Compute the product$$ \left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right) \cdots $$
For $N \geq 2$, define$$ a_{N}=\left(1-\frac{4}{1}\right)\left(1-\frac{4}{9}\right)\left(1-\frac{4}{25}\right) \cdots\left(1-\frac{4}{(2 N-1)^{2}}\right) . $$The problem asks us to find $\lim _{N \rightarrow \infty} a_{N}$. The defining product for $a_{N}$ telescopes as follows:$$ \begin{aligned} a_{N} &=\left[\left(1-\frac{2}{1}\right)\left(1+\frac{2}{1}\right)\right]\left[\left(1-\frac{2}{3}\right)\left(1+\frac{2}{3}\right)\right] \cdots\left[\left(1-\frac{2}{2 N-1}\right)\left(1+\frac{2}{2 N-1}\right)\right] \\ &=(-1 \cdot 3)\left(\frac{1}{3} \cdot \frac{5}{3}\right)\left(\frac{3}{5} \cdot \frac{7}{5}\right) \cdots\left(\frac{2 N-3}{2 N-1} \cdot \frac{2 N+1}{2 N-1}\right)=-\frac{2 N+1}{2 N-1} . \end{aligned} $$Hence the infinite product is equal to$$ \lim _{N \rightarrow \infty} a_{N}=-\lim _{N \rightarrow \infty} \frac{2 N+1}{2 N-1}=-1 . $$