<s>[INST] A sequence is given by $a_1 = 1$, $a_{n+1} = \sqrt{1+2a_n}$ for $n=1,2,3,4...$ Show that the sequence is increasing and bounded by 3.
Find out if it converges, and find its limit.
So far i think the limit is 
$$
1+\sqrt{2},
$$
and that i should use induction to find that its bounded by 3.
I have solved everything except the convergence part. Can someone show me, and relieve my headache?
thanks for any help! 
 [/INST] Assume that, as $n\to\infty$, the limit of $a_n$ is $u$, hence $a_{n+1}=a_n=u$.
$$\begin{align}
a_{n+1}&=\sqrt{1+2a_n}\\
u&=\sqrt{1+2u}\\
u^2-2u-1&=0\\
\because u>0\therefore u&=1+\sqrt{2}
\end{align}$$

NB: 
$$\begin{align}
a_1&=1\\
a_2&=\sqrt{1+2}=\sqrt{3}\\
a_3&=\sqrt{1+2\sqrt{3}}\\
a_4&=\sqrt{1+2\sqrt{1+2\sqrt{3}}}\\
a_5&=\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{3}}}}\\
\vdots \\
u=\lim_{n\to\infty}{a_n}&=\sqrt{1+2\underbrace{\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\cdots}}}}}_{u}}\\u&=\sqrt{1+2u}\\
u^2-2u-1&=0\\
\because u>0\therefore u&=1+\sqrt{2}
\end{align}$$
</s>