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Let $\{\alpha_k\}\subset \mathbb{R}$ be a positive sequence satisfying $ \lim_{k\rightarrow\infty}\alpha_k=0, \quad \sum_{k=0}^{\infty}\alpha_k=+\infty. $ Put $\displaystyle S_k=\prod_{i=0}^{k}(1-\alpha_i)$. Find the limits (if exist)

  • $\displaystyle\lim_{k\rightarrow\infty}S_k.$

  • $\displaystyle\lim_{k\rightarrow\infty}\sqrt[k]{S_k}.$

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    [This question](http://math.stackexchange.com/questions/106188/calculating-the-limit-lim-limits-k-to-infty-prod-limits-i-1k1-alpha) is similar to the first part of your question.2012-04-04

1 Answers 1

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I will assume that $\alpha_i\in(0,1)$.


Note that $0\le\prod_{i=0}^k (1-\alpha_i) \le \prod_{i=0}^k \frac1{1+\alpha_i} = \frac1{\prod_{i=0}^k(1+\alpha_i)} \le \frac 1{1+\sum_{i=0}^k \alpha_i}$

For $k\to \infty$ the RHS tends to zero, so you get $\lim\limits_{k\to\infty} S_k=0$.

We have used $1-x\le \frac1{1+x}$, which follows from $(1-x)(1+x)=1-x^2\le 1$.


For the second part, let us try to use this result: $\liminf_{n\to\infty}\frac{c_{n+1}}{c_n}\leq\liminf_{n\to\infty}\sqrt[n]{c_n} \le \limsup_{n\to\infty}\sqrt[n]{c_n}\leq\limsup_{n\to\infty}\frac{c_{n+1}}{c_n}$ which is true for any positive sequence $(c_n)$, see e.g. this answer and this question.

If we apply this to the sequence $(S_k)$, we get $\liminf_{k\to\infty} (1-\alpha_{k+1}) \le \liminf_{k\to\infty} \sqrt[k]{S_k} \le \limsup_{k\to\infty} \sqrt[k]{S_k} \le \limsup_{k\to\infty} (1-\alpha_{k+1}),$ which implies $\lim\limits_{k\to\infty} \sqrt[k]{S_k}=1$.