$\prod_{n=1}^{+\infty} \left(1-\frac{1}{2n^2}\right)>0$? How to prove that the infinite product $\prod_{n=1}^{+\infty} \left(1-\frac{1}{2n^2}\right)$ is positive ?

Thanks
Hint: Only the lower limit is a problem. Look e.g. at the product for $n\geq 2$, you could prove:
$$ \prod_{n\geq 2} (1-\frac{1}{2 n^2}) \geq  \prod_{n\geq 2} (1-\frac{1}{ n^2})=\prod_{n\geq 2} \frac{(n-1)(n+1)}{ n \cdot n}=\frac12$$
(the last being a telescopic product). So your product is $\geq \frac14$.
$1-\dfrac{1}{n^2}&lt;1-\dfrac{1}{2n^2}&lt;\left(1-\dfrac{1}{n^2}\right)^2$,So you can see Its limit exist.

Sorry For I cannot use LaTex and My poor English.