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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 $\lambda_k\rightarrow0$?

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Is it not true that $\{\lambda_{k}\}$ is strictly decreasing since we always multiply by $(1-\alpha_{k})\in(0,1)$? – Daniel Littlewood Nov 27 '12 at 21:32
@DanielLittlewood: that doesn't imply that its limit is $0$. – WimC Nov 27 '12 at 21:37
It does not imply "directly" that its limit is 0, but don't you know a result about bounded decreasing sequences which could help? – Sebastien B Nov 27 '12 at 21:40
up vote 5 down vote accepted

$\log(\lambda_k) = \sum_{j<k} \log(1-\alpha_j) \leq \sum_{j<k}-\alpha_j$.

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Argh! You beat me to it. Anyhow, this is a special case of the more general result that $\prod(1+a_k)$ is absolutely convergent if and only if $\sum a_k$ is. (Here, a product is called absolutely convergent if it converges to a finite nonzero number, and the product is invariant under permutation of the indices.) The proof is built on the same basic idea, except you need to use $\log(1+a_k)=a_k+O(a_k^2)$. Otherwise, the inequality goes the right way only for negative $a_k$. – Harald Hanche-Olsen Nov 27 '12 at 21:46

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, $\lim_{n\rightarrow\infty}{\lambda_n}=0$.

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