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Let ${a_n}$ be the sequence where $a_1 =\sqrt{2}$, and $a _{n+1} = \sqrt{2+ \sqrt{a_n}}$ for all n ∈ N. Prove that ${a_n}$ is bounded above by 2, and then prove that ${a_n}$ converges.

My thought is to generally employ the monotonic convergence theorem:

Proof.

Because $a_1= \sqrt{2}$, and the sequence is increasing, we see that it is bound below by $\sqrt{2}$. And, if we take the limit as n approaches infinity, we see the limit equals 2, so it is bounded above by 2.

Further by the monotonic convergence theorem, as we see that the sequence is monotonically increasing and bounded above and below, it follows through the M.C.T. that it converges. QED

My question really has to do with showing that it is monotonically increasing, which I believe I need to use induction for, however, I am not sure how exactly to do that, as my book glosses over the subject in a generic review of the concept, i.e. noting the notion that you do $s(n)$ and then show that $s(n+1)$ is also true for all n, but I'm fuzzy on the subject. Moreover, am I correct with my bound assumptions?

Greg Martin
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Bad at algebra and proofs
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  • Since you have to prove that the sequence is convergent, you cannot use "limit" to check the bound. For your own understanding it is alright! – Aniruddha Deshmukh Jun 20 '18 at 06:00

3 Answers3

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My question really has to do with showing that it is monotonically increasing

Hint:   $\,a_{n+1}-a_n\,$ and $\,a_{n}-a_{n-1}\,$ have the same sign, so in the end all differences between consecutive terms have the same sign as $\,a_2-a_1\,$:

$$\require{cancel} \begin{align} a _{n+1} - a_{n} &= \left(\sqrt{2+ \sqrt{a_n}} - \sqrt{2+ \sqrt{a_{n-1}}}\right) \cdot \dfrac{\sqrt{2+ \sqrt{a_n}} + \sqrt{2+ \sqrt{a_{n-1}}}}{\sqrt{2+ \sqrt{a_n}} + \sqrt{2+ \sqrt{a_{n-1}}}} \\ &= \dfrac{(\cancel{2}+\sqrt{a_n})- (\cancel{2}+\sqrt{a_{n-1}})}{\sqrt{2+ \sqrt{a_n}} + \sqrt{2+ \sqrt{a_{n-1}}}} \\ &= \dfrac{\sqrt{a_n} - \sqrt{a_{n-1}}}{\sqrt{2+ \sqrt{a_n}} + \sqrt{2+ \sqrt{a_{n-1}}}} \end{align} $$

dxiv
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Clearly $a_2=\sqrt{2+\sqrt{a_1}}>\sqrt2=a_1$. Assume inductively that $a_{n+1}>a_n$. Since $a_n>0$, $$\sqrt{a_{n+1}}>\sqrt{a_n},$$ $$2+\sqrt{a_{n+1}}>2+\sqrt{a_n},$$ $$\sqrt{2+\sqrt{a_{n+1}}}>\sqrt{2+\sqrt{a_n}},$$ $$a_{n+2}>a_{n+1}.$$

Angina Seng
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Hint.

$a_\infty=\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}$

$a_\infty=\sqrt{2+a_\infty}$

Chris2018
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  • This doesn't help unless you first prove that the limit $,a_\infty,$ does in fact exist, which is precisely what OP's question asks for help with. – dxiv Jun 20 '18 at 06:17