TITLE: Evaluating a strange sequence
QUESTION [5 upvotes]: I'm a first year undergrad maths student and I've come across this particular sequence, which I can't find a limit for.
$$\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10\cdots}}}}}_{n \text{ times}}$$
I know this may be embarrassingly easy, but I can't think of a way to start. Can you share any hints?

REPLY [2 votes]: We have the recurrence relationship
$$x_{n+1}=\sqrt{10x_n}\implies x_{n+1}^2=10x_n$$
If $x_0=0$, then $x_n=0$ for all $n$ and $\lim_{n\to \infty}x_n=0$.
If $x_0=10$, then $x_n=10$ for all $n$ and $\lim_{n\to \infty}x_n=10$.
If $10>x_0>0$, then clearly $0<x_n<x_{n+1}$ and $x_n$ is increasing.  And since $x_0<10$, the sequence is clearly bounded above by $10$.  Since $x_n$ is monotonically increasing and bounded above it converges.  Let's call this limit $L$.  Thus, $\lim_{n\to \infty}x_{n+1}=L=\lim_{n\to \infty}\sqrt{10x_n}=\sqrt{10L}\implies L=10$
If $10<x$, then clearly $x_n>x_{n+1}>0$ and $x_n$ is decreasing.  And since $x_n>0$, the sequence is clearly bounded below by $0$.  Since $x_n$ is monotonically decreasing and bounded below it converges.  Let's call this limit $L$.  Thus, $\lim_{n\to \infty}x_{n+1}=L=\lim_{n\to \infty}\sqrt{10x_n}=\sqrt{10L}\implies L=10$.
Putting it all together, we have
$$\lim_{n\to \infty}x_n=
\begin{cases}
10&,x_0>0\\\\
0&,x_0=0
\end{cases}
$$
and we are done!