Given an arbitrary sequence $\{\alpha_k\}_{k=0}^\infty:\alpha_k\in(0,1),\sum_{k=0}^\infty\alpha_k=\infty$.
Define $\lambda_0=1,\lambda_{k+1}=(1-\alpha_k)\lambda_k$.
How to show $\lambda_k\rightarrow0$?
Given an arbitrary sequence $\{\alpha_k\}_{k=0}^\infty:\alpha_k\in(0,1),\sum_{k=0}^\infty\alpha_k=\infty$.
Define $\lambda_0=1,\lambda_{k+1}=(1-\alpha_k)\lambda_k$.
How to show $\lambda_k\rightarrow0$?
$\log(\lambda_k) = \sum_{j
We see $\lambda_n=\prod_{i=0}^{n-1}{(1-\alpha_i)}$. Note this is a bounded monotonically decreasing sequence, so $\lim_{n\rightarrow\infty}{\lambda_n}$ exists. If $\lambda_n\rightarrow\gamma>0$, we get a contradiction: assume $\lim_{n\rightarrow\infty}{\lambda_n=\gamma}>0$. Then, by Cauchy Convergence Criterion, we see $\prod_{i=j}^k{(1-\alpha_i)}\rightarrow 1$, which implies $\sum_{i=j}^k{\alpha_i}\rightarrow 0$. However, this shows $\sum_{i=0}^\infty{\alpha_i}<\infty$ (this shows the sum is Cauchy). Therefore, the limit $\gamma\leq 0$. It is clear the limit is not less than $0$, and since the limit must exist, $\lim_{n\rightarrow\infty}{\lambda_n}=0$.