Does $x_{n+1}=1+\dfrac{2}{x_n+1}$ converge? Question: Define $(x_n)_{n\geq 1}$ as $x_1=1$, and $x_{n+1}=1+\dfrac{2}{x_n+1}$ for $n\geq 1$. Investigate the convergence and the limits of $(x_n)_{n\geq 1}$.
My Process: I can guess that this sequence converges and its limit is $\sqrt{3}$. It is obvious that $1\leq x_n\leq 2$, however, it is neither non-increasing nor non-decreasing. So, I could not use Monotone Convergence Theorem.
 A: You can use the fixed point convergence test, a proof of which can be found here; it is a baby version of a more general theorem called the Banach Fixed Point Theorem, a.k.a the contraction mapping theorem.

If $f(x)$ is differentiable on $[a,b]$, and if $|f'(x)|\lt1,\,\forall x\in[a,b]$, the iterative scheme defined by $x_{n+1}=f(x_n)$ will converge in $[a,b]$.

Our $f,f'$ here are:
$$f(x)=1+\frac{2}{x+1},\,f'(x)=-\frac{2}{(x+1)^2}$$
And $|f'(x)|\lt1$ for all $x\gt1$, so we are assured convergence.
A: I have a different proof to what FSrike has given. One that is less general but require less knowledge. Let $f:[1,2]\rightarrow[1,2]$ be given by $f(x)=1+\frac{2}{1+x}$. Observe that for $x,y\in [1,2]$,
$$|f(x)-f(y)|=\bigg|\frac{2(y-x)}{(1+x)(1+y)}\bigg|\le \frac{1}{2}|x-y|.$$
Then the sequence $(x_n)$ is Cauchy since
$$|x_{n+2}-x_{n+1}|\le \frac{1}{2}|x_{n+1}-x_n|\le\frac{1}{2^n}|x_2-x_1|.$$
Then $(x_n)$ is convergent. Since $f$ has one fixed point $x=\sqrt 3$, $x_n\rightarrow \sqrt 3$.
A: Add $1$ to both sides as follows
$$x_{n+1}=1+\dfrac{2}{x_n+1} \\
x_{n+1} + 1 = 2 + \frac{2}{x_n + 1}$$
Substituting $y_n = x_n + 1$, you can find what $\lim y_n$ is as follows:
$$y_{n+1} = 2 + \frac{2}{y_n} = 2 + \frac{2}{2+\dfrac{2}{y_{n-1}}} \to 2 + \frac{2}{2 + \dfrac{2}{2 + \dfrac{2}{\ddots}}}$$
Now, let's actually evaluate the last continuous fractions. Let that be equal to $y$. Then $y$ satisfies
$$y = 2 + \frac{2}{y} \implies y =1+\sqrt 3$$
Therefore $$\lim x_n = y - 1 = \sqrt 3$$
