$X_1=2^{1/2}$ 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?


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.

  • $\begingroup$ 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
  • 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.

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.