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.

  • 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 $1$?

Answer

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.

Attribution
Source : Link , Question Author : Community , Answer Author : t.b.

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