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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 is similar to the first part of your question. –  Martin Sleziak Apr 4 '12 at 15:59
    
Thanks Martin Sleziak for your helping. –  drmath Apr 4 '12 at 19:52
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1 Answer 1

up vote 2 down vote accepted

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 $\lim\limits_{k\to\infty} \sqrt[k]{S_k}=1$.

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Thank you for your nice solution. –  drmath Apr 4 '12 at 19:49
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