Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)

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

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

share|cite|improve this question
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

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

share|cite|improve this answer
Thank you for your nice solution. –  drmath Apr 4 '12 at 19:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.