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$X_1=2^{1/2}$ and $X_{n+1}=(2+X_n)^{1/2}$ for each $n\ge 1$.

Find the supremum of the set $\{ X_n \}$ and prove it.

I figured that the supremum should be $2$ by just examining the elements of the set and from the fact that each element is less than $2$. Is there a proper way to find and prove the supremum of this set?

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We can reduce finding the supremum of $x_n$ to finding the limit. This follows because this is an increasing sequence, given the initial condition.

Firstly, as you mentioned, $x_n < 2$ for all $n$. This can be demonstrated by induction. Moreover, all the terms are positive.

Now we would like $$x_{n+1} > x_n$$ this gives us the equation: $$0 \ge (x_n +1)(x_n -2).$$ This is satisfied provided that $$-1 < x_n < 2,$$ which we have already established.

We need to find the limit of this sequence. It is a bounded monotonic sequence, so there is a limit. Let $L$ be this limit. We can find an algebraic equation that this limit must satisfy:

$$L=\lim_{n\to\infty} x_n = \lim_{n\to\infty} (2+x_n)^{1/2} = (2+L)^{1/2}$$ the last step follows by continuity of $(2+x)^{1/2}.$ Finally this means that $L$ satisfies: $$L^2 - L - 2 = 0$$ and hence $L=-1$ or $L=2$. $L\neq -1$ since it is the limit of a sequence of positive real numbers, and $[0,\infty)$ is closed. Therefore $L=2$. Since this sequence is increasing to $L$, that is our supremum.

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  • $\begingroup$ If <Xn> is a monotonically increasing bounded above sequence then <Xn> converges to an element in R which in fact the supremum of the set {Xn}.Is that the theorem that we have to use? and If we actually find the limit of <Xn> we can conclude the fact that it is in fact the supremum of the set {Xn} .And also if we know the supremum of the set {Xn} we can conclude that its the limit of the sequence. Am i right? $\endgroup$ – Razor1692 May 7 '15 at 15:55
  • $\begingroup$ Yes that is correct. First of all, $R$ bounds the set from above. Thus $\sup x_n < R$. If the supremum was less than $R$, say $S$, then there is an $\epsilon > 0$ for which $S < R-\epsilon < R$. Since $x_n$ converges to $R$, there is an $n$ for which $R-\epsilon < x_n < R$, and thus $S$ cannot be the supremum, a contradiction. $\endgroup$ – Joel May 7 '15 at 15:58
  • $\begingroup$ Suppose that we have to prove that the supremum is 2.If im going to use the definition of the supremum that supXn =L iff 1)for each n an element of N , Xn<=L and 2) For each epsilon>0 ,There exists Xm an element of Xn such that Xm>L-epsilon. I can show the first condition but when i try to prove the second condition im stucked in finding a particular Xm such that Xm>2-epsilon... how can i proceed with this? $\endgroup$ – Razor1692 May 7 '15 at 16:03
  • $\begingroup$ Honestly, I don't believe you will be asked to find a particular $m$. You can demonstrate that one exists through the convergence of the sequence. That is sufficient. $\endgroup$ – Joel May 7 '15 at 16:08
  • $\begingroup$ Yeah i guess that is sufficient since it is difficult to show that there exists a particular m.Well since there is a supremum i guess that it should be possible to show that there exists a particular m.Do you have any idea of a way to show that?.Anyways thank you for the answer it helped a lot :) $\endgroup$ – Razor1692 May 7 '15 at 16:13
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  • Prove that $X_n \leq 2$ for all $n$. This is immediate from induction.
  • Prove that $X_n$ is monotone.
  • Hence, by monotone sequence theorem, $\lim_{n} X_n = L$ exists.
  • This gives us $L = \sqrt{2+L} \implies L = 2$.
  • Conclude that $2$ is the supremum.
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