TITLE: Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$?
QUESTION [55 upvotes]: It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } \text{ ?} $ The answer might be $-1$, but I'm not sure as $i$ is not a real number. Can anyone help?

REPLY [10 votes]: The two highest-voted answers (as of writing) are incorrect because they do not show that the expression even has a limit, only that if the limit exists then it is −1. – Rahul

We have the sequence
$$ a_0 = i,\quad a_{n+1} = i \sqrt{a_n}. $$
I think the other answers have sufficiently covered that the argument of each element of the sequence lies in $(0, \pi)$, so we can be sure that we're always taking the principal square root. They also have shown clearly enough (except for off-by-one errors) that
$$ a_n = \exp(i\pi - i\pi/2^n).$$
I assert that the limit of this sequence is $-1$. For any $\varepsilon>0$ there exists $N$ such that $|-1 - a_n| < \varepsilon$ when $n>N$.
$$| -1 - a_n |=| -1 - \exp(i\pi)\exp(-i\pi/2^m) |=| -1+\cos(\pi/2^n)+i\sin(\pi/2^n)|$$ which is less than or equal to
$$ |1 - \cos(\pi/2^n)| + |\sin(\pi/2^n)|. $$
$1-\cos(x)<x$ for all $x>0$, as is $\sin(x)<x$. So the above is less than $2\pi/2^n$. Then for $N > \log_2(2\pi/\varepsilon)$ we have $|a_n+1|<\varepsilon$.