Let the probability of success in the ith trial be $p_i$ and let $\sum_{i=0}^\infty p_i=\infty$ Show that the probability of having no success , $N$ is zero
N is a subset of the event that there are no successes in the first n $$P(N)\le\prod_{i=1}^n(1-p_i)$$
Taking logs $$\begin{align} \log P(N)&\le\sum_{i=1}^n\log(1-p_i)\\ &\le \sum_{i=1}^n(-p_i) \tag{*} \end{align}$$
I believe that (*) is wrong. I took $n= 10$ and $p_i=0.5$. Can someone confirm. How would you prove it. I know that the first product tends to zero we can bound it by the maximum of the $(1-p_i)$ Would that be enough of a proof.