The convergence of a recursive sequence 
The sequence $(a_n)$ is defined by $a_1 = 1$ and
  $a_{n+1}=a_{n}+\sqrt{1+a_{n}^{2}}$. A sequence $(b_n)$ is defined by
  $b_n = \dfrac{a_n}{2^n}$. It is easy to see $(b_n)$ is monotone but is
  it convergent?

Suppose $(b_n)$ be bounded. Then $(b_n)$ is convergent by Monotone Convergence Theorem. Say, $b_n$ converges to a limit $b$, then $(b_{n+1}) \to b$.
We have $a_{n+1}=a_{n}+\sqrt{1+a_{n}^{2}}$. Dividing $2^{n+1}$ in both sides of the equation,
$\Rightarrow \dfrac{a_{n+1}}{2^{n+1}}=\dfrac{a_{n}+\sqrt{1+a_{n}^{2}}}{2^{n+1}}$
$\Rightarrow b_{n+1} = \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}$
$\Rightarrow \lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} \dfrac{b_n}{2} + \lim_{n \to \infty} \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}$
$\Rightarrow b = \dfrac{b}{2} + \sqrt{0 + \dfrac{b^2}{4}}$
$\Rightarrow b = b$.
No decisive conclusions could be made. 
However numerical evidences suggest that $ b_n \approx 0.63$ whenever $100\le n \le 200$. Does it diverge to infinity but sure, it diverges pretty slow? I tried to prove otherwise that $b_n$ is bounded above by $1$ but it didn't work but neither can find an counterexample to it. How far do I need to go in the sequence to surpass $1$?

$$
 b_{n+1} = \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}}
$$
where $b_1 = 1/2$. If you try to bound the relation, you get
$$
 \dfrac{b_n}{2} + \sqrt{\dfrac{1}{2^{2n+2}} + \dfrac{b_n^2}{4}} \le b_n + x_n
$$$$
\dfrac{1}{2^{2n+2}}  \le b_nx_n + x_n^2
$$
so $x_n = 2^{-n}$ is enough. This proves that 
$$b_{n+1}\le b_n + 2^{-n}\le b_1 + 1$$
so the sequence is bounded, and the monotone sequence converges.
