Does anyone know how to evaluate the infinite product
$$ \left(1 - \frac{4}{1}\right) \prod_{k = 3}^{\infty} \left( 1 - \frac{4}{k^2} \right) $$
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Sign up to join this communityFor $k \ge 3$:
$$\prod_{k=3}^{\infty} \left ( 1-\frac{4}{k^2}\right )$$
This is equal to
$$\frac{1\cdot 5}{3 \cdot 3} \frac{2\cdot 6}{4 \cdot 4} \frac{3\cdot 7}{5 \cdot 5} \cdots$$
With cancellations: note that only $1\cdot 2$ survives in the numerator, and a single $3 \cdot 4$ survives in the denominator. Thus, the product is $1/6$. The front factor produces a $-3$, so the stated product is $-1/2$.
Here is an alternative proof that does not use telescoping. We have that $$\frac{\sin(\pi z)}{\pi z}=\prod_{k=1}^\infty \left(1-\frac{z^2}{k^2}\right).$$ This can be proven by using the Weierstrass product for the Gamma function combined with Euler's reflection formula. Dividing both sides by $1-\frac{z^2}{4}$, we see that $$\prod_{k\neq 2}\left(1-\frac{4}{k^2}\right)=\lim_{z\rightarrow 2} \frac{4\sin(\pi z)}{\pi z(2-z)(2+z)}.$$ Taylor expanding $\sin(\pi z)$ around $z=2$, we are able to conclude that $$\prod_{k\neq 2}\left(1-\frac{4}{k^2}\right)=-\frac{1}{2}.$$