# 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 :)

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 $a_n = \prod_{k=1}^n (2-2^{1/k}).$ Take the log to get $$\log(a_n) = \sum_{k=1}^n \log(2-2^{1/k}).$$ This is a typical thing to do when you have a product. Note that for $k$ large $2-2^{1/k}$ is near $\sqrt{2}-1$ and $\log(\sqrt{2}-1)<0$. Thus the series diverges and $\log(a_n) \to -\infty$. This means $a_n \to 0$.