Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to show that if $u_{0}=u_{1}=u_{2}=0$, and if, when $n>1$, $$u_{2n-1}=\dfrac {-1} {\sqrt {n}}, u_{2n}=\dfrac {1} {\sqrt {n}}+\dfrac {1} {n}+\dfrac {1} {n\sqrt {n}}$$ then $\prod \limits_{n=0}^{\infty }\left( 1+u_{n}\right) $ converges.

We observe that $\sum \limits_{n=0}^{\infty }u_{n}$ and $\sum \limits_{n=0}^{\infty }u_{n}^{2}$ are divergent by ratio test. I am unsure how to proceed from here. Any help would be much appreciated. Could we argue possibly since $\lim _{n\rightarrow \infty }u_{n}=0$ that's why $\prod \limits_{n=0}^{\infty }\left( 1+u_{n}\right) $ converges ?

share|cite|improve this question
up vote 3 down vote accepted

Combine terms to get

$$\prod_{n=2}^N \left(1-\frac{1}{\sqrt{n}}\right)\left(1+\frac{1}{\sqrt{n}}+\frac{1}{n}+\frac{1}{n\sqrt{n}}\right)=\prod_{n=2}^N \left(1-\frac{1}{n^2}\right)=\frac{N+1}{2N}\to\frac{1}{2}.$$

Also see here.

I should add that no, $\lim\limits_{n\to\infty}u_n=0\;$ does not alone establish convergence. Consider

$$\prod_{n=1}^N\left(1+\frac{1}{n}\right)>\sum_{n=1}^N\frac{1}{n}\to\infty \quad \text{yet}\quad \lim_{n\to\infty}\frac{1}{n}=0.$$

Some form of alternating test would work, but is computationally overdoing it in my opinion.

share|cite|improve this answer

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.