How do I show that for any starting $n$? (I am not really sure it is $0$, but I think it is).
$$\prod_{i=n}^\infty \left[1-\frac 1 i\right]=\prod_{i=n}^\infty \left[\frac{i-1}{i}\right]=0$$
I tried taking logarighm, then I get a sum. $\sum_{i=n}^\infty \ln\left(\frac{i-1}{i}\right)$. But then I must show that this sum goes to $-\infty$ can I show that in some way?
$$\sum_{i=n}^\infty \log \frac{i-1}{i} = \sum_{i=n}^\infty (\log(i-1)-\log(i))=\lim_{N\to\infty} [\log(n-1)-\log(N)]$$