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Let $a_1=0$ and $$a_{n+1}= \dfrac{a_n^2+1}{2 (a_n+1)}$$ $\forall n> 1.$ Show that sequence $a_n$ convergent.

I tried to prove $a_n$ is less than 1 by looking at few terms. But i failed to prove my claim.

Martin Sleziak
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Sophie Clad
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    What did you try already ? – Claude Leibovici Jan 13 '15 at 12:00
  • @Claude I tried to p r ove $a_n $ is less than 1 by looking at few terms .but i failed toprove my claim which is why i posted , please donot downvote – Sophie Clad Jan 13 '15 at 12:03
  • I shall not downvote, be sure ! And now, you received an answer. So, I stop thinking !! Cheers :-) – Claude Leibovici Jan 13 '15 at 12:11
  • @SophieClad try to prove that it is decreasing. – pointer Jan 13 '15 at 12:13
  • I shall try perhaps later to write an answer. Right now MathJax is making problems and every two second the answer gets uncoded, making it unbelievably annoying and hard to write decently. – Timbuc Jan 13 '15 at 12:27
  • I used the quotient ceiterium and got: $ 1/2 * ( 1/(a+1) + 1/(a^2 +a)) $ from there I would not know how to go further to prove this term is always less than 1.. Can someone offer a solution solution (?) – Imago Jan 13 '15 at 13:30
  • $$a_{n+1} - a_n = \frac{a_n^2 +1}{2(a_n + 1)} -a_n = -\frac{a_n^2 + 2a_n -1}{2(a_n + 1)}$$ So this iteration is the Newton method to solve $f(x) = x^2 + 2x -1 = 0$. Then verify the convergence condition for this method – Petite Etincelle Jan 13 '15 at 14:51
  • @SophieClad: Note that $$a_{n+1}+1=\dfrac{1}{2}\Big((a_n+1)+\dfrac{2}{(a_n+1)}\Big)$$ Now you can take another solution from here. – Bumblebee Jan 14 '15 at 13:01

3 Answers3

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Clearly $a_n\ge 0\forall n.$ $$a_{n+1}= \dfrac{a_n^2+1}{2 (a_n+1)}$$ $$a_n^2-2a_na_{n+1}+(1-2a_{n+1})=0$$ This is a quadratic equation of $a_n.$ Since $a_n\in\mathbb{R}$ we have $$4a_{n+1}^2-4(1-2a_{n+1})\ge 0$$ $$(a_{n+1}+1)^2\ge 2$$ Hence $$a_{n+1}\ge \sqrt{2}-1.$$ Now consider $$a_{n+1}-a_n=\dfrac{-a_n^2-2a_n+1}{2 (a_n+1)}=-\dfrac{(a_n+1+\sqrt{2})(a_n+1-\sqrt{2})}{2 (a_n+1)}\le 0.$$ Therefore $a_n$ is decreasing and bounded below for $n\ge 2$.

Bumblebee
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since $$a_{n+1}-(\sqrt{2}-1)=\dfrac{a^2_{n}+1-2(\sqrt{2}-1)(a_{n}+1)}{2a_{n}+2}=\dfrac{[a_{n}-(\sqrt{2}-1)]^2}{2a_{n}+2}\tag{1}$$ simaler $$a_{n+1}+\sqrt{2}+1=\dfrac{(a_{n}+\sqrt{2}+1)^2}{2a_{n}+2}\tag{2}$$ $\dfrac{(1)}{(2)}$ we have

$$\dfrac{a_{n+1}-(\sqrt{2}-1)}{a_{n+1}+\sqrt{2}+1}=\left(\dfrac{a_{n}-(\sqrt{2}-1)}{a_{n}+\sqrt{2}+1}\right)^2=\cdots=\left(\dfrac{a_{0}-(\sqrt{2}-1)}{a_{0}+\sqrt{2}+1}\right)^{2^n}\to 0$$ beause $$\dfrac{a_{0}-(\sqrt{2}-1)}{a_{0}+\sqrt{2}+1}<1$$ so $$\lim_{n\to\infty}a_{n}=\sqrt{2}-1$$

math110
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1) Clearly (add a little explanation here) the sequence is non-negative for all indexes

2) For $\;n\ge 2\;$ , the sequence is monotonic decreasing, since:

$$a_{n+1}\le a_n\iff \frac{a_n^2+1}{2(a_n+1)}\le a_n\iff a_n^2+2a_n-1\ge 0\iff$$

$$\iff (a_n+1+\sqrt2)(a_n+1-\sqrt2)\ge 0\iff\begin{cases}a_n\le -1-\sqrt2\\{}\\or\\{}\\a_n\ge-1+\sqrt2\end{cases}$$

The first case above is impossible, and the second one follows from

$$a_{n+1}\ge-1+\sqrt2\iff a_n^2+1\ge (2\sqrt2-2)a_n+2\sqrt2-2\iff$$

$$\iff a_n^2-2(\sqrt2-1)a_n-(2\sqrt2-3)\ge 0\iff (a_n+1-\sqrt2)^2\ge 0$$

and we're done since the last inequality is trivially true.

Now you can easily deduce the sequence converges, and its limit $\;x\;$ fulfills

$$x=\frac{x^2+1}{2(x+1)}\implies x=-1+\sqrt2$$

Timbuc
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  • How did you know sequence is decreasing before proving it .How to think about thesetyoes of questions – Sophie Clad Jan 14 '15 at 09:41
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    @SophieClad I didn't before the proof: I calculate the first few elements and by taking into account the form of the sequence one makes "and educated guess". Next, the proof must come in. – Timbuc Jan 14 '15 at 12:14
  • but relation is for n greater than 1 .we cannot put 1 in it to get next terms – Sophie Clad Jan 17 '15 at 13:05
  • @SophieClad Yes... so? I don't get your point./ – Timbuc Jan 17 '15 at 13:06
  • Point is how you calculated few terms of sequence – Sophie Clad Jan 17 '15 at 13:07
  • @SophieClad We're given $$a_1=0\implies a_2=\frac12\implies a_3=\frac{\left(\frac12\right)^2+1}{2\left(\frac12+1\right)}=\frac{\frac54}{2 \cdot\frac32}=\frac5{12};,;;\text{and etc.}$$ Still, I can't see your point...? – Timbuc Jan 17 '15 at 13:45
  • How have u got a2 = 1\2 ¿ , recurrence relationis valid for n greater than 1 not for 1 – Sophie Clad Jan 17 '15 at 14:21
  • @SophieClad But you wrote $;a_1=0;$ and then the recursive formula for $;n>1;$...read your own question again! And it obviously must be true also for $;n+1=2;$ , otherwise you have no sequence at all as you can not calculate even the first elements! – Timbuc Jan 17 '15 at 14:55