How I compute $(2-\sqrt{2})(2-\sqrt[3]{2})\cdots(2-\sqrt[n]{2})$

Question

The product has only positive factors so it has zero as lower bound. Also the product is decreasing as all its factors are less than one. In conclusion the series must have a limit. I also compute the first 100 values of the product and I got around 0.03. I also look at link but I can't use the same trick. –

I am pretty sure that the limit is zero but I do not know how to prove it.

Answer 1

As the arithmetic mean is greater than or equal to the geometric mean, we know that $$ \frac{2^{1/n}+2^{-1/n}}{2} \geq \sqrt{2^{1/n}\cdot 2^{-1/n}} = 1 $$ Therefore $2-2^{1/n} \leq 2^{-1/n}$ and $$ (2-2^{1/2})(2-2^{1/3})\cdot \ldots \cdot (2-2^{1/n}) \leq 2^{-\left( \frac{1}{2} +\frac{1}{3} +\ldots +\frac{1}{n} \right) } $$ As the harmonic series diverges, the limit of the original product is $0$.

Answer 2

Hint: The infinite product $\prod_{n=1}^\infty a_n$ (with $a_n > 0$) converges to a nonzero value if and only if the series $\sum_{n=1}^\infty \log(a_n)$ converges.