Infinite product of the form $2-2^{1/k}$

Question

How can I show that $$\lim_{n\to\infty} \prod_{k=2}^{n} (2-2^{1/k})=0$$ This is an exercise from a college admission exam, and the answer is given as 0. I don't understand how infinitely many positive numbers can have a product equal to $0$. I tried to take the logarithm of the product and use the fact that $$\lim_{x\to0} \frac{\ln(1+x)}{x}=1$$ and get rid of it, but I am left with $$\sum_{k=2}^{\infty} (1-2^{1/k})$$ Is it correct? Any other approaches would be appreciated.

Answer 1

Given $1+x\leq e^x$, we have that

$$1+\frac{\ln 2}{k} \leq 2^{1/k}$$

$$1-\frac{1}{2k}\geq 1-\frac{\ln 2}{k} \geq 2-2^{1/k}.$$

So, it suffices to prove that the infinite product

$$\prod_{k=2}^{\infty} \frac{2k-1}{2k}$$

tends to $0$. Can you take it from here?

Answer 2

Indeed it is equivalent to prove that $\sum_{k=2}^\infty (1-2^{1/k})=-\infty$. You can find a general statement about logarithms of infinite products at ProofWiki

To conclude, use the Mean value theorem to have $2^\alpha-1 = 2^\alpha-2^0 \geq \log 2 \cdot \alpha$ for $\alpha \geq 0$. This shows that our series goes to infinity at least as fast as the harmonic series.