I have multiple questions regarding one problem: For a positive real $a$ what's the value of $\sqrt{a\sqrt{a\sqrt{a...}}}$

Question

I'm reading a book, the book is aimed at high schoolers so it is perhaps not the most rigorous book. The intended solutions the book provides for the problem are the following:

First assume the expression has a value and call it $x$, so $x = \sqrt{a\sqrt{a\sqrt{a...}}}$ Then notice that $x = \sqrt{ax}$, then $x^2 = ax$, then $x(x-a) = 0$, then since $a$ is positive the solution is $x = a$ and that's the value of the whole expression.

They also offer a second solution that goes like this: If $x = \sqrt{a\sqrt{a\sqrt{a...}}}$ then we can also express it as: $x = a^{\frac{1}{2}}a^{\frac{1}{4}}a^{\frac{1}{8}}...$, then $x = a^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}$, and since $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... = 1$, then $x = a$ once again.

My first question is, how valid are any of these solutions at the university level? Say if this was a real analysis class, would any of those two answers be considered valid?

Also, although I haven't taken a real analysis class yet, I have heard of the $\epsilon-N$ for sequences and series, I assume if we want to do the same for this particular problem we would have to look at the sequence $\sqrt{a}, \sqrt{a\sqrt{a}}, \sqrt{a\sqrt{a\sqrt{a}}}, ...$, then check if it converges using $\epsilon-N$ and if it does then that's the value of he expression $\sqrt{a\sqrt{a\sqrt{a...}}}$ Second question is, is my speculation correct? Would this way of doing it with $\epsilon-N$ constitute a valid way of solving the problem if this was a real analysis class?

Lastly, my last question is, how would you solve the problem in a way that is valid for a real analysis class? Either using $\epsilon-N$ or whatever else, please show me in a way someone who hasn't taken a real analysis class before can understand.

Answer 1

To answer your question on how to prove it more rigourously this is how I will prove it.
Just take into account that I am just a student so beware with my solution.

I/Prove that if: $\lim_{n \to \infty} a_n = a$ and that $\lim_{n \to \infty} b_n = b$, then $\lim_{n \to \infty} (a_n) ^{b_n}= a^b$
1- We rewrite $(a_n) ^{b_n} = e^{log((a_n) ^{b_n})} = e^{b_nlog(a_n)}$.

2- More over we know that the function $e^x$ is continuous everywhere and that $log(x)$ is continuous on $x>0$.
Hence according to Heine theorem if $\lim_{n \to \infty} a_n = a \Rightarrow \lim_{n \to \infty} log(a_n) = log(a)$

3- Now according to direct application of limit arithmetic we have that $\lim_{n \to \infty} b_nlog(a_n) = b log(a)$

4- We continue and we use again the Heine theorem for the sequence of points $x_n = b_nlog(a_n)$. Because $e^x$ is continuous you have that if $\lim_{n \to \infty} x_n =x \Rightarrow \lim_{n \to \infty} e^{x_n} = e^x$ and so here $\lim_{n \to \infty} (a_n)^{b_n} = e^{b_nlog(a_n)} = e^{blog(a)} = a^b$

II/ Applying this theorem to your specific question.
Just write here $b_n = \sum_{1 \leq k \leq n} \frac{1}{2^k} \underset{n \to \infty}{\rightarrow} 1$ and $a_n = a $ after it is simply a direct application of the theorem we writte in "I/"

Q.E.D.

I hope this helped you.

Rem: $a,b$ here are consiedered to be finite real numbers

Answer 2

your first approach seems more reasonable and simple, although it is not complete. Stopping the solution where you did is wrong: you assumed the limit existed, but never proved it. So in order to complete the proof, tou need to actually verify that the limit does exist.

A possible way to go about it is this. Define the sequence $\{x_n\}_n$ recursively as $$ x_1=\sqrt{a}\\ x_{n+1}=\sqrt{ax_n} $$

We prove the sequence is convergent (i.e. admits a limit) by showing that it increasing and bounded from above.

It is indeed bounded by $a$. By induction

$$ x_1 = \sqrt{a}\le a\\ x_{n+1}=\sqrt{ax_n} \le \sqrt{a^2} = a $$

For the monotonicity instead we notice that

$$ x_n < x_{n+1} \iff x_n < \sqrt{ax_n} \iff x_n^2 -ax_n<0 \iff 0<x_n <a $$

which is true because the sequence is bounded by $a$.

Since now we proved the limit exests, the answer is complete

---

As pointed out in the comments, the reasoning works only if $a>1$.

In case $a<1$, the original limit is $1/\sqrt{a\sqrt{a \dots}}$, which means the conclusion follows with the result $1/a$