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.

  • 2
    $\begingroup$ see this answer math.stackexchange.com/a/955216/136544 $\endgroup$
    – daw
    Commented Dec 14, 2023 at 16:13
  • $\begingroup$ @daw Thanks! People do seem to be using epsilon-N when talking about the problem so I guess I was right about that, however that thread does not answer my question about whether or not the two answers I provided would be considered valid at the university level. $\endgroup$
    – zlaaemi
    Commented Dec 14, 2023 at 16:29
  • $\begingroup$ @zlaaemi The first one would be considered wrong because you've only proved that if the expression has a limit, then that limit is $a$. Second one debatable, it depends on how much the listener values your ability to understand the underlying details. Obviously, if you've decided for using the dots, then you can't neglect them whenever you feel like it. $\endgroup$ Commented Dec 14, 2023 at 16:35
  • $\begingroup$ @SassatelliGiulio Thanks, I've added the missing dots. What exactly do you mean by "depends on how much the listener values your ability to understand the underlying details"? $\endgroup$
    – zlaaemi
    Commented Dec 14, 2023 at 16:43
  • $\begingroup$ It means that somebody might be interested in testing whether or not you can prove that $a^{1/2}a^{1/4}a^{1/8}\cdots=a^{\frac12+\frac14+\frac18+\cdots}$, or, which is slightly different, whether or not you can prove that the limit of the sequence defined recursively as $x_0=1$, $x_{n+1}=\sqrt{ax_n}$ is $a^{\frac12+\frac14+\frac18+\cdots}$, despite them not being usually interested in that. $\endgroup$ Commented Dec 14, 2023 at 16:50

2 Answers 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 $1/a$

  • $\begingroup$ Thanks for your reply, it does make sense that showing it's an increasing function and bounded would be enough, but I don't understand why you have picked that specific sequence. If I understand it correctly, we would have $x_1 = a$, $x_2 = \sqrt{a^2}$, $x_3 = \sqrt{a\sqrt{a^2}}$ and so on, but don't we want the sequence: $x_1 = \sqrt{a}$, $x_2 = \sqrt{a\sqrt{a}}$, $x_3 = \sqrt{a\sqrt{a\sqrt{a}}}$ and so on...? $\endgroup$
    – zlaaemi
    Commented Feb 24 at 16:50
  • $\begingroup$ certanly, I didn't see the square root was missing. thanks for spotting it. I edited my reply $\endgroup$
    – finch
    Commented Feb 24 at 16:56
  • $\begingroup$ I'm sorry but I don't understand what you mean. why would you say the argument doesn't work? $\endgroup$
    – finch
    Commented Feb 24 at 19:11
  • $\begingroup$ The base of induction in wrong. $\sqrt a \le a $ only if $a\ge 1$. $\endgroup$
    – jjagmath
    Commented Feb 25 at 11:45
  • $\begingroup$ The sequence is increasing or decreasing depending on whether $a\ge 1$ or $a \le 1$. $\endgroup$
    – jjagmath
    Commented Feb 25 at 11:54

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/"


I hope this helped you.

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


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