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!

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: $\;1-\sqrt2<0\;$ .
