# Find the limit of $(\sqrt2-\root3\of2 )(\sqrt2 -\root4\of2 )...(\sqrt 2-\root{n}\of2 )$ without using the squeezing principle

$$\mathop {\lim }\limits_{n \to \infty } (\sqrt 2 - \root 3 \of 2 )(\sqrt 2 - \root 4 \of 2 )...(\sqrt 2 - \root n \of 2 )$$

I was able easily to find the limit of this series using the squeezing principle, but how can you find it without it? I'm guessing it's an algebra trick :)

• The tag (limit-theorems) is not a good fit for this questions, see the tag-wiki. Nov 29 '13 at 8:06
• In this case, the answer to that other question implicitly uses the squeeze theorem, though it calls it the comparison test. @MartinSleziak Nov 29 '13 at 18:39
• It seems that I was too quick on the trigger @ThomasAndrews. I am voting to reopen in that case. Nov 30 '13 at 8:43
• Other questions about the same limit: math.stackexchange.com/questions/452173/…, math.stackexchange.com/questions/549111/… Dec 1 '13 at 6:44

I see no trick nor algebra, just the simple observation that for every $k \geq 3$ we have $2^{1/2} > 2^{1/k} > 1$, so that $$0 \leq \prod_{k=3}^n \left(2^{1/2}-2^{1/k}\right) \leq \left(\sqrt{2}-1\right)^{n-2}$$ The r.h.s. being a geometric sequence with $0 < \sqrt{2} - 1 < 1$, the sequence converges (very quickly) towards $0$.
Our sequence is $$a_n = \prod_{k=1}^n (2^{1/2}-2^{1/k}).$$ Take the log to get $$\log(a_n) = \sum_{k=1}^n \log(2^{1/2}-2^{1/k}).$$ This is a typical thing to do when you have a product. Note that for $$k$$ large $$2^{1/2}-2^{1/k}$$ is near $$\sqrt{2}-1$$ and $$\log(\sqrt{2}-1)<0$$. Thus the series diverges and $$\log(a_n) \to -\infty$$. This means $$a_n \to 0$$.