Sine of a product

Calculus Level 5

sin ⁡ ( ∏ n = 1 ∞ [ ( 2 n ) 2 ( 2 n − 1 ) ( 2 n + 1 ) ] ) = K \sin \left( \displaystyle \prod_{n=1}^{\infty} \left[ \dfrac{(2n)^2}{(2n-1)(2n+1)}\right] \right) = \mathfrak{K}

Find ⌊ 1 0 3 K ⌋ \left \lfloor 10^3 \mathfrak{K} \right \rfloor .

Details and assumptions:

  • The argument of sine is given in radians.


The answer is 1000.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

First Consider only the Product :

P = ∏ n = 1 ∞ ( 2 n ) 2 ( 2 n − 1 ) ( 2 n + 1 ) = ∏ n = 1 ∞ 2 n ( 2 n − 1 ) ∏ n = 1 ∞ 2 n 2 n + 1 \large \displaystyle \mathfrak{P}= \prod_{n=1}^{\infty} \frac{(2n)^2}{(2n-1)(2n+1)} = \prod_{n=1}^{\infty} \frac{2n}{(2n-1)} \prod_{n=1}^{\infty} \frac{2n}{2n+1}

The famous formulae of John Wallis's states that ,

P = π 2 \large \mathfrak{P} = \frac{\pi}{2}

K = s i n ( π 2 ) ⟹ ⌊ 1 0 3 K ⌋ = 1000 \large \mathfrak{K} = sin(\frac{\pi}{2})\implies \boxed{\lfloor 10^3\mathfrak{K} \rfloor = 1000}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

You can't split the product like that !!!

∏ n = 1 + ∞ 2 n 2 n − 1 = 0 \displaystyle\prod_{n=1}^{+\infty}\dfrac{2n}{2n-1}=0

And,

∏ n = 1 + ∞ 2 n 2 n + 1 = + ∞ \displaystyle\prod_{n=1}^{+\infty}\dfrac{2n}{2n+1}=+\infty

Proof by taking the natural logarithm and with the fact that the harmonic series diverges.

Théo Leblanc - 1 year, 10 months ago

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