real analysis - Why is $sqrt{2sqrt{2sqrt{2cdots}}} = 2$?

Why is $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}}}$ equal to 2? Does this work for other numbers?

Answer

The number in question is simply

$$2^{1/2+1/4+1/8+\cdots} = 2^1 = 2$$

Yes, this works for other numbers. More interesting is if the numbers are not equal inside the radicals. For example, say you have a positive sequence element $a_n$ inside the $n$th radical. If we assume the expression converges to some value $P$, then

$$\log{P} = \sum_{k=1}^{\infty} \frac{\log{a_n}}{2^n}$$