I was studying for some quizzes when wild question appears. It looks like this:

Find $\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)$

My work

I think it's a repeated multiplication of the expression $1-\frac{1}{k^2}$. It looks like this:

$$\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right) = \left(1-\frac{1}{(2)^2}\right)\left(1-\frac{1}{(3)^2}\right)\left(1-\frac{1}{(3)^2}\right)\left(1-\frac{1}{(4)^2}\right).....$$

I barely had any experience evaluating these new summation...How do evaluate $\prod_{k=2}^{+\infty}\left(1-\frac{1}{k^2}\right)$?

  • 1
    $\begingroup$ Try to find out what happens if $k=6$ (a small value of $k$). Much of the staff you have is canceled. $\endgroup$ – Konstantinos Gaitanas Sep 25 '17 at 8:43
  • $\begingroup$ @KonstantinosGaitanas What do you mean "Much of the staff you have is canceled"? additional information is lost when I do the operation above? $\endgroup$ – Palautot Ka Sep 25 '17 at 8:52
  • $\begingroup$ Look at the numerators and the denominators in Gerhard's answer below. $\endgroup$ – Konstantinos Gaitanas Sep 25 '17 at 8:53
  • $\begingroup$ @KonstantinosGaitanas What topic in math do I learn this truth? because I just found this question during our review for some licensure exams.... $\endgroup$ – Palautot Ka Sep 25 '17 at 9:27

\begin{eqnarray*} &=&\prod_{k=2}^{+\infty}\left(1-\frac1{k^2}\right)\\ &=&\prod_{k=2}^{+\infty}\frac{(k-1)(k+1)}{k^2}\\ &=&\frac{1\cdot3}{2\cdot 2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot 4}\cdot\ldots\\ &=&\lim_{k\to+\infty}\frac{k+1}{2\cdot k}\\ &=&\frac12 \end{eqnarray*}

  • $\begingroup$ wow! what topic in math deals with questions like mine? $\endgroup$ – Palautot Ka Sep 25 '17 at 8:49

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