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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 ?

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1 Answer 1

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.

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