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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


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.

  • $\begingroup$ The idea is fully correct. In an analysis course, you would probably be expected to write it up in the style of user17762. $\endgroup$ – André Nicolas Oct 3 '13 at 4:40
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    $\begingroup$ 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. $\endgroup$ – anon Oct 3 '13 at 4:42
  • $\begingroup$ @anon : yes, yes.. Now i understand the gap.. Thank you :) $\endgroup$ – user87543 Oct 3 '13 at 4:45
  • $\begingroup$ see mathoverflow.net/questions/27592/why-is-frac-pi212-ln2-not-true about issues with rearrangement that are pretty close to this problem $\endgroup$ – Will Jagy Oct 3 '13 at 4:50

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

  • $\begingroup$ Is it more correct to replace your last = with --> ? In other words, this sum approaches 1/2, not equals 1/2. $\endgroup$ – 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.


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 $1/2$.

  • $\begingroup$ take a look at mathoverflow.net/questions/27592/why-is-frac-pi212-ln2-not-true $\endgroup$ – Will Jagy Oct 3 '13 at 4:49
  • $\begingroup$ @Marie : could you please give some link for that euler discovered product.. $\endgroup$ – user87543 Oct 3 '13 at 4:50
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    $\begingroup$ @WillJagy Cool! :) $\endgroup$ – Bruno Joyal Oct 3 '13 at 4:52
  • $\begingroup$ @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. $\endgroup$ – Bruno Joyal Oct 3 '13 at 12:04

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

  • $\begingroup$ 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 :) $\endgroup$ – user87543 Oct 3 '13 at 4:42
  • $\begingroup$ OK. Then the general method to consider product is take log and transform to summation. But it's quite the same. $\endgroup$ – Shuchang Oct 3 '13 at 4:49
  • $\begingroup$ "Then the general method to consider product is take log and transform to summation." Not "the" general method, as exemplified by other answers on this page. $\endgroup$ – Did Jul 6 '16 at 11:19