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I would like to specify an upper bound of a sequence where $a_1=\sqrt2$ and each subsequent term is given by $a_{n}=\sqrt{2+ a_{n-1}}$

I know this sequence converges to 2 from a separate proof and also that it is monotonically increasing (one can prove this with induction), so any number larger than 2 works for an upper bound. That being said, I would like to approach this by showing a contradiction in the negation statement for boundedness. In other words, I want to explicitly state a $u$ so that there does not exist a $n$ where the $n^{th}$ term of the sequence is larger than $u$, and show that this is the true with some sort of algebraic manipulation.

I have looked at a few similar questions on this site and elsewhere, but I have not yet found an instance where one shows boundedness in this way.

EDS
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1 Answers1

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Use induction. (I've taken $u = 4$ below.)

We have $a_1 = \sqrt{2} < 4$. Now, let $n \in \mathbb{N}$, and suppose $a_n < 4$. Then $$ a_{n+1} = (2+\sqrt{a_n}\,)^{1/2} < (2 + \sqrt{4} \,)^{1/2} = 2 < 4. $$ Therefore, $a_n < 4$ for all $n$, by mathematical induction.

(I could have taken $2$ for $u$, but I chose $4$ for $u$ to make things nicer.)

Mark Twain
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  • Exactly what I was looking for. With the edited definition of the sequence, for u=4, we have a_{n+1}<sqrt(2+4)<4 and reach the same conclusion. – EDS Dec 18 '17 at 02:02