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Show that the sequence is increasing and bounded by 3. Find out if it converges, and find its limit. So far i think the limit is $$ 1+\sqrt{2}, $$ and that i should use induction to find that its bounded by 3. I have solved everything except the convergence part. Can someone show me, and relieve my headache? thanks for any help!

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  • $\begingroup$ I gave you a freebie in terms of edits. Please check the edits in particular the title to see if what I did makes sense.. $\endgroup$
    – Chinny84
    Nov 13, 2014 at 12:43
  • $\begingroup$ Have you used induction? The hint is very clear. $\endgroup$
    – John
    Nov 13, 2014 at 12:47
  • $\begingroup$ I have problem understanding induction on more advanced sequences. if the sequence is increasing, why does it have a limit, i can't grasp that. $\endgroup$
    – alf
    Nov 13, 2014 at 14:26
  • $\begingroup$ Because it is bounded above. All bounded above increasing sequences have a limit. You may use induction to prove this sequence both bounded above and increasing. $\endgroup$
    – Alistair
    Nov 13, 2014 at 14:41
  • $\begingroup$ thanks Alistair, but i still don't understand how to apply induction. $\endgroup$
    – alf
    Nov 13, 2014 at 17:28

2 Answers 2

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Assume that, as $n\to\infty$, the limit of $a_n$ is $u$, hence $a_{n+1}=a_n=u$.

$$\begin{align} a_{n+1}&=\sqrt{1+2a_n}\\ u&=\sqrt{1+2u}\\ u^2-2u-1&=0\\ \because u>0\therefore u&=1+\sqrt{2} \end{align}$$


NB: $$\begin{align} a_1&=1\\ a_2&=\sqrt{1+2}=\sqrt{3}\\ a_3&=\sqrt{1+2\sqrt{3}}\\ a_4&=\sqrt{1+2\sqrt{1+2\sqrt{3}}}\\ a_5&=\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{3}}}}\\ \vdots \\ u=\lim_{n\to\infty}{a_n}&=\sqrt{1+2\underbrace{\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\cdots}}}}}_{u}}\\u&=\sqrt{1+2u}\\ u^2-2u-1&=0\\ \because u>0\therefore u&=1+\sqrt{2} \end{align}$$

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  • $\begingroup$ This argument assumes the limit exists. $\endgroup$ Nov 13, 2014 at 16:25
  • $\begingroup$ Yes the answer is for the second part, finding the limit. $\endgroup$ Nov 13, 2014 at 16:29
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Let us see if we can provide an answer to your first two question.

Use induction to show that the sequence $a_n$ is bounded by $3$. Remember that induction goes as follows. Presume that the statement is correct for some $n$, then demonstrate that it is also true for $(n+1)$. Apply this line of reasoning to the starting value. From this one can conclude the statement is true for all $n$.

Okay, let us assume that $a_n<3$ for some arbitrary $n$. It follows that $a_{n+1} < \sqrt {7}$. Since $\sqrt{7} < 3$, we can conclude that $a_{n+1} < 3$. For the initial value $a_1 = 1$ the statement is certainly true. Hence all $a_n$ in the sequence are bounded by $3$. [In fact the sequence is bounded by $1+\sqrt{2}$]

To show that the sequence is increasing, consider the difference $D_n$ between two consecutive terms, $a_{n+1}$ and $a_n$. We find $D_n = a_{n+1} - a_n = \sqrt{2a_n+1} - a_n$. If you examine this function, you will see that it is positive in the interval $(0, 1+\sqrt {2})$. And since $a_n$ are bounded by $1+\sqrt 2$, it follows that the sequence is indeed monotonically increasing.

UPDATE As hypergeometric has shown, the limit of the sequence must be $u = 1 + \sqrt 2$. Let us examine analytically whether the series indeed converges to this value. Define $D(n)$ as the difference between $u$ and $a(n)$. We already know that $D(n)$ is positive, because the sequence is increasing and hence approaches $u$ from below. Therefore: $u - D(n+1) = \sqrt{1 + 2 (u - D(n))}$. Taking both sides to the power $2$ yields after cancellation: $2uD(n+1) - (D(n+1))^2 = 2D(n)$. From this it follows that $D(n+1) < \frac {D(n)}u$. The sequence converges indeed. And in fact it does so very quickly. At every step the difference between $u$ and $a(n)$ decreases by at least a factor $2.414$.

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  • $\begingroup$ What about convergence? how do i solve it? $\endgroup$
    – alf
    Nov 14, 2014 at 13:39
  • $\begingroup$ alf - hypergeometric has provided this answer very clearly $\endgroup$
    – 123
    Nov 14, 2014 at 15:47
  • $\begingroup$ @mathtastic, hypergeometric only answered the second part, about finding the limit, assuming it exists. The OP wants to know why the sequence converges. $\endgroup$ Nov 14, 2014 at 15:53
  • $\begingroup$ Yes! Mr OP wants to know why the sequence converges, not just that it do. $\endgroup$
    – alf
    Nov 14, 2014 at 19:59

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