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! 
 A: 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: 
$$\begin{align}
a_1&=1\\
a_2&=\sqrt{1+2}=\sqrt{3}\\
a_3&=\sqrt{1+2\sqrt{3}}\\
a_4&=\sqrt{1+2\sqrt{1+2\sqrt{3}}}\\
a_5&=\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{3}}}}\\
\vdots \\
u=\lim_{n\to\infty}{a_n}&=\sqrt{1+2\underbrace{\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{\cdots}}}}}_{u}}\\u&=\sqrt{1+2u}\\
u^2-2u-1&=0\\
\because u>0\therefore u&=1+\sqrt{2}
\end{align}$$
A: Let us see if we can provide an answer to your first two question.
Use induction to show that the sequence $a_n$ is bounded by $3$. Remember that induction goes as follows. Presume that the statement is correct for some $n$, then demonstrate that it is also true for $(n+1)$. Apply this line of reasoning to the starting value. From this one can conclude the statement is true for all $n$. 
Okay, let us assume that $a_n<3$ for some arbitrary $n$. It follows that $a_{n+1} < \sqrt
{7}$. Since $\sqrt{7} < 3$, we can conclude that $a_{n+1} < 3$. For the initial value $a_1 = 1$ the statement is certainly true. Hence all $a_n$ in the sequence are bounded by $3$. [In fact the sequence is bounded by $1+\sqrt{2}$]
To show that the sequence is increasing, consider the difference $D_n$ between two consecutive terms,  $a_{n+1}$ and  $a_n$. We find $D_n = a_{n+1} - a_n = \sqrt{2a_n+1} - a_n$. If you examine this function, you will see that it is positive in the interval $(0, 1+\sqrt {2})$. And since $a_n$ are bounded by $1+\sqrt 2$, it follows that the sequence is indeed monotonically increasing. 
UPDATE
As hypergeometric has shown, the limit of the sequence must be $u = 1 + \sqrt 2$. Let us examine analytically whether the series indeed converges to this value. Define $D(n)$ as the difference between $u$ and $a(n)$. We already know that $D(n)$ is positive, because the sequence is increasing and hence approaches $u$ from below. Therefore: $u - D(n+1) = \sqrt{1 + 2 (u - D(n))}$. Taking both sides to the power $2$ yields after cancellation: $2uD(n+1) - (D(n+1))^2 = 2D(n)$. From this it follows that $D(n+1) < \frac {D(n)}u$. The sequence converges indeed. And in fact it does so very quickly. At every step the difference between $u$ and $a(n)$ decreases by at least a factor $2.414$.
