nth term of $2,2+\frac{1}{2},2+\frac{1}{2+\frac{1}{2}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}...$ What is the nth term of the sequence: $$2,2+\frac{1}{2},2+\frac{1}{2+\frac{1}{2}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}},2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}}...$$
in terms of $s_{n-1}$.
I have tried for some time to calculate this but to no avail. Hopefully somebody with a better understanding of the world of sequences could help/point me in the right direction!
 A: In my answer to this question, I detailed the steps for solving a first-order rational difference equation such as
$${ a_{n+1} = \frac{ma_n + x}{a_n + y} }=m+\frac{x-m y}{a_n+y}$$
For your case $m=2$, $x=1$ and $y=0$. So, using the initial condition,
$$a_n=\frac{\left(1+\sqrt{2}\right)^n-\left(1-\sqrt{2}\right)^n } { \left(1+\sqrt{2}\right) \left(1-\sqrt{2}\right)^n+\left(\sqrt{2}-1\right)\left(1+\sqrt{2}\right)^n}$$
Edit
In the documentation of sequence $A000129$ in $OEIS$, there is superb formula given by Peter Luschny in year $2018$. It write
$$a_n=\frac 1{\sqrt{2}}\, e^{\frac{i \pi  n}{2}}\,\sinh \left(n \cosh ^{-1}(-i)\right)$$
A: Assuming you already know the sequence converges, and that is not trivial as the subsequence $\;\left\{x_{2n}\right\}_{n=0}^\infty\;$ is monotonic descending whereas the subsequence $\;\left\{x_{2n+1}\right\}_{n=0}^\infty\;$ is monotonic ascending ( yet both sequences are appropiatedly bounded and converge to the same non-zero finite limit ), and if we put $\;\alpha=\lim\limits_{n\to\infty} x_n\;$  , we can then use arithmetic of limits and get
$$\alpha=\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty}\left(2+\frac1{x_n}\right)=2+\frac1\alpha\implies\alpha^2-2\alpha-1=0$$
and solving the above quadratic we get that the only plausible limit is $\;1+\sqrt2\;$ ,since the other root of the quadratic is negative: $\;1-\sqrt2<0\;$ .
