Evaluating a strange sequence 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?
 A: Well..: 
$$\sqrt{10\sqrt{10\sqrt{10\sqrt{10\cdots}}}} =  10^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots}$$
A: 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!
A: Hint: Let $x = \lim_{n \to \infty}\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10}}}}...}_{n \text{ times}}\ $.
Now observe that $$\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10}}}}...}_{n \text{ times}} = \sqrt{10\underbrace{\sqrt{10\sqrt{10\sqrt{10}}}...}_{n-1 \text{ times}}}$$
so that in the limit, you have $x = \sqrt{10 x}$.
A: Let $x=\sqrt{10\sqrt{10\sqrt{10\cdots}}}$.  Then, $x^2=10\sqrt{10\sqrt{10\sqrt{10\cdots}}}=10x$.  Therefore, $x^2=10x$, so $x=10$.
