Divergence of $a_{n+1}=\sqrt{2a_n+3}$? I am wondering what I am missing from my proof. I would like to show that the limit of the sequence $$a_{n+1}=\sqrt{2a_n+3},\,\, a_1=4,$$ goes to $\infty$, as  $n \rightarrow \infty$. Is there anything that I should add to make my argument more clear?
Proof: Let $M \in \mathbb{R}$. Choose $n_0 \in \mathbb{N}$ such that $a_{n_0} > {(M^2-3)/2}$. Then, for any $M$, we have $\sqrt{2a_n+3} > M$ for $n \geq n_0$.
QED
 A: This sequence does not tend to infinity and it is not unbounded. 
$$
\lim_{n\to\infty}a_n=3.
$$
Proof. First we prove that the sequence is decreasing. Clearly, $0<a_2=\sqrt{2\cdots 4+3}<4=a_1$. Inductively, if $a_k<a_{k-1}$, then $\,0<a_{k+1}=\sqrt{2a_{k}+3}<\sqrt{2a_{k-1}+3}=a_{k}$.
Hence, $\{a_n\}_{n\in\mathbb N}$ is decreasing and lower bounded by  $0$, and thus it converges, say to $a$. But 
$$
a_n\to a\quad\Longrightarrow a_{n+1}=\sqrt{2a_n+3}\to \sqrt{2a+3},
$$
and hence $a=\sqrt{2a+3}$ or $a^2-2a-3=0$ or $(a+1)(a-3)=0$, and as $a_n>0$, then $a\ge 0$ and hence $a=3$.
A: Your proof can only prove that $\{a_n\}$ is unbounded.
Let $a_n=n\cos \frac{n\pi}{2}$. Although it is unbounded, however it don't tends $\infty$.
A: What you are missing in your proof: "Let $M \in \mathbb{R}$. Choose $n_0 \in \mathbb{N}$ such that $a_{n_0} > {(M^2-3)/2}$."
Provided such an $n_0$ exists ! It usually doesn't as the sequence is decreasing and  $a_n\le4$.
Assuming that $n_0$ can be found amounts to declaring the sequence divergent, a circular argument.
