Prove that the sequence $(_)$ defined below converges, and find its limit $_1= 2, _{n+1} = \frac{2}{1 + _n}$, for all $n\in \mathbb{N}$.

Question

I tried to use the MCT monotone convergence theorem but it failed since it is not monotone. So I’m not sure how to prove that it convergence, I thought about using Cauchy definition. I found that the limit is -2 and 1 but I’m not sure which one to select since it is not monotone which value it converges to.

Answer 1

Let X=$$\begin{array}{l}\lim_{n\to\infty}x_n=\frac{2}{1+\frac{2}{1+\frac{2}{...\infty}}}\end{array}$$

Implies X=2/(1+x)

X^2+X-2=0

(X+2)(X-1)=0

Since X is obviously>0

X=1

Answer 2

if $x_{n}$ has limit, so does $x_{n+1}$. Let's denote $\lim\limits_{n \to \infty} x_{n} = a \Rightarrow a = \frac{2}{1+a} \Rightarrow a^2 + a - 2 = 0$. The roots of this equation are 1 and -2, but -2 can't be the limit, because $x_{n} > 0 \ \forall n$, thus $\lim\limits_{n \to \infty} x_{n} = 1$

Answer 3

Claim: $$ x_n =\frac{2^{n+1}+2(-1)^{n+1}}{2^{n+1}+(-1)^n} $$ For $n=1$, we have $x_1=6/3=2$. Further, $$ x_{n+1} (1+x_n)= \frac{2^{n+2}+2(-1)^{n}}{2^{n+2}+(-1)^{n+1}}\cdot\left(1+\frac{2^{n+1}+2(-1)^{n+1}}{2^{n+1}+(-1)^n}\right )=2 $$This clearly gives $\lim\limits_{n\to\infty}x_n=1$.