Limit of a product sequence for infinite terms

Question

I tried solving an infinite limit like this and got the answer as 0.5 I think there's a flaw in my approach. Please suggest a better approach.

Answer 1

HINT:

Note that

$$\begin{align} \log\left(\prod_{n=2}^N \left(1-\frac1{n^2}\right)\right)&=\sum_{n=2}^N \log\left(1-\frac1{n^2}\right)\\\\ &=\sum_{n=2}^N \left(\log(n+1)-\log(n)\right)+\sum_{n=2}^N \left(\log(n-1)-\log(n)\right) \end{align}$$

Now, exploit the telescoping series.

Answer 2

With shorter notation

$$\prod_{n=2}^\infty\left(1-\frac1{n^2}\right)=\prod_{n=2}^\infty\left(1-\frac1n\right)\left(1+\frac1n\right)$$

But

$$\left(1+\frac1n\right)\left(1-\frac1{n+1}\right)=\frac{n+1}n\cdot\frac n{n+1}=1$$

and thus all the factors after the first one cancel with each other and you indeed get $\;\cfrac12\;$ , so: good work!