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.
 A: A sequence which is monotonic and bounded is convergent.
Boundedness
We claim that $a_n<1+\sqrt{2}$ for all natural numbers $n$. Check that $n=1$ satisfies this. Now, suppose that the result holds upto some natural number $k$. Then, $a_{k+1}=\sqrt{1+2a_k}<\sqrt{1+2+2\sqrt{2}}=\sqrt{3+2\sqrt{2}}=\sqrt{2}+1$ so indeed the sequence is bounded above.
Monotonicity
We claim that $\{a_n\}$ is monotonically increasing. Consider $a_2=\sqrt{3}>1=a_1$ and assume the result holds upto some natural number $k$. Then, $a_{k+1}-a_k=\sqrt{1+2a_k}-a_k>0$ iff $a_k^2-2a_k-1<0$ implying $1-\sqrt{2}<a_k<1+\sqrt{2}$. But by the proof of boundedness of $\{a_n\}$ above, and that $a_n>0$, we can easily conclude the desired inequality is true. Thus the sequence is monotonic increasing.
Limit
Now we are in a position to say that the limit exists, say $l$. We have from $a_{n+1}=\sqrt{1+2a_n}$ that as $n\to \infty$, $\lim a_{n+1}=\sqrt{1+2\lim {a_n}}$ i.e. $l=\sqrt{1+2l}$. Solving, we get $l=1+\sqrt{2}$.
A: Hint: See if this helps
$$x = \sqrt{1 + 2\sqrt{1 + 2\sqrt{1 + 2\sqrt{...}}}} \Rightarrow x ^2 = 1 + 2x$$
You have a monotonic limited sequence, what can you take from that?
A: 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
$$\displaystyle\lim_{n\to \infty}a_{n}=l$$ $\therefore \displaystyle\lim_{n\to \infty}a_{n+1}=\displaystyle\lim_{n\to \infty}\sqrt{1+2a_{n}} \implies l=\sqrt{1+2l} \implies l^2-2l=1 \implies l=1\pm \sqrt{2}$
$${\boxed{l\neq 1-\sqrt{2}\ (?) \implies l=1+\sqrt{2}}}$$
