Q:

The limit of iterated square root with multiplication under the root, $\sqrt{ a \sqrt{ a \sqrt{a \cdots}}}$

$$
\sqrt{ a  \sqrt{ a  \sqrt{a \cdots}}}=\text{ ?}
$$
options were given as 

$0$
$-a$
$a$
$1$

i did not know how to solve it or what it was related to. Could anyone please explain the concept and/or provide helpful references or links? 

A:

That's just $a^{1/2}\cdot a^{1/4} \cdots = a^{1/2+1/4+1/8+\cdots}$.
You can use the summation $\sum_{i=1}^{\infty} \frac{1}{2^i} = 1$ to get the answer.

A:

Let $x_1=\sqrt{a}$, and $x_n=\sqrt{ax_{n-1}}$, for $n\ge 2$ if the limit $$x:=\lim_{n\to \infty}x_n$$
If that limit exist. We have
\begin{align*}
x^2&=a\lim_{n\to\infty}x_n\\
&=ax
\end{align*}
So, $x=0$ or $x=a$.

