<s>[INST] 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?
 [/INST] You have taken $a_1 = 1$. For this value $ { a_{n+1}} $ keeps on increasing. But had you taken $ a_1$ greater than $ 1\, + \, √2, a_{n+1}$  decreases. Now, for all values of $ a_1$  less than $1\, +√2, \, a_{n+1} $ increases. And as long as $a_n$ is less than $ 1+√2,$ the $a_{n+1}$  is also less than $1+√2.$ Thus for $a_1$ less than $1+√2,$ sequence is increasing and bounded by $ 1+√2.$
</s>