I want to prove that $\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}$ limits to 2.

Let $a_0$ = $\sqrt{2}$

$a_n$= $\sqrt{2+a_{n-1}}$.

Then, proving that $\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}$ limits to 2 is equivalent to proving that lim $a_n$ =2.

Write $a_n$ = $e^{log(\sqrt{(2+a_{n-1})}}$

Then lim $a_n$= $$lim (e^{log(\sqrt{(2+a_{n-1})}})$$ and, by continuity of the exponential function $$=e^{lim(log(\sqrt{(2+a_{n-1})})},$$

$$=e^{lim(\frac{1}{2}log({2+a_{n-1})})},$$ and, by continuity of the log function $$=e^{\frac{1}{2}log(lim({2+a_{n-1})})},$$ $$=e^{\frac{1}{2}log({2+ lim(a_{n-1})})},$$ $$=e^{\frac{1}{2}log({2+ lim(a_n)})},$$ $$=(2+ lim(a_n))^{\frac{1}{2}},$$

which implies that $$(lim a_n)^2 = 2+ lim(a_n)$$ $$=(lim a_n)^2- lim(a_n)-2 = 0$$

So, the limit, if it exists, must satisfy the above quadratic equation.

The roots are 2 and -1, but since the sequence is positive for all n, we must have that the limit exists and is equal to 2.

Is this correct?

I know that there are many proofs of this on MSE, but I am wary of those proofs that seem to assume that the limit already exists, and that all that is needed is to label lim$a_n$ = L and then solve an algebraic equation and arrive at L=2. Is this way not rigorous and perhaps not allowed in a real analysis exam (unlike, say, calculus I)?


  • $\begingroup$ DO you mean the nested limit $\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}$? $\endgroup$ – GohP.iHan Jul 2 '15 at 1:03
  • $\begingroup$ Yeah, this is correct but not as simple as it could be. $\endgroup$ – JacksonFitzsimmons Jul 2 '15 at 1:03
  • $\begingroup$ Yes, sorry, I'll change it, @GohP.iHan. $\endgroup$ – User001 Jul 2 '15 at 1:03

No, you should prove that the limit exists and finite in the first place. This is a common error that students make.

How to prove that the limit exist and finite?

1) We prove by induction that $a_n\in (0,2)$ for any $n\in \mathbb{N}$ : if $2>a_n>0$ then $$2=\sqrt{2+2}>\sqrt{2+a_n}=a_{n+1}=\sqrt{2+a_n}>0.$$
and clearly $a_n\in (0,2)$.

2) Now, we show that $(a_n)$ is increasing, note that $$a_{n+1}-a_n=\sqrt{a_n+2}-a_n=\frac{a_n+2-a_n^2}{\sqrt{a_n+2}+a_n}=\frac{(a_n+1)(2-a_n)}{\sqrt{a_n+2}+a_n}>0.$$ Now, $(a_n)$ is increasing and bounded therefore convergent to some $L\in [0,2]$, the rest follows.

Can you give me an example where this does not work?

take for example $a_0=0$ and $a_{n+1}=2a_n+1$. I haven't much clever counter-example, if I find one, I'll post it.

  • $\begingroup$ Hi @aziiri, I was afraid of that, too. BTW, what if we didn't know the limit was 2? I don't really understand your induction proof then to show the boundedness of $a_n$. $\endgroup$ – User001 Jul 2 '15 at 1:20
  • $\begingroup$ Can we bound this sequence without knowledge of its limit being 2? $\endgroup$ – User001 Jul 2 '15 at 1:21
  • 1
    $\begingroup$ you can bound it with any real number $>2$ but $2$ is the best upper bound you can have, the induction is straight forward : Assume that $a_n>0$ then $a_{n+1}>0$, assume that $a_n<2$ then $a_{n+1}<\sqrt{2+2}=2$, and $a_0\in (0,2)$. Done. as for finding the limit, generally you should try to study the behavior of the sequence or even calculate some terms. $\endgroup$ – aziiri Jul 2 '15 at 1:27
  • 1
    $\begingroup$ yes, if a sequence is bounded and monotonic in $\mathbb{R}$ then it converges & you're welcome. $\endgroup$ – aziiri Jul 2 '15 at 1:37
  • 1
    $\begingroup$ yes it is correct. $\endgroup$ – aziiri Jul 2 '15 at 1:42

You don't have to pass through the exponential.

$$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \sqrt{2 + a_{n-1}} = \sqrt{2 + \lim_{n\to\infty} a_{n-1}} = \sqrt{2 + \lim_{n\to\infty} a_{n}} .$$

Let $A = \lim_{n\to\infty} a_n$, then $$ A = \sqrt{2 + A}$$ So $A = 2$.

P.S. @aziiri is right. To complete the proof you have to prove that the limit exists and is finite. The idea is to prove that all the $a_n$s locate in $(0,2)$ and they are increasing.

  • 1
    $\begingroup$ This only works if the limit exist and finite. $\endgroup$ – aziiri Jul 2 '15 at 1:05
  • 1
    $\begingroup$ Yeah you are right. I just point out the unnecessary part of his proof. $\endgroup$ – corindo Jul 2 '15 at 1:10
  • $\begingroup$ Ok, sorry, I have misread your answer. $\endgroup$ – aziiri Jul 2 '15 at 1:11
  • $\begingroup$ Ok, got it - thanks Corindo. I am currently trying to bound the sequence, without knowledge that its limit is 2. (In case this question shows up on an exam, but instead of 2's in the square roots, there are 3's in the square roots... ) $\endgroup$ – User001 Jul 2 '15 at 1:27
  • 2
    $\begingroup$ Of course you can bound the sequence. A common way is to set a bound $M$ and then, supposing that $a_{n-1} < M$, prove that $a_{n} < M$ too. And the choice of $M$ is left to good sense. $\endgroup$ – corindo Jul 2 '15 at 1:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.