Given the recursively defined sequence
$$ a_1 = 0, a_{n+1} = \frac 1{2+a^n} $$
Show it converges. I'm working with Cauchy sequences, and proved in a previous question that any sequence of real numbers $(a_n)$ satisfying:
$$|a_n - a_{n+1}| \leq \frac1{2^n} $$
is convergent (by showing $(a_n)$ was Cauchy), and was told that it may be helpful to use induction to prove that the above recursive sequence satisfies the inequality and is therefore convergent.
Work:
Base Case: When $n = 1 , a_1 = 0, a_2 = \frac12,$ so $$ \left|0-\frac12\right| \leq \frac1{2^{1}}$$ and the base case checks out
Inductive Step: Want to show $$|a_{n+1} - a_{n+2}| \leq \frac1{2^{n+1}} $$
So $$\left|\frac1{2+a_n} - \frac1{2+a_{n+1}}\right| \leq \frac1{2^{n+1}}$$
And I'm not sure where to proceed from here. Any help would be greatly appreciated, thanks.
Continuing as Thomas Andrews suggested, $a_{n+1}-a_{n+2} =\frac1{2+a_n}-\frac1{2+a_{n+1}} =\frac{(2+a_{n+1})-(2+a_n)}{(2+a_n)(2+a_{n+1})} =\frac{a_{n+1}-a_n}{(2+a_n)(2+a_{n+1})} $ so $|a_{n+1}-a_{n+2}| =\big|\frac{a_{n+1}-a_n}{(2+a_n)(2+a_{n+1})}\big| <|\frac{a_{n+1}-a_n}{4}| $.
From this, $|a_{n+k}-a_{n+k+1}| <|\frac{a_{n+1}-a_n}{4^k}| $. Putting $n = 0$, $|a_{k}-a_{k+1}| <|\frac{a_{1}-a_0}{4^k}| $ which is more than enough to get convergence.
It is interesting that this shows that the convergence is at least $\frac1{4^k}$, not just $\frac1{2^k}$.
To find the limit: $|a_{n+1}-a_{n}| =|a_n-\frac1{2+a_n}| =|\frac{a_n(2+a_n)-1}{2+a_n}| =|\frac{a^2_n+2a_n-1}{2+a_n}| =|\frac{(a_n+1)^2-2}{2+a_n}| $.
Since $a_n$ converges, $(a_n+1)^2-2 \to 0 $ or $a_n \to \sqrt{2}-1$.
To find the true rate of convergence, since $a_n \to \sqrt{2}-1$, $|a_{n+1}-a_{n+2}| =\big|\frac{a_{n+1}-a_n}{(2+a_n)(2+a_{n+1})}\big| \approx \big|\frac{a_{n+1}-a_n}{(2+\sqrt{2}-1)(2+\sqrt{2}-1)}\big| = \big|\frac{a_{n+1}-a_n}{(\sqrt{2}+1)^2)}\big| = \big|\frac{a_{n+1}-a_n}{3+\sqrt{2}}\big| $, so the convergence is like $\frac1{(3+\sqrt{2})^k} $.