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I've been trying to solve this exam question on an exam in real analysis. Thus, only such methods may be used. The problem is as follows.

Show that the sequence $a_n$ defined by $a_1=1$ and $a_{n+1}=(3+a_n)^{-1}$ for $n=1,2,\dots$ converges and determine the limit.

I'm not really able to find a closed form for the sequence. I guess that computing that and then showing that it's a Cauchy sequence works for showing convergence.

I just have to find that closed form first, which I'm unable to do. If there's a general strategy I'd love to see how that works as I've been unable to find any.

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    $\begingroup$ Hint: monotonicity. Maybe monotonicity for subsequences. $\endgroup$ – xbh Aug 19 '18 at 19:56
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Recursively $a_n>0$, $f(x)={1\over{x+3}}$, $f'(x)=-{1\over{(x+3)^2}}$, implies that $|a_{n+2}-a_{n+1}|=|f'(c_n)||a_{n+1}-a_n|\leq {1\over 3^2}|a_{n+1}-a_n|$, $c_n$ is an element of $(a_n,a_{n+1})$ or $(a_{n+1},a_n)$. You deduce that the sequence is a Cauchy sequence. (recursively you can show that $|a_{n+1}-a_n|\leq {1\over 3^{2(n-2)}}|a_2-a_1|$.

so it converges, the limit verifies $f(l)=l$ which implies that $l^2+3l-1=0$ ans is the positive root of this quadratic equation.

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There need not be such a closed form for the sequence. If a limit $l$ exists it must be that$$l=\dfrac{1}{l+3}$$which yields to $$l=-1.5\pm\sqrt{3.25}$$where $l=-1.5-\sqrt{3.25}$ is not possible. To show that the sequence converges to $\sqrt{3.25}-1.5$ lets define $$e_n=a_n-(\sqrt{3.25}-1.5)$$therefore $$e_{n+1}=a_{n+1}-(\sqrt{3.25}-1.5)=\dfrac{1}{a_n+3}-(\sqrt{3.25}-1.5)=\dfrac{1-(e_n+1.5+\sqrt{3.25})(\sqrt{3.25}-1.5)}{a_n+3}=\dfrac{-e_n(\sqrt{3.25}-1.5)}{a_n+3}$$therefore $$|e_{n+1}|<\dfrac{|e_n|(\sqrt{3.25}-1.5)}{3}$$which shows that $e_n\to 0$ and $$a_n\to \sqrt{3.25}-1.5$$

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  • $\begingroup$ +1. I had thought of the same approach. Do you have something in mind to show that the convergence happens without having to assume that the limit exists first? $\endgroup$ – Vizag Aug 19 '18 at 20:18
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    $\begingroup$ Well! Consider $$a_{n+2}=\dfrac{1}{3+a_{n+1}}=\dfrac{a_n+3}{3a_n+10}<\dfrac{4}{10}a_n$$ therefore we can conclude what we want $\endgroup$ – Mostafa Ayaz Aug 19 '18 at 20:23
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Monotonicity method.

Rewrite the recursive relation as $$ a_{n+2} = \frac 1 {3+ \dfrac 1 {3 + a_n}} = \frac {3+a_n} {3(3+a_n) + 1} = \frac {a_n + 3} {3 a_n + 10}, $$ or $$ a_{n+2} = \frac 13 - \frac 1{9a_n + 30} \leqslant \frac 13. $$ Since $$ a_{n+4}-a_{n+2} = \frac 1 {9a_n + 30} - \frac 1{9a_{n+2} +30} = \frac {9(a_{n+2}-a_n )} {(9a_n +30) (9a_{n+2}+30)}, $$ and clearly $a_n \geqslant 0$, we conclude that subsequences $a_{2k}, a_{2k+1}$ are monotonic [not necessarily the same type]. Since $a_1 = 1, a_2 = 1/4, a_3 = 4/13, a_4=13/34, a_1 > a_3, a_2 < a_4$, we know that $a_{2k} < a_{2k+2} \leqslant 1/3, 1 \geqslant a_{2k+1} > a_{2k+3}$. Hence $a_{2k}, a_{2k+1}$ both converge [since both are monotonic and bounded]. By the recursive relation, they converge to the same limit $L > 0$.

Let $n \to \infty$ in the first formula, $$ L = \frac {L + 3}{3L + 10}, $$ or simply, $$ L(L+3) =1. $$

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Here is a somewhat systematic approach. The crux of the method is to separate the numerator and denominator. Write $a_n = p_n / q_n$. Can we work out $p_{n+1}, q_{n+1}$? We have $$x_{n+1} = \frac{1}{3 + p_n/q_n} = \frac{q_n}{3q_n + p_n}$$ So $p_{n+1} = q_n$ and $q_{n+1} = 3q_n + p_n = 3q_n + q_{n-1}$. Now we can see that the important value is the denominator. We are lucky because our second equation is a second order recurring relation for $q_n$, so we can solve for the general form using the following theorem:

Let $(x_n)_{n \ge 0}$ be a sequence such that there exists $(a,b)\in \mathbb{C}^2$ $x_{n+2} = ax_{n+1} + bx_n$. Then the sequence $(u_n)_{n \ge 0}$, $u_n = r^n$ is a solution iff $r$ satisfies the characteristic equation: $$r^2 - ar - b = 0$$ Furthermore if the characteristic equation admits two distinct roots then any solution is of the form $$\alpha r_1^n + \beta r_2 ^n$$

We don't mention the case of repeated roots. In our case the characteristic equation is $$r^2 - 3r - 1 = 0$$

The two roots are $j_1 = \frac{3 + \sqrt{13}}{2}, j_2 = \frac{3-\sqrt{13}}{2} = $. The best part is that we don't need to solve for $\alpha$ and $\beta$ (though you do need to check $\alpha \ne 0$, which is easily done); notice that $|j_2| < 1$. Now we can write $$x_n = \frac{p_n}{q_n} = \frac{q_{n-1}}{q_n} = \frac{\alpha j_1^{n-1} + \beta j_2^{n-1}}{\alpha j_1^{n} + \beta j_2^{n}}$$ Since $|j_2|<1$, as $n\to \infty$, this goes to $1/j_1$. You can check that $1/j_1$ satisfies $L = 1/(3 + L)$

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