Show convergence and find the limit of the sequence given by $a_1=1$ and $a_{n+1}=\frac{1}{3+a_n}$

Question

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.

Answer 1

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.

Answer 2

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 $$a_n\to \sqrt{3.25}-1.5$$

Answer 3

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, $$ L(L+3) =1. $$

Answer 4

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 $L = 1/(3 + L)$