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I need a little bit of help (just a hint, please) with an induction proof on this sequence, which I need to prove is bounded above by 3. $$ a_1 = \sqrt{2} $$ $$ a_{n+1} = \sqrt{2 + a_n} $$

My attempt: $$ a_k < 3 $$ $$ a_k + 2 < 5 $$ $$ \sqrt{a_k + 2} < \sqrt{5} $$ $$ a_{k+1} < \sqrt{5} $$ ... and I don't know where to go from here.

If I were to find a limit of this sequence, which way would I have to go? Should I try to rewrite the sequence into a formula?

user35443
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3 Answers3

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Once you have $a_{k+1} < \sqrt{5}$, you can use that $\sqrt{5} < 3$ to prove that $a_{k+1} < 3$. Hence by induction all terms of this sequence are bounded by $3$.

Now for the limit part, your sequence is bounded above, if you can show that it is an increasing sequence then it follows (see a theorem about monotone convergence) that the sequence should have a limit. Once that is established you can assume that $\lim_{n \to \infty}a_n=a$. Now you have $$\lim_{n \to \infty}a_{n+1}=\sqrt{2+\lim_{n \to \infty}a_{n}}.$$ Solve for $a$.

Anurag A
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  • And how to you solve for $a$ here? @Anurag A – Andrej Nov 16 '18 at 21:18
  • Because if you let $L = \lim_{n \rightarrow \infty}a_{n+1} = \lim_{n \rightarrow \infty}a_n$ you get the equation $L = \sqrt{2 + L}$ and then $L^2 = L + 2 = L^2 - L - 2 = 0$ and then $(L - 2)(L + 1) = 0$, so you get $L_1 = 2$ and $L_2 = -1$. But what to do with $L_2 = -1$? – Andrej Nov 16 '18 at 21:24
  • @user124942 Since the sequence consists of all positive terms so the limit cannot be negative. Thus your limit is $L_1=2$. – Anurag A Nov 17 '18 at 02:07
  • Ok. Cool. ;) Problem solved. ;) @Anurag A – Andrej Nov 17 '18 at 11:22
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How to find a limit of recursive sequence $$a_{n+1} = \dfrac{a_n^2 + 1}{2}$$? @Anurag A

Andrej
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Then you can write $$a_{k+1}<\sqrt 5<3$$therefore $a_k<3$ means $a_{k+1}<3$. Since $a_1=\sqrt 2<3$ therefore $a_2<3,a_3<3,a_4<3,\cdots $ and $a_k<3$ for all $k$.

Mostafa Ayaz
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