Is $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$ [duplicate]

Question is to check if $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1$

we have $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}{3}.\frac{2}{3})(\frac{5}{4}.\frac{3}{4})...$

In above product we have for each term $\frac{a}{b}$ a term $\frac{b}{a}$ except for $\frac{1}{2}$.. So, all other terms gets cancelled and we left with $\frac{1}{2}$.

So, $\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}$.

I would be thankful if some one can assure that this explanation is correct/wrong??

I am solving this kind of problems for the first time so, it would be helpful if some one can tell if there are any other ways to do this..

Thank you

Solution 1:

One of the easiest ways to deal with infinite sums/products, is to stop at a finite value, say $N$, and look at what happens to the finite sum/product and then let $N \to \infty$. (In fact, this is the typical way infinite sums/products are to be understood/interpreted.)

Hence, in your case, let us look at \begin{align} S_N & = \prod_{n=2}^N \left(1-\dfrac1{n^2}\right) = \dfrac{1}{2}\cdot\dfrac{3}{2}\cdot \dfrac23 \cdot \dfrac43 \cdot \dfrac34 \cdot \dfrac54 \cdots \dfrac{N-2}{N-1} \cdot \dfrac{N}{N-1} \cdot \dfrac{N-1}{N} \cdot \dfrac{N+1}N\\ & = \dfrac12 \cdot \dfrac{N+1}N = \dfrac{N+1}{2N} \end{align} Now let $N \to \infty$ to conclude that $$\prod_{n=2}^{\infty} \left(1-\dfrac1{n^2}\right) = \dfrac12$$

Solution 2:

Your explanation as is is not sufficient. The part where you say that everything but $1/2$ gets cancelled needs more rigorous verification.

To demonstrate, let's "prove" that the infinite product $1\cdot1\cdot1\cdot\ldots$ equals $1/2$. Indeed, $$ 1\cdot 1 \cdot 1 \cdot \ldots = \left(\frac{1}{2} \cdot 2\right) \cdot \left(\frac{1}{2} \cdot 2\right) \cdot \left(\frac{1}{2} \cdot 2\right) \cdot \cdots. $$ The $2$ in the first factor cancels the $\frac{1}{2}$ in the second, the $2$ in the second factor cancels the $\frac{1}{2}$ in the third, and so on. Everything but the very first $\frac{1}{2}$ gets canceled, so the infinite product equals $\frac{1}{2}$.

This is clearly wrong. For a correct explanation you should look at partial products, like in user17762's answer.

Solution 3:

In fact, Euler discovered that

$$\frac{\sin \pi z}{\pi z} = \prod_{n=1}^\infty (1-z^2/n^2)$$

which we can rearrange to

$$­\frac{\sin \pi z}{\pi z (1-z^2)} = \prod_{n=2}^\infty (1-z^2/n^2).$$

By comparison of both sides at $z=1$, your product is $1/2$.

Solution 4:

Obviously, the product of any positive numbers that less than 1 is still less than one.