Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anyone know how to evaluate the infinite product

$$ \left(1 - \frac{4}{1}\right) \prod_{k = 3}^{\infty} \left( 1 - \frac{4}{k^2} \right) $$

share|improve this question
2  
$k=2 \to 1 - \frac{4}{k^2}=0$ –  hhsaffar Dec 22 '13 at 23:49
1  
Do you possible mean to start the index at $k = 3$? –  user61527 Dec 22 '13 at 23:50
1  
Modify the start. There is pleasant telescoping. –  André Nicolas Dec 22 '13 at 23:50
    
Oops... I mistyped the question. I have now fixed it. –  Geoff Smith Dec 22 '13 at 23:51

2 Answers 2

For $k \ge 3$:

$$\prod_{k=3}^{\infty} \left ( 1-\frac{4}{k^2}\right )$$

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 $-1/2$.

share|improve this answer
1  
Without the guesswork you could also go like this : \begin{align*} \prod_3^n (1-\frac{4}{k^2}) =&\ \prod_3^n \frac{k^2-4}{k^2}= \prod_3^n \frac{(k+2)(k-2)}{k^2}\\ =&\ \prod_3^n (k+2) \ \prod_3^n (k-2) \ \frac{1}{\prod_3^n k^2}\\ =&\ \frac{(n+2)!}{4!} \ (n-2)! \ \frac{1}{(n!/2)^2}\\ =&\ \frac{n!(n+1)(n+2)}{24} \ \frac{n!}{n(n-1)} \ \frac{4}{n!^2}= \frac{(n+1)(n+2)}{6 n (n-1)} \rightarrow 1/6 \end{align*} –  A.G. Dec 23 '13 at 3:15
1  
@A.G.: there's no guesswork - it's a matter of observing which numbers appear on top or bottom once and which ones twice (or not at all). That's it. But sure, your way does the job; I just tried to keep it as simple as possible. –  Ron Gordon Dec 23 '13 at 5:44

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 $$\prod_{k\neq 2}\left(1-\frac{4}{k^2}\right)=-\frac{1}{2}.$$

share|improve this answer
2  
[Comment about how this is overkill] –  Pedro Tamaroff Dec 23 '13 at 0:34
5  
[Comment defending alternative methods of evaluation and the importance of complex analysis] –  Pedro Tamaroff Dec 23 '13 at 0:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.