Show sequence is convergent and the limit Given the sequence 
$$\left\{a_n \right\}_{n=1}^\infty $$
which is defined by
$$a_1=1 \\
a_{n+1}=\sqrt{1+2a_n} \ \ \ \text{for} \ n\geq 1 $$
I have to show that the sequence is convergent and find the limit.
I am quite stuck on this, hope for some help.

Monotonicity
$$a_2>a_1$$ $\boxed{\therefore\ a_{n+1}> a_n \implies 1+2a_{n+1}>1+2a_{n} \implies \sqrt{1+2a_{n+1}}>\sqrt{1+2a_{n}} \implies a_{n+2}>a_{n+1}}$
Boundedness
$$a_1<4$$ $$\boxed{\therefore\ a_n<4 \implies 1+2a_n<9 \implies \sqrt{1+2a_n}<3 \implies a_{n+1}<4}$$
Limit
$$\displaystyle\lim_{n\to \infty}a_{n}=l$$ $\therefore \displaystyle\lim_{n\to \infty}a_{n+1}=\displaystyle\lim_{n\to \infty}\sqrt{1+2a_{n}} \implies l=\sqrt{1+2l} \implies l^2-2l=1 \implies l=1\pm \sqrt{2}$
$${\boxed{l\neq 1-\sqrt{2}\ (?) \implies l=1+\sqrt{2}}}$$
