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

Question

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

1. $0$
2. $-a$
3. $a$
4. $1$

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?

Answer 1

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 $x=a$.

Answer 2

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.

Answer 3

Supposing that $x=\sqrt{a\sqrt{a\sqrt{a\sqrt{\cdots}}}}$, then squaring both sides, we have $x^2 = a\sqrt{a\sqrt{a\sqrt{a\sqrt{\cdots}}}} = ax$. If we assume $x$ is nonzero, we may divide both sides by $x$ yielding $x=a$. Otherwise, $x=0$; this occurs when $a=0\;(=x)$, so both possible cases are encapsulated in the response $x=a$.