Is my proof for this limit correct?

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)?

Thanks,

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

• 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 $a_n$. – User001 Jul 2 '15 at 1:20
• Can we bound this sequence without knowledge of its limit being 2? – User001 Jul 2 '15 at 1:21
• 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. – aziiri Jul 2 '15 at 1:27
• yes, if a sequence is bounded and monotonic in $\mathbb{R}$ then it converges & you're welcome. – aziiri Jul 2 '15 at 1:37
• yes it is correct. – 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.

• This only works if the limit exist and finite. – aziiri Jul 2 '15 at 1:05
• Yeah you are right. I just point out the unnecessary part of his proof. – corindo Jul 2 '15 at 1:10
• 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. – corindo Jul 2 '15 at 1:30