# 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.

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!

• +1 for showing all the work you did --- a nicely-asked question! (And no, I don't actually have anything useful to provide as an answer, alas.) Feb 8, 2019 at 15:25
• To avoid headaches and trivialize massively all this and every similar question, my advice would be to draw the graph of the function $f:x\mapsto x^2+3x+1$ and the line $y=x$ on the same figure, and to plot the first values of the sequence by the well-known cobweb plot associated to $f$. You should see the desired results literally pop up from the figure... In addition, the asymptotics of sequences starting from any $x_0$ in $(-1,\infty)$ or $[-2,-1]$ or $(-\infty,-2)$ should become obvious as well.
– Did
Feb 8, 2019 at 15:36
• +1, can I know how you have written the code in Desmos? Feb 8, 2019 at 15:44
• @taritgoswami I've just manually printed the equations for the first 10 terms and then added a table to display the points. If you expand the 'terms' folder you'll see a list of equations. Feb 8, 2019 at 15:47

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