TITLE: Study convergence of $x_{n+1} = x_n^2 + 3x_n + 1$, where $x_1 = a$, and $a$ takes different values and find its limit.
QUESTION [7 upvotes]: Given a recurrence relation:
  $$
x_{n+1} = x_n^2 + 3x_n + 1 \\
x_1 = a\\
n\in\Bbb N
$$
  Figure out whether this sequence has a limit (either finite or infinite) and find it for:
  $$
\begin{align*}
a = -{5\over 4}\tag1 \\
a = -{3\over 4}\tag2
\end{align*}
$$


Start with case $(1)$. It took some time to notice but seems like the sequence is monotonically increasing no matter what initial conditions are given. That is because:
$$
x_{n+1} = x_n^2 + 3x_n + 1 \iff x_{n+1}-x_n = x_n^2 + 2x_n + 1 = (x_n+1)^2>0
$$
Than means:
$$
x_{n+1} - x_n > 0 \iff x_{n+1} > x_n
$$
That observation is crucial for all the next steps. In $(1)$ we are given that:
$$
x_1 = a = -{5\over 4} > -2
$$
By monotonicity of $x_n$:
$$
\forall n\in\Bbb N : x_n > -2
$$
Let's suppose the limit exists. Then by finding fixed points of the recurrence we may get an insight of what that limit might be:
$$
L = L^2 + 3L + 1 \iff (L+1)^2 = 0 \iff L = -1
$$
Thus the only possible finite limit in $\Bbb R$ is $L=-1$. Let's try to bound $x_n$ above. Using induction:
$$
x_1 < x_2 = -{19\over 16}  < -1
$$
Suppose $x_n < -1$. Then:
$$
x_n \in (-2; -1) \implies \underbrace{(x_n + 1)^2 + x_n}_{x_{n+1}} \in (-2, -1)
$$
Thus it follows that $x_{n+1} < -1$. Now by monotone convergence theorem a monotonic bounded sequence has a limit. Therefore:
$$
\boxed{\lim_{n\to\infty}x_n = -1}
$$

This case is more of a headache. Given $a = -{3\over 4}$ makes the sequence diverge to $+\infty$. But to show this I had to calculate the value for $6$ first terms. It follows that:
$$
\forall n \ge 6: x_n > 0
$$
Moreover:
$$
\forall n \ge 7: x_n > 1
$$
So:
$$
\boxed{\lim_{n\to\infty}x_n = +\infty}
$$
Does there exist a more elegant way to solve for case $(2)$?
Also is this argumentation enough to show what's requested in question section? I have doubts about the second case. Because formally I should have shown that the sequence is not bounded, not sure how to do it. And the solution is ugly. Here is a sandbox I've been using to play around with the recurrence.
Could you please verify the above and point to the mistakes just in case? Thank you!

REPLY [5 votes]: Does there exist a more elegant way to solve for case $(2)$? Yes there is ! You don't need to compute the first six terms at all ! The sequence is nondecreasing, so it either converges to a finite value or diverges to $+\infty$. It if converges, the limit can only be $-1$ as you have shown, but this is impossible since your sequence, being nondecreasing, will always be $\geq x_0=-\frac{3}{4}$.