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!
Best Answer
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 $\geq x_0=-\frac{3}{4}$.