Convergence (Real Analysis)

Question

$X_1 = 2$ and $X_{n+1} = \frac{3}{5+2X_n}$ for $n\in\mathbb{N}$. Prove that $X_n$ is convergent and find $\lim_{n \to \infty}(X_n)$. How do I prove that the sequence is convergent?

Answer 1

(1). If it converges to $L,$ let $X_n=L+d_n.$ $$\text {Then }\quad L+d_{n+1}=x_{n+1}=3/(5+2X_n)=3/(5+2L+2d_n).$$ $$\text {So }\quad 2L^2+5L-3=-d_{n+1}(5+2L+2d_n).$$ Since the RHS above tends to $0$ as $n\to \infty,$ we have $2L^2+5L-3=0,$ so $L=1/2$ or $L=-3.$ By induction on $n,$ if $X_1\geq 0$ then $X_n\geq 0$ for all $ n.$ So if the limit $L$ exists then $L=1/2.$

(2). Let $X_n-1/2=d_n$. Then $$d_{n+1}=X_{n+1}-1/2=3/(5+2X_n)-1/2=3/(6+2d_n)-1/2=-d_n/(6+2d_n).$$ If $|d_n|<1$ then $|6+2d_n|>4$ so $$|d_{n+1}|=|d_n|/|6+3d_n|\leq |d_n|/4.$$ We have $|d_2|=1/6<1.$ By induction on $n$ we have $|d_n|\leq 4^{2-n}/6$ for $n\geq 2.$ So $d_n\to 0$ and $X_n\to 1/2.$

Remarks: The sequence converges to $1/2$ whenever $X_1>-5/2.$ It suffices that $X_n\geq 0$ for some $n$ and that $|d_n'|<1$ for some $n'\geq n.$ If $X_1>-5/2$ then $X_2>0.$ If $0<X_2<3/2$ then $|d_2|<1.$ If $X_2>3/2$ then $|d_3|<1.$

Answer 2

I will introduce the fixed point method. You can generalize this result directly. As given sequence is positive, we can construct a differentiable map $f:(0,\infty) \rightarrow \mathbb{R}$ such that $f(x)=\frac{3}{5+2x}$. We can check that $|f'|$ is bounded by some number $M\in(0,1)$. $f$ has a fixed point $x_0$, $f(x_0)=x_0$. By computations, $x_0=\frac{1}{2}$. Now, we have to show the convergence.

Mean value theorem implies following result.

(i) $|x_2-x_0|=|f'(c_1)(x_1-x_0)|$ for some $c_1$ between $x_0$ and $x_1$.

So, $|x_2-x_0|\le M|x_1-x_0 |$.

(ii) Likewise, $|x_3-x_0|\le M^2|x_1-x_0|$. Inductively, we can show that $|x_n-x_0|\le M^{n-1}|x_1-x_0|$. Hence, we are done.

Answer 3

It is interesting to note that you can obtain the convergence value by assuming it converges, and then using algebra. Suppose it converges to X.

Then X should be a stable value under the iterative step:

$X = \frac{3}{5+2X}$

In other words, we can guess that X is a stable attractor, and then use that to work out a proof.

Solving for X, you get $\frac{1}{2}$ or -3

It should be pretty easy to show that $X_n$ is never negative, and so, the value it converges to should be $\frac{1}{2}$.

So, what happens when we perturb $X = \frac{1}{2}$ by some small value, $\epsilon$, and then apply the iterative step?

Let's find out.

$\frac{3}{5+2(X+\epsilon)} - X = \frac{3}{5+2(X+\epsilon)} - \frac{3}{5+2X}=\frac{6\epsilon}{(5+2X)^2 + 2\epsilon} < \epsilon$

So, you see that if you perturb X by epsilon, and then apply the iterative step, you get another value that is even closer to X.

Now, all you have left is to put this into proof form.