Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$ While trying some problems along with my friends we had difficulty in this question.


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*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 


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*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,


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*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 $1$?

 A: 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.
A: 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.
A: 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 $1/2$.
A: $\displaystyle\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\bigg)$ 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 $\dfrac{1}{2}$)
A: You could use the infinite product for the sine function which says that:
$$\sin(x)=x\prod_{n=1}^\infty (1-\frac{x^2}{\pi^2n^2})$$
Or:
$$\frac{\sin(x)}{x(1-\frac{x^2}{\pi^2})}=\prod_{n=2}^\infty (1-\frac{x^2}{\pi^2n^2})$$
Then, for your problem, you can just add $\lim_{x\rightarrow\pi}$ to the equation and solve.
A: 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.
