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I was given the following exercise as homework: find the limit of $b_{n+1} = \sqrt{2 + b_n}$, $b_1 = \sqrt{2}$, with a hint that $b_n < 2 \forall n \in \mathbb{N}$. I have proven that $b_n$ is strictly monotonically increasing and strictly bounded above by $2$ by using induction over $\mathbb{N}$ (which means that it has a limit, by the monotone convergence principle). However, now I need to prove that

$$2 = \text{sup}\{b_n : b \in \mathbb{N}\}$$

In order to prove that the limit is indeed $2$. By definition, I would only have to prove that $\forall \epsilon > 0 \exists n \in \mathbb{N} : 2 - \epsilon < b_n$, which is intuitively true, but I have been so far unable to prove it.

Any hints will be appreciated.

Servaes
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Misguided
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1 Answers1

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You have proved that the limit exists, so let's set $$b:=\lim_{n\to\infty}b_n.$$ Given the recurrence relation $b_{n+1}^2=2+b_n$ we can take limits on both sides to find $$\lim_{n\to\infty}b_{n+1}^2=\lim_{n\to\infty}(2+b_n)=2+\lim_{n\to\infty}b_n=2+b.$$ Now on the left-hand side we have $$\lim_{n\to\infty}b_{n+1}^2=\left(\lim_{n\to\infty}b_{n+1}\right)^2=b^2.$$ This gives us the relation $b^2=2+b$. Solving for $b$ shows that either $b=-1$ or $b=2$. As you've shown that the sequence is increasing with $b_1=\sqrt{2}$, we conclude that $$\lim_{n\to\infty}b_n=2.$$

Servaes
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  • Could you rephrase the argument in a more formal way? it seems way too hand-wavy – Misguided Aug 28 '15 at 20:15
  • What part of it seems hand-wavy to you? – Servaes Aug 28 '15 at 20:17
  • The fact that you are using continuity as an argument for exchanging the limit with function application. We can't use continuity yet, since we haven't even defined the limit of a function. – Misguided Aug 28 '15 at 20:19
  • I'm not taking the limit of a function. I'm taking the limit of a sequence. Being able to take a limit inside a function is precisely what the continuity of a function means. But let me rephrase my answer. – Servaes Aug 28 '15 at 20:22
  • Right, but continuity is defined in terms of function limits. The point is, using continuity is not allowed, since we haven't defined it yet, which means that putting the limit inside the function application is applying a theorem we a priori do not know – Misguided Aug 28 '15 at 20:26
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    I've rephrased to use only that limits distribute over multiplication and addition, whenever both/all limits are finite. Is this okay to use for you? Otherwise these facts are certainly worth proving when working with limits! – Servaes Aug 28 '15 at 20:27