Show that $\{a_n\}$ is an increasing sequence and bounded from above Question
Let $a_1$ = 1 and $a_{n+1} = \sqrt{1+2a_n}$. Show that $\{a_n\}$ is increasing and bounded from above.
Attempted solution
showing that the sequence is increasing:
\begin{align*}
a_n &\geq 1 \quad \forall n = {1,2,3,...} \\
\Rightarrow 1+2a_n &\geq 3 \\
\Rightarrow \sqrt{1+2a_n} &\geq \sqrt{3} \\
\Rightarrow a_{n+1} &\geq \sqrt{3} \quad \forall n = {1,2,3,...} \\ 
\Rightarrow a_{n+1} &\geq a_n
\end{align*} Is this a valid approach?
Assuming that I have shown that the sequence is increasing, I show that it is bounded by doing the following
\begin{align*}
a_{n+1} &\geq a_n \\
\Rightarrow \sqrt{1+2a_n} &\geq a_n  \\
\Rightarrow 1+2a_n &\geq a^2_n \\ 
\Rightarrow (a_n+1)^2 - a^2_n &\geq a^2_n \\
\Rightarrow \frac{a_n + 1}{a_n} &\geq \sqrt{2} \\  
\Rightarrow a_n &\leq 1+\sqrt{2} \quad \forall n = 1,2,3,...
\end{align*} 
Is there another more straightforward way to show that this sequence is bounded?

We prove by induction that the generalized sequence
$$a(1)=1,a(n+1)=\sqrt{2 c \;a(n)+1},\;c>0$$
is     
(i) increasing.
The quotient of two successive terms of the sequence is
$$q(n+1)=\frac{a(n+1)}{a(n)}$$
Obviously
$$q(2)=\sqrt{2 c+1}>1$$
Now
$$q(n+1)=\sqrt{\frac{2 c a(n-1) q(n)+1}{2 c a(n-1)+1}}>\sqrt{\frac{2 c a(n-1)+1}{2 c a(n-1)+1}}=1$$
QED.
(ii) the sequence is bounded.
In fact we have just proven that
$$a(n+1)=\sqrt{2 c a(n)+1}>a(n)$$
Hence 
$$2 c a(n)+1>a(n)^2$$
$$1>a(n)^2-2 c a(n)=(a(n)-c)^2-c^2$$
or
$$c^2+1>(a(n)-c)^2$$
Which means that $\left| a(n)\right|$ cannot grow unbounded.         
QED.
(iii) the sequence converges to one of the fixed points, i.e. solutions of
$$a^2=2 a c+1$$
giving
$$\text{a1}=\sqrt{c^2+1}+c,\text{a2}=c-\sqrt{c^2+1}$$
Since $a(1)>0$ and the sequence increases it canot become negative. Hence the limit is $\text{a1}$.      
In the case of the OP we have $c=1$ and hence $a1 = 1+\sqrt{2}$.
