While trying some problems along with my friends we had difficulty in this question.

  • True or False: The value of the infinite product $$\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)$$ is $1$.

I couldn't do it and my friend justified it by saying that since the terms in the product have values less than $1$, so the value of the product can never be $1$. I don't know whether this justification is correct or not. But i referred to Tom Apostol's Mathematical Analysis book and found a theorem which states, that

  • The infinite product $\prod(1-a_{n})$ converges if the series $\sum a_{n}$ converges.

This assures that the above product converges. Could anyone help me in finding out where it converges to? And,

  • Does there exist a function $f$ in $\mathbb{N}$ ( like $n^{2}$, $n^{3}$) such that $\displaystyle \prod\limits_{n=1}^{\infty} \Bigl(1-\frac{1}{f(n)}\Bigr)$ has the value $1$?

We have \begin{align*} p_{k} &= \prod_{n = 2}^{k} \left( 1 - \frac{1}{n^{2}} \right) = \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^{2}} \\ & = \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2\cdot 4}{3 \cdot 3} \cdot \frac{3\cdot 5}{4 \cdot 4} \cdot \cdots \cdot \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot \frac{(k-1)(k+1)}{k\cdot k}\\ & = \frac{1}{2}\left(1 + \frac{1}{k}\right) \end{align*} because all but the first and last numerators and denominators cancel. Therefore \[ \prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) = \lim_{k\to\infty} p_{k} = \frac{1}{2}. \]

To put this into a little context: Euler has shown that \[ \sin{(\pi z)} = \pi z \prod_{n=1}^{\infty} \left( 1 - \frac{z^{2}}{n^{2}} \right) \] for all $z \in \mathbb{C}$.

We would like to plug in $z = 1$ on both sides. This doesn't work because we get $0 = 0$. But a simple trick yields what we want: \[ \pi \prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) = \lim_{z \to 1} \frac{\sin{(\pi z)}}{(1- z^2)} = \lim_{z \to 1} \frac{\pi \cos{(\pi z)}}{-2z} = \frac{\pi}{2} \] and cancelling on both sides with $\pi$ gives $\frac{1}{2}$ for the infinite product.

There is a lot of fun that you can have with Euler's product. For instance $z = \frac{1}{2}$ yields the Wallis product formula \[ \frac{\pi}{2} = \frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdot \cdots = \prod_{n=1}^{\infty} \frac{2n}{2n -1}\frac{2n}{2n +1} \] and $z = i$ yields \[ \frac{\sin{(i\pi)}}{i \pi} = \frac{e^{\pi} - e^{-\pi}}{2\pi} = \prod_{n = 1}^{\infty} \left( 1 + \frac{1}{n^{2}} \right). \] A thorough treatment of these and many other topics involving infinite products can be found in chapters 1 and 2 of Remmert, Classical topics in complex function theory, Springer GTM 172, 1998. The title of the German original is simply Funktionentheorie 2.

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    $\begingroup$ +1: I was thinking of adding an answer with the product formula of sin :-) $\endgroup$ – Aryabhata Jan 19 '11 at 19:27
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    $\begingroup$ @Moron: Yeah, it would have been a pity if nobody had mentioned it. $\endgroup$ – t.b. Jan 19 '11 at 19:40

The product can never be 1. Recall that the product is defined as the limit of the partial products $$ \prod_{n\le k}\left(1-\frac1{f(n)}\right)=A_k. $$ Now, if $f(n)=1$ then the product is 0. I assume you know how to handle "divergence to 0"? To simplify, let's assume $f(n)\ne 1$ for all $n$.

Then we have that $A_k>A_{k+1}$ for all $k$, since $1-\epsilon<1$ for any positive $\epsilon$ (say, $\epsilon=1/f(k+1)$). Since $A_1=1-1/f(1)<1$, our sequences will converge to a positive number smaller than 1, or diverge to 0, but it most certainly cannot converge to 1.

Here is a hint to evaluate $$\prod_{n=2}^\infty\left(1-\frac1{n^2}\right) :$$ Note that this is a telescoping product, since $1-1/n^2=(n-1)(n+1)/n^2$. Now play with the first few terms to see the emerging pattern.


Your friend is right: the partial products are clearly $\le 3/4$, and since the infinite product converges (by the test you showed), its value must also be $\le 3/4$. To find the actual value, note that $1 - 1/n^2 = (n-1)(n+1)/n^2$, and so the infinite product is $$ \prod_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right) = \frac{1}{2}\frac{3}{2}\cdot\frac{2}{3}\frac{4}{3}\cdot\frac{3}{4}\frac{5}{4}\cdot\frac{4}{5}\frac{6}{5}\cdot ... $$ All terms cancel in pairs except the first; so the value of the product is $1/2$.

  • $\begingroup$ oh no. I missed such a simple trick. Couldn't see the pattern. $\endgroup$ – anonymous Jan 19 '11 at 18:42
  • $\begingroup$ It's misleading to say that the friend was right. The friend's argument was that each partial product is smaller than one, so the infinite product can not be one. That statement is 100% false. What you are saying is that the partial products are bounded away from one. $\endgroup$ – Dan Petersen May 18 '11 at 6:50
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    $\begingroup$ @DanPetersen: The friend said "the terms in the product" - that is, the numbers being multiplied together - have values less than $1$, and therefore the value of the product can never be $1$. This is correct. An infinite product of positive terms that converges to $1$ must have some terms (not partial products, but terms) that are at least $1$. $\endgroup$ – mjqxxxx Oct 8 '12 at 21:21

If $a_m = \displaystyle \prod_{n=2}^{m}(1 - \frac{1}{n^2})$, then $a_{m+1} = (1 - \frac{1}{(m+1)^2})a_m < a_m$.

So we have $a_m < a_2 = (1 - \frac{1}{4}) = \frac{3}{4}$, $\forall m > 2$.

Hence, $\displaystyle \lim_{m \rightarrow \infty} a_m < a_2 = \frac{3}{4}$.

Hence the statement is false.


$\displaystyle\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\bigg)$ converges $\iff$ $\displaystyle \sum_{n=2}^{\infty}\ln \bigg(1-\frac{1}{n^2}\bigg)$ converges.

One can show that for $x \in (1, \infty),$ we have $\ln(x) \leq \dfrac{x-1}{\sqrt{x}}.$

$\Rightarrow \displaystyle \ln \bigg(1-\frac{1}{n^2}\bigg) \leq \dfrac{-1}{n\sqrt{n^2 -1}}$ and since $\displaystyle \sum_{n=2}^{\infty}\dfrac{1}{n\sqrt{n^2 -1}}$ converges by the integral test (since $f(x) =\displaystyle \dfrac{1}{x\sqrt{x^2 -1}}$ is continuous, non-negative and monotone decreasing on $[2, \infty]$ and $ \displaystyle\int_{2}^{\infty} \dfrac{dx}{x\sqrt{x^2 -1}} = \sec^{-1}(x)\bigg|_{2}^{\infty} = \dfrac{\pi}{6}$) 0r by the comparison test (since $\dfrac{1}{n\sqrt{n^2 -1}} \leq \dfrac{\sqrt{2}}{n^2}$) it follows that $\displaystyle\sum_{n=2}^{\infty}\dfrac{-1}{n\sqrt{n^2 -1}}$ converges and hence so does $\displaystyle \sum_{n=2}^{\infty}\ln \bigg(1-\frac{1}{n^2}\bigg)$.

Let $L = \displaystyle\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\bigg)$. Then the conclusion of the integral test tells us that $\ln(L) < \dfrac{-\pi}{6}$ which implies $L <1.$ (In fact $e^{-\pi/6} = 0.5923$ is a decent approximation of $\dfrac{1}{2}$)


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