Convergence of $\sum_{n=1}^{\infty}(x_{n+1}-x_n)$ if $x_{n+1}=\frac{k+x_n}{1+x_n}$

Question

Problem

Assume that $x_{n+1}=\dfrac{k+x_n}{1+x_n},k>1,x_1 \geq 0.$

1) Prove $\sum\limits_{n=1}^{\infty}(x_{n+1}-x_n)$ is absolutely convergent.

2) Evaluate $\sum\limits_{n=1}^{\infty}(x_{n+1}-x_n)$.

Attempt

By induction, we can obtain $x_n\geq0$, namely $1+x_n\geq 1$. Thus $$x_{n+1}=\frac{k+x_n}{1+x_n}=\frac{1+x_n+(k-1)}{1+x_n}=1+\frac{k-1}{1+x_n}\leq 1+(k-1)=k.$$

Consider the function $$f(x)=\frac{k+x}{1+x}, x \in [0,k].$$ We obtain $$f'(x)=\frac{1-k}{(1+x)^2}.$$ Therefore $$|f'(k)|\leq k-1.$$ But this can not satisfy the condition such that the contraction mapping principle holds.

Answer 1

Here's my solution. Please check.

It's obvious that，if $0\leq x_n<\sqrt{k}$，then $x_{n+1}>\sqrt{k}$；if $x_n>\sqrt{k}$，then $0\leq x_{n+1}<\sqrt{k}$；if $x_n=\sqrt{k}$，then $x_{n+1}=\sqrt{k}$. Thus, by induction, we obtain

1）if $0\leq x_1<\sqrt{k}$，then $\forall n:x_{2n}>\sqrt{k},0\leq x_{2n+1}<\sqrt{k}$；

2）if $x_1>\sqrt{k}$，then $\forall n:0\leq x_{2n}<\sqrt{k},x_{2n+1}>\sqrt{k}$；

3）if $x_1=\sqrt{k}$，then $\forall n: x_n \equiv \sqrt{k}$.

When $1)$ holds，then $$x_{2n+1}-x_{2n-1}=\frac{2k+(k+1)x_{2n-1}}{k+1+2x_{2n-1}}-x_{2n-1}=\frac{2(k-x_{2n-1}^2)}{k+2x_{2n-1}+1}>0,$$ and $$x_{2n+2}-x_{2n}=\frac{2k+(k+1)x_{2n}}{k+1+2x_{2n}}-x_{2n}=\frac{2(k-x_{2n}^2)}{k+2x_{2n}+1}<0,$$ which implies the subsequence $\{x_{2n-1}\}$ is increasing and upward bounded, and the one $\{x_{2n}\}$ is decreasing and downward bounded. Therefore, both of them are convergent. Denote the limit of them as $a,b$ respectively. Taking the limits of the equalities $$ x_{2n+1}=\frac{2k+(k+1)x_{2n-1}}{k+1+2x_{2n-1}},~~~x_{2n+2}=\frac{2k+(k+1)x_{2n}}{k+1+2x_{2n}},$$ we obtain $$a=\frac{2k+(k+1)a}{k+1+2a},~~~b=\frac{2k+(k+1)b}{k+1+2b}.$$ Solve to get $$a=\sqrt{k},b=\sqrt{k}.$$ This shows that $\{x_{2n-1}\}$ and $\{x_{2n}\}$ converges to the same limit. Thus, $\{x_n\}$ converges to it as well, namely $$\lim_{n \to +\infty}x_n=\sqrt{k}.\tag{*}$$

Moreover, if $2)$ holds, the reasoning is similar；if $3)$ holds，it's a trivial case. It follows that $(*)$ holds for all cases.

Now, consider $$f(x)=\frac{k+x}{1+x},(x\geq 0,k>1).$$ We have $$|f'(x)|=\frac{k-1}{(1+x)^2},$$ which is decreasing over $[0,+\infty)$. Thus $$\forall x \in I:= (\sqrt{k}-1,\sqrt{k}+1):|f'(x)|<|f'(\sqrt{k}-1)|=1-\frac{1}{k}<1.$$ Denote $|f'(\sqrt{k}-1)|=p$. Since $x_n \to \sqrt{k}(n \to \infty)$， then $$\exists N \in \mathbb{N},\forall n>N:|x_n-\sqrt{k}|<1,$$ namely $$x_{N+1},x_{N+2},\cdots \in I.$$ By Lagrange's MVT，for all $q=1,2,\cdots$, it holds that \begin{align*} |x_{N+q+2}-x_{N+q+1}|&=|f(x_{N+q+1})-f(x_{N+q})|\\&=|f'(\xi)||x_{N+q+1}-x_{N+q}|\\ &<p|x_{N+q+1}-x_{N+q}|. \end{align*} By induction, we have $$|x_{N+q+2}-x_{N+q+1}|<p^q|x_{N+2}-x_{N+1}|.$$ Therefore， for a large sufficiently $n$， \begin{align*} \sum_{k=1}^{n}|x_{k+1}-x_k|&=\sum_{k=1}^{N}|x_{k+1}-x_k|+\sum_{k=N+1}^{n}|x_{k+1}-x_k|\\ &<\sum_{k=1}^{N}|x_{k+1}-x_k|+(1+p+p^2+\cdots+p^{n-N-1})|x_{N+2}-x_{N+1}|\\ &<\sum_{k=1}^{N}|x_{k+1}-x_k|+\frac{1}{1-p}|x_{N+2}-x_{N+1}|\\ &<+\infty, \end{align*} which shows the partial sum $\sum\limits_{k=1}^{n}|x_{k+1}-x_k|$ is bounded. Thus $\sum\limits_{k=1}^{\infty}|x_{k+1}-x_k|$ is convergent. As a result,$\sum\limits_{k=1}^{\infty}(x_{k+1}-x_k)$ is abosolutely convergent， which shows that $(1)$ holds.

Meanwhile，$(2)$ is simple，since $$\sum_{n=1}^{+\infty}(x_{n+1}-x_n)=\lim_{n \to \infty}\sum_{k=1}^{n}(x_{k+1}-x_k)=\lim_{n \to \infty}(x_{n+1}-x_1)=\sqrt{k}-x_1.$$

Answer 2

Consider $f\circ f$. It has bounded image (as $f$ has) and is increasing. It follows that the subsequences $(x_{2n})$ and $(x_{2n-1})$ are convergent. Their limit is $\sqrt{k}$ (as pointed out in the comments), so $(x_n)$ converges to $\sqrt{k}$.