# What is $\lim\limits_{n\to\infty}(\sqrt2-\sqrt[3]2)\cdots(\sqrt2-\sqrt[n]2)$? How to approach? [duplicate]

$$\lim\limits_{n\to\infty}(\sqrt2-\sqrt[3]2)(\sqrt2-\sqrt[4]2)(\sqrt2-\sqrt[5]2)\cdots(\sqrt2-\sqrt[n]2)$$

Could you tell me how to approach this kind of question? How do I find the limit of this sequence?

I know that for very large $n$ the each bracket is more than $1$, so my guess is its going to infinity, how do I prove such a thing?

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## marked as duplicate by Martin Sleziak, Norbert, Hanul Jeon, TZakrevskiy, Daniel Robert-NicoudNov 30 '13 at 12:55

What happens when you compare this limit to $\lim_{n\to\infty}(\sqrt 2-\sqrt[n]{2})^n$? What is the numeric value of $\sqrt 2-1$? – abiessu Nov 2 '13 at 16:25
this will obviously go to infinity as each bracket is more then 1 to the power of n, but thats not the original sequence was... – guynaa Nov 2 '13 at 16:28
I think you have overestimated the value of $\sqrt 2 -1$... – abiessu Nov 2 '13 at 16:34
oops, i got it, using the sandwich theorem i can prove that the series you provided is approaching 0 that proves it, right? – guynaa Nov 2 '13 at 16:37
That is correct :-) – abiessu Nov 2 '13 at 16:56

Formally, we can say that $\sqrt 2\gt\sqrt[3]2\gt\sqrt[4]2\gt\dots\gt\sqrt[n]2\gt 1$ as $n\to\infty$, and $0\lt\sqrt 2-\sqrt[3] 2\lt\sqrt 2-\sqrt[4] 2\lt\dots\lt\sqrt 2-\sqrt[n] 2\lt\sqrt 2-1\lt \frac 12,$ therefore
$$0\le\lim_{n\to\infty}\prod_{i=3}^n(\sqrt 2-\sqrt[i] 2)\le\lim_{n\to\infty}\left(\frac 12\right)^n=0$$