# Evaluate this infinite product: $\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)$

$$\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)\;=\;\text{???}$$

I took the LCM and split the numerator as $(n+2)(n-2)$ and then took the product of the numerator and the denominator separately but I was not able to get the answer from that so can you please help me in what to do next.

Note that, by induction, we have $$F(K)=\prod _{n=3}^K \left(1-\frac{4}{n^2}\right) = \frac{(K+1) (K+2)}{6 K (K-1)}$$ for $K\geq3$.

It's them simple to see that $$\lim_{K\to\infty}F(K) = \frac{1}{6}.$$

A different solution. You can use the fact that $$\sin(\pi x)=\pi x \prod_1^\infty \left(1-\frac{x^2}{n^2} \right).$$

So your product is e qual to

$$\lim_{x \to 2} \frac{\sin(\pi x)}{\pi x (1-x^2)(1-x^2/4)}=1/6$$

$$\prod_{n=3}^{\infty} \left( 1 - \frac{4}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{(n-2)(n+2)}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{n-2}{n} \right)\prod_{n=3}^{\infty} \left( \frac{n+2}{n} \right) = \lim_{n\rightarrow \infty} \frac{2^2}{4!} \frac{(n-2)!}{n!} \frac{(n+2)!}{n!}$$

So now we pull out Stirling's Approximation

$$\frac{1}{3!} \lim_{n \rightarrow \infty} \frac{2 \pi \sqrt{n^2-4}\left( \frac{n-2}{e} \right)^{n-2}\left( \frac{n+2}{e} \right)^{n+2} }{2 \pi n \left( \frac{n}{e} \right)^{2n}}$$

Dividing out common terms we have:

$$\frac{1}{3!} \lim_{n \rightarrow \infty} \frac{\sqrt{n^2-4}\left( \frac{n-2}{e} \right)^{n-2}\left( \frac{n+2}{e} \right)^{n+2} }{n\left( \frac{n}{e} \right)^{2n}}$$

We can take a natural logarithm and then exponentiate our answer to recover the original. So now consider:

$$\lim_{n \rightarrow \infty} \frac{1}{2} \ln \left(\frac{n^2 - 4}{n} \right) + n \ln \left( \frac{(n+2)^2}{n^{n}(n-2)^2} \right)+n - \ln(3!)$$

Our problem then simplifies to:

$$\lim_{n \rightarrow \infty} n \ln \left( \frac{(n+2)^2}{(n-2)^2} \right)+n - \ln(3!) - n \ln(n)$$

Which at limit yields:

$$\lim_{n \rightarrow \infty} O(n) - O(n \ln n)$$

And so obviously tends to $-\infty$. This means that the original tends to 0.

So we conclude that the product tends to 0 (using our approximation).

An easier method.

Note that

$$\frac{1 \times 2 ... \times n-2}{3 \times 4 ... \times n} = \frac{2}{(n-1)n}$$

On the other hand we have:

$$\frac{5 \times 6 ... \times n+2}{3 \times 4 ... \times n} = \frac{(n+1)(n+2)}{3 \times 4}$$

So their combined product is

$$\frac{2(n+1)(n+2)}{12(n-1)n} = \frac{1}{6} \frac{(n+1)(n+2)}{n(n-1)}$$

Now it's easy to prove $\lim_{n\rightarrow \infty} \frac{(n+1)(n+2)}{n(n-1)} = 1$ and you get the desired result.

• But the answer is 1/6 – Gem Apr 20 '16 at 10:04
• That means the approximation was too rough. – frogeyedpeas Apr 20 '16 at 10:05
• Oh I see where it was too rough – frogeyedpeas Apr 20 '16 at 10:06
• What is Stirling approximation – Gem Apr 20 '16 at 10:07
• It's a way to approximate the factorial function. Sometime's it's useful when you want to get an asymptotic approximation of the factorial without being super accurate (such as in a limit). The problem is your multiplication meant that even the smaller parts that I was neglecting by the approximation affected the answer, and hence I made a mistake by using it. If you're curious you can read more about it here: en.wikipedia.org/wiki/Stirling%27s_approximation – frogeyedpeas Apr 20 '16 at 10:13