Does anyone know how to evaluate the infinite product
$$ \left(1 - \frac{4}{1}\right) \prod_{k = 3}^{\infty} \left( 1 - \frac{4}{k^2} \right) $$
$\begingroup$
$\endgroup$
0
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
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 $-1/2$.
answered Dec 22, 2013 at 23:55
-
4$\begingroup$ 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*} $\endgroup$– A.G.Dec 23, 2013 at 3:15
-
2$\begingroup$ @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. $\endgroup$ Dec 23, 2013 at 5:44
$\begingroup$
$\endgroup$
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 $$\prod_{k\neq 2}\left(1-\frac{4}{k^2}\right)=-\frac{1}{2}.$$
answered Dec 23, 2013 at 0:04
-
4
-
8$\begingroup$ [Comment defending alternative methods of evaluation and the importance of complex analysis] $\endgroup$– Pedro ♦Dec 23, 2013 at 0:34