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Question

Let $a_1$ = 1 and $a_{n+1} = \sqrt{1+2a_n}$. Show that $\{a_n\}$ is increasing and bounded from above.

Attempted solution

showing that the sequence is increasing: \begin{align*} a_n &\geq 1 \quad \forall n = {1,2,3,...} \\ \Rightarrow 1+2a_n &\geq 3 \\ \Rightarrow \sqrt{1+2a_n} &\geq \sqrt{3} \\ \Rightarrow a_{n+1} &\geq \sqrt{3} \quad \forall n = {1,2,3,...} \\ \Rightarrow a_{n+1} &\geq a_n \end{align*} Is this a valid approach?

Assuming that I have shown that the sequence is increasing, I show that it is bounded by doing the following \begin{align*} a_{n+1} &\geq a_n \\ \Rightarrow \sqrt{1+2a_n} &\geq a_n \\ \Rightarrow 1+2a_n &\geq a^2_n \\ \Rightarrow (a_n+1)^2 - a^2_n &\geq a^2_n \\ \Rightarrow \frac{a_n + 1}{a_n} &\geq \sqrt{2} \\ \Rightarrow a_n &\leq 1+\sqrt{2} \quad \forall n = 1,2,3,... \end{align*}

Is there another more straightforward way to show that this sequence is bounded?

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    $\begingroup$ I don't get the conclusion $a_{n+1} \ge \sqrt{3} \implies a_{n+1} \ge a_n$. Do you guess that $a_n \le \sqrt 3$ for every $n$? If you do, you should include that in your induction proof... $\endgroup$
    – user251257
    Oct 27 '16 at 17:02
  • $\begingroup$ It would be false to assume $a_n \leq \sqrt{3}$ for every $n$ since, for example, $a_3 = \sqrt{1+2\sqrt{3}}$ which is greater than $\sqrt{3}$. My reasoning is that since $a_{n+1}$ and $a_n$ are always bounded below by $\sqrt{3}$ and 1, respectively, it should follow that $a_{n+1}$ must always lie above $a_{n}$ for every $n$ $\endgroup$
    – Joey
    Oct 27 '16 at 17:09
  • $\begingroup$ $2\ge \sqrt{3}$ and $3\ge \sqrt{3}$ does not imply $2 \ge 3$ :) $\endgroup$
    – user251257
    Oct 27 '16 at 17:11
  • $\begingroup$ I see your point. Have you got any advice on how to prove this sequence is increasing? Would an induction approach be the best way? $\endgroup$
    – Joey
    Oct 27 '16 at 17:14
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Hints:

  1. Let us assume that the statement is true, that is, $a_n$ increases and is bounded. Then, it converges to its supremum $s$ and by continuity we have $$ s = \sqrt{1 + 2s}. $$
  2. By solving that equation we obtain a guess / conjecture for the supremum of $a_n$.
  3. Try to prove that it is indeed the supremum of $a_n$. Prove by induction that for every $n\in\mathbb N$ we have $$ 1 \le a_n \le s$$ and $$ a_n \le a_{n+1}. $$
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  • $\begingroup$ Shouldn't it be $s = \sqrt{1+2s}$? $\endgroup$
    – Joey
    Oct 27 '16 at 18:39
  • $\begingroup$ @Joey thx. Was a typo $\endgroup$
    – user251257
    Oct 27 '16 at 18:43
  • $\begingroup$ Thank you! Think I've got it :) $\endgroup$
    – Joey
    Oct 27 '16 at 18:43
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We prove by induction that the generalized sequence

$$a(1)=1,a(n+1)=\sqrt{2 c \;a(n)+1},\;c>0$$

is

(i) increasing.

The quotient of two successive terms of the sequence is

$$q(n+1)=\frac{a(n+1)}{a(n)}$$

Obviously

$$q(2)=\sqrt{2 c+1}>1$$

Now

$$q(n+1)=\sqrt{\frac{2 c a(n-1) q(n)+1}{2 c a(n-1)+1}}>\sqrt{\frac{2 c a(n-1)+1}{2 c a(n-1)+1}}=1$$

QED.

(ii) the sequence is bounded.

In fact we have just proven that

$$a(n+1)=\sqrt{2 c a(n)+1}>a(n)$$

Hence

$$2 c a(n)+1>a(n)^2$$

$$1>a(n)^2-2 c a(n)=(a(n)-c)^2-c^2$$

or

$$c^2+1>(a(n)-c)^2$$

Which means that $\left| a(n)\right|$ cannot grow unbounded.

QED.

(iii) the sequence converges to one of the fixed points, i.e. solutions of

$$a^2=2 a c+1$$

giving

$$\text{a1}=\sqrt{c^2+1}+c,\text{a2}=c-\sqrt{c^2+1}$$

Since $a(1)>0$ and the sequence increases it canot become negative. Hence the limit is $\text{a1}$.

In the case of the OP we have $c=1$ and hence $a1 = 1+\sqrt{2}$.

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You have taken $a_1 = 1$. For this value $ { a_{n+1}} $ keeps on increasing. But had you taken $ a_1$ greater than $ 1\, + \, √2, a_{n+1}$ decreases. Now, for all values of $ a_1$ less than $1\, +√2, \, a_{n+1} $ increases. And as long as $a_n$ is less than $ 1+√2,$ the $a_{n+1}$ is also less than $1+√2.$ Thus for $a_1$ less than $1+√2,$ sequence is increasing and bounded by $ 1+√2.$

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