Consider the following sequence:

$$ \sqrt{2},\sqrt{2+\sqrt{2}},\sqrt{2+\sqrt{2+\sqrt{2}}} \cdots $$

a) Prove by induction that all terms of the sequence are bounded above by two.

b) Show that this sequence converges.

c) Calculate it's limit.

I'm having trouble with (c). How can I do it using the definition of a limit of a sequence?

Here are my answers for (a) and (b):


The sequence can be defined as: $$ x_1 = \sqrt{2}\\ x_{n+1} = \sqrt{2+x_n} $$ Now let's prove by induction. The base case, where $n=1$, we have immediately that $\sqrt{2}<2$. Let's then suppose that $x_n < 2$. Then we have:

$$ x_n + 2 < 2 + 2 = 4\\ x_{n+1} = \sqrt{2+x_n} =\sqrt{x_n+2} < \sqrt{4} = 2 $$

(b) Since $ x_{n+1} = \sqrt{2 + x_n} $ we can see that $x_{n+1}>x_n$ therefore that's a monotonic sequence. In (a) we showed that it is bounded above, and because $x_{n+1}>x_n$ we can say that $|x_n| < 2$, hence the sequence is bounded above and below. By Monotone Convergence Theorem, the sequence converges.

(c) From definition of limit of a sequence:

$$ \forall \epsilon > 0, \exists n_0 \in \mathbb{N} \text{ such that } \forall n > n_0 \rightarrow |x_n - L| < \epsilon $$

Now I need to find L... How can I apply that definition to find the limit?


marked as duplicate by dxiv, Michael Lee, Jack D'Aurizio real-analysis Apr 3 '18 at 4:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @dxiv It's different... I'm asking about the limit, not the convergence... I've showed it. $\endgroup$ – Bruno Reis Apr 3 '18 at 2:53
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    $\begingroup$ @BrunoReis I know you showed the convergence, in fact yours is a very well asked question and I +1'd it for that. But if you read the answers under the linked post you'll find the respective limit derived in a number of different ways there. For that reason, I am maintaining my close-as-duplicate vote. $\endgroup$ – dxiv Apr 3 '18 at 2:56

In the last point you should not use the $\epsilon$–$\delta$ definition. You know that the limit exists. Let it be $L$. Then $$ \sqrt{2+L}=L $$ or $2+L=L^2$, $L^2-L-2=0$, $(L-2)(L+1)=0$. But $L>0$, so $L=2$.

  • $\begingroup$ Why can you assume that $\sqrt{2+L} = L$ just by knowing that the limit exist? $\endgroup$ – Bruno Reis Apr 3 '18 at 2:55
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    $\begingroup$ @BrunoReis Beacause $x_{n+1}=\sqrt{2+x_n}$, and $x_n$ (and all its subsequences) tends to $L$. $\endgroup$ – Przemysław Scherwentke Apr 3 '18 at 3:01
  • $\begingroup$ $x_{n+1} = f(x_n)$ for $f$ continuous. If $x_n\to L$, then $f(x_n) \to L$ and, of course, $x_{n+1} \to L$. $\endgroup$ – Clement C. Apr 3 '18 at 3:02
  • $\begingroup$ @PrzemysławScherwentke Ohhh that's right. If a sequence converges to L all of it's subsequence converges to L too.. Thank you :) $\endgroup$ – Bruno Reis Apr 3 '18 at 3:05

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