Prove that the sequence $a_{n+1}=\sqrt{2a_n+3},$ $a_1=1$, is bounded. 
Prove that the sequence $a_{n+1}=\sqrt{2a_n+3},$ $a_1=1$, is bounded. 

Proof: it's increasing and bounded above by $2$. 
Is that right?
 A: $3$ is an attracting fixed point of the function $f(a)=\sqrt{2a+3}$.
Unrolling that assertion explains how the bounds work.  If you start the iteration with $a_1 \geq 3$, the sequence decreases down to $3$, and all values are in $[3,a_1]$.   If $a_1 < 3$ the sequence increases and stays in $[a_1,3]$. 
A: Lets try to work backwards. Suppose that we already prove that the sequence $a_n$ converges say to $a$. Let's calculate the limit of this sequence. It must follow that $a_{n+1}$ must converge to $a$. Thus $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{2a_n+3}=\sqrt{2\lim_{n\to\infty}a_n+3}\implies a^2=(2a+3)\implies a^2-2a-3=0\implies (a-3)(a+1)=0\implies a=3$$ or $a=-1$. But $a=-1$ cannot happen since the inside of a square root must always be greater than or equal to 0. Hence $a=3$. Now work in reverse. First show that the sequence is increasing and you already know it much be bounded below by 0. Show that it is bounded above by $3$. 
A: Assume that you series converges and let 
$$\lim_{n\to\infty}a_n=a$$
then 
$$a=\sqrt{2a+3}\Rightarrow a^2-2a-3=0\Rightarrow a=3$$
we have ignored the other root as $a_n>0$ for all $n$. Now we claim that $a_n\leq3$ for all $n$. Assume otherwise than there must exists some $N$ such that $a_N\geq3$ which would imply 
$$a_N=\sqrt{2a_{N-1}+3}\geq3\Rightarrow a_{N-1}\geq3\Rightarrow a_{N-2}\geq3...\Rightarrow a_1\geq3$$ which clearly is a contradiction as $a_1=1$. Therefore $a_n\leq3$ for all $n$.
Now 
$$a_{n+1}-a_n=\sqrt{2a_n+3}-a_n=\frac{2a_n+3-a_n^2}{\sqrt{2a_n+3}+a_n}=\frac{(3-a_n)(1+a_n)}{\sqrt{2a_n+3}+a_n}\geq0$$
Hence the sequence is increasing and bounded by $3$. Indeed the sequence converges to $3$ as shown above.
