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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

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marked as duplicate by Martin Sleziak, Did real-analysis May 16 '15 at 9:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The idea is fully correct. In an analysis course, you would probably be expected to write it up in the style of user17762. – André Nicolas Oct 3 '13 at 4:40
To add to Andre's comment: formally it might deserve more rigor. In general you are not allowed to regroup terms in an infinite series or infinite product $-$ you must reason on the level of partial sums or partial products. – anon Oct 3 '13 at 4:42
@anon : yes, yes.. Now i understand the gap.. Thank you :) – Praphulla Koushik Oct 3 '13 at 4:45
see about issues with rearrangement that are pretty close to this problem – Will Jagy Oct 3 '13 at 4:50
up vote 14 down vote accepted

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 $$\prod_{n=2}^{\infty} \left(1-\dfrac1{n^2}\right) = \dfrac12$$

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Is it more correct to replace your last = with --> ? In other words, this sum approaches 1/2, not equals 1/2. – Matt Cremeens Aug 28 '15 at 16:03

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.

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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 $1/2$.

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@Marie : could you please give some link for that euler discovered product.. – Praphulla Koushik Oct 3 '13 at 4:50
@WillJagy Cool! :) – Bruno Joyal Oct 3 '13 at 4:52
@PraphullaKoushik Sorry I just saw your comment. I don't have a specific source but googling 'euler sine product' turns up a lot of results. – Bruno Joyal Oct 3 '13 at 12:04

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

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This would be easiest way to check if it is correct or not, but i even wanted to know what is its exact value.. This comment is useful though :) – Praphulla Koushik Oct 3 '13 at 4:42
OK. Then the general method to consider product is take log and transform to summation. But it's quite the same. – Shuchang Oct 3 '13 at 4:49

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