) ) + log ⁡ ( 2 x π ) = ∑ n = 1 ∞ ∫ π 2 x 2 ( − 1 ) n z z 2 − ( π n ) 2 d z + ∫ π 2 x 1 z d z = ∫ π 2 x ( 2 z ∑ n = 1 ∞ ( − 1 ) n z 2 − ( π n ) 2 + 1 z ) d z = ∫ π 2 x 1 sin ⁡ ( z ) d z = log ⁡ ( sin ⁡ ( x 2 ) ) − log ⁡ ( cos ⁡ ( x 2 ) ) \displaystyle\sum _{n=1}^{\infty } \left((-1)^n \log \left(x^2-\pi ^2 n^2\right)-(-1)^n \log \left(\frac{\pi ^2}{4}-\pi ^2 n^2\right)\right)+\log \left(\frac{2 x}{\pi }\right)=\displaystyle\sum _{n=1}^{\infty } \int_{\frac{\pi }{2}}^x \frac{2 (-1)^n z}{z^2-(\pi n)^2} \, dz+\int_{\frac{\pi }{2}}^x \frac{1}{z} \, dz=\displaystyle\int_{\frac{\pi }{2}}^x \left(2 z \displaystyle\sum _{n=1}^{\infty } \frac{(-1)^n}{z^2-(\pi n)^2}+\frac{1}{z}\right) \, dz=\displaystyle\int_{\frac{\pi }{2}}^x \frac{1}{\sin (z)} \, dz=\log \left(\sin \left(\frac{x}{2}\right)\right)-\log \left(\cos \left(\frac{x}{2}\right)\right)

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