Infinite Product – Evaluating the Infinite Product of (1-1/k^2)

Question

Evaluate the infinite product

$$\lim_{ n\rightarrow\infty }\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ).$$

I can't see anything in this limit , so help me please.

Answer 1

Note that $$1-\frac1{k^2}=\left(1-\frac1k\right)\left(1+\frac1k\right)=\frac{k-1}{k}\frac{k+1}{k}=\frac{a_k}{a_{k-1}}$$ with $a_k= \frac{k+1}k$, hence this is a telescoping product, i.e. $$ \prod_{k=2}^n\left(1-\frac1{k^2}\right)=\frac{a_2}{a_1}\frac{a_3}{a_2}\cdots\frac{a_n}{a_{n-1}}=\frac{a_n}{a_1}=\frac{n+1}{2n}.$$