What does this series converge to?

$$ \sqrt{3\sqrt {5\sqrt {3\sqrt {5\sqrt \cdots}}}} $$

and also this?

$$ \sqrt{6+\sqrt {6+\sqrt {6+\sqrt {6+\sqrt \cdots}}}} $$

And, generally speaking, how should one approach these kind of questions?

  • 2
    $\begingroup$ Hint:use log function $\endgroup$
    – wwbb90
    Oct 4, 2013 at 18:50
  • 1
    $\begingroup$ Standard approach: Identify a part of the expression that is similar to the whole expression. Denote this $x$, replace it and equate the new expression with $x$. Solve the equation. $\endgroup$
    – Daniel R
    Oct 4, 2013 at 18:56
  • $\begingroup$ @DanielR Again, that is if we know that it converges. The interesting part is showing that it does so. $\endgroup$
    – chubakueno
    Oct 4, 2013 at 18:59
  • $\begingroup$ @chubakueno Yes, showing convergence is of course important. $\endgroup$
    – Daniel R
    Oct 4, 2013 at 19:00

3 Answers 3


Here's for $$ \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{\cdots}}}}. $$ First define a sequence recursively by $a_1 := \sqrt{6}$ and $a_n := \sqrt{6+a_{n-1}}$ ($n \geq 2)$. Then we wish to compute $\lim_{n \to \infty} a_n$. In order to find this limit, we need, of course, to know that it exists. But instead of trying to show right away that it does, we will assume that it does and try to obtain candidates for its value. This will help us to show convergence (as it is often the case... in fact, most of the time, when we want to show that a sequence converges to some limit $L$, we need to first guess the value of $L$, which is not always simple... Cauchy sequences are useful justly because we don't have to guess a limit in order to prove convergence).

So, assuming the limit exists, put $\lim_{n \to \infty} a_n := L$. Then by definition of $a_n$ we have $$ \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{6+a_n}, $$ which gives $$ L = \sqrt{6+L}. $$ In this case W.W. showed that $L \in \{-2, 3\}$ and correctly observed that $L = -2$ is impossible, since $a_n \geq 0$ $\forall n$. Thus, the only candidate for $L$ is $3$.

Now it is easily seen that $a_{n+1} \geq a_n$, so the sequence is increasing. In order to show convergence we know it suffices to exhibit an upper bound for $a_n$. But we might as well try to show that $a_n \leq 3$, in view of our candidate for $L$. This we can do by induction :

  1. Base step : Clearly $a_1 = \sqrt{6} \leq \sqrt{9} = 3$ is true.
  2. Induction step : Assume $a_{n-1} \leq 3$ for some $n \geq 2$. Then $a_n = \sqrt{6+a_{n-1}} \leq \sqrt{6+3} = 3$.

By the principle of mathematical induction, we have shown that $a_n \leq 3$ $\forall n$, which is what we were seeking.

We may conclude that $a_n$ converges and that its limit is 3.

  • $\begingroup$ What do you mean ? $\endgroup$
    – Amateur
    Oct 4, 2013 at 20:41
  • $\begingroup$ From my cellphone it got prematurely posted and I couldn't change it for the five minutes thing. Here it goes again: Note:In this case it works because it is monotone increasing, but we must be careful with induction and infinity, since $\infty \notin \mathbb{N}$(Although here we are taking the limit).If so, we could argue that $\pi$ is rational :) . $\endgroup$
    – chubakueno
    Oct 4, 2013 at 20:48
  • $\begingroup$ Very Through... $\endgroup$
    – Bak1139
    Oct 5, 2013 at 11:43

Well, define

$$x=\sqrt{3\sqrt{5\sqrt3\ldots}}=\sqrt{3\sqrt{5x}}\stackrel{\text{square}}\implies x^2=3\sqrt{5x}\stackrel{\text{square again}}\implies x^4=45x\implies$$

$$x^3=45\iff x=\sqrt[3]{45}$$

But...the above follows at once from arithmetic of limits, on the sequence


and we must first prove the above sequence converges finitely, so


== Prove your sequence is monotone ascending

== Prove your sequence is bounded above


For $$\sqrt{6 + \sqrt{6 + \sqrt{6 +\dots}}}:$$

Let \begin{align*} x &= \text{the given equation}\\ &= \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}} \end{align*}

Since the series is infinite, we can write $x = \sqrt{6 + x}$, or $$x^2 - x - 6 = 0.$$

Therefore, $x = 3$ or $x = -2$

Since the answer cannot be negative, reject $x = -2$. Therefore, $x = 3$.

Do the same thing for the first series.

  • $\begingroup$ It is the same with the first question. Given convergence. $\endgroup$
    – chubakueno
    Oct 4, 2013 at 18:53
  • $\begingroup$ The series can't be equal to two values. How would you produce a negative value by using the square root function, which only returns positive values? $\endgroup$
    – Daniel R
    Oct 4, 2013 at 18:54
  • 1
    $\begingroup$ @W.W I tried to edit your answer for LaTeX formatting, but you reedited it in the meantime. Please use LateX for clarity. $\endgroup$
    – chubakueno
    Oct 4, 2013 at 18:58
  • 1
    $\begingroup$ You also need to prove that it converges in the first place. This will likely use induction (and the fact that monotone bounded sequences converge) $\endgroup$ Oct 4, 2013 at 19:03
  • 1
    $\begingroup$ @chubakueno LaTeX is nice :D $\endgroup$
    – W.W.
    Oct 4, 2013 at 19:04

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