# Recursive successions limit.

Good morning everyone, I tried to solve an exercise and I want to ask you what do you think about my solution (if legit, clear, if I could have done it better) and if you can help me going forward.

Let $$x_{n+1}=\frac{|x_n^3-x_n|}{n+x_n^2} \qquad x_0=2017$$ Compute $\lim_{n\to \infty}x_n$ and $\lim_{n\to \infty}2^nx_n$.

That's what I've done: For the first question we first notice that $x_n \geq 0 \ \forall n \in \mathbb{N}$ (easy proof by induction). Suppose $x_n$ has a limit $l \in \mathbb{R}$; taking the recurrence we have:

$$\lim_{n\to \infty} x_{n+1}=\lim_{n\to \infty} \frac{|x_n^3-x_n|}{n+x_n^2}$$

So as $x_{n+1}$ and $x_n$ have the same limit we have that it must be $0$. So the unique possibilities for the limit are $0$ and $+\infty$. Let's now prove that $x_{n+1} \leq x_{n}$ so we can say that the limit exists and is $0$ as $x_n$ is decreasing and bounded.

$$x_{n+1}\leq x_n \iff x_n(n+x_n^2)\geq |x_n^3-x_n| \iff -nx_n-x_n^3\leq x_n^3-x_n \leq x_n^3+nx_n$$

and by using the fact that $x_n \geq 0$ the conclusion is clear as $n\to \infty$. So for the first point we have that $\lim_{n\to \infty}x_n=0$.

Now I'm stuck at the second point because I actually have no ideas how to proceed. I thought that I may be searching for an asymptotic behaviour for my recurrence but I can't find any so I tried to do root test but that's what I found:

$$\lim_{n \to \infty} (\frac{2^n|x_n^3-x_n|}{n+x_n^2})^{\frac{1}{n}}= 2\lim_{n \to \infty}e^{\frac{1}{n}\log\left(\frac{|x_n^3-x_n|}{n+x_n^2}\right)}$$

but I don't know how to compute the last one. Any help? Thanks in advance. Add: What about $\lim n!x_n$? I think that the limit is $\infty$ and I already proved that is monotonically increasing. How can I say that it can't go towards a real limit?

Here is another solution, progressively improving the asymptotics of $x_n$.

• We know that $(x_n)$ is non-negative for all $n$. Also, for $n \geq 1$,

$$x_{n+1} = \frac{\lvert x_n^2 - 1\rvert}{x_n^2 + n} \cdot x_n \leq \frac{x_n^2 + 1}{x_n^2 + n} \cdot x_n \leq x_n.$$

So $(x_n)$ is monotone decreasing and hence converges.

• Since $(x_n)$ is bounded, the same is true for $\lvert 1 - x_n^2\rvert$. Let $C$ be a bound of $\lvert 1 - x_n^2\rvert$. Then

$$x_{n+1} = \frac{\lvert x_n^2 - 1\rvert}{x_n^2 + n} \cdot x_n \leq \frac{C}{n} \cdot x_n \leq \cdots \leq \frac{C^n}{n!}x_1.$$

So $2^n x_n$ converges to $0$ by the squeezing theorem. In particular, $x_n$ converges to $0$ as well.

• If $x_N = 1$ for some $N$, then $x_n = 0$ for all $n > N$ and hence $n!x_n \to 0$ as $n\to\infty$. So it suffices to consider the case where $x_n \neq 1$ for all $n$. Then by noting that

$$\log \Bigg( \frac{\left| 1 - x_n^2 \right| }{1 + \frac{x_n^2}{n}} \Bigg) = \mathcal{O}(x_n^2) = \mathcal{O}\left( \frac{C^{2n-2}}{(n-1)!^2} \right) \quad \text{as } n\to\infty,$$

it follows that $\sum_{n=1}^{\infty} \log \Big( \frac{\left| 1 - x_n^2 \right| }{1 + \frac{x_n^2}{n}} \Big)$ converges. Let $A$ denote the value of this sum. Then

$$(n-1)!x_n = x_1 \prod_{k=1}^{n-1} \frac{\left| 1 - x_k^2 \right| }{1 + \frac{x_k^2}{k}} \xrightarrow[n\to\infty]{} x_1 e^{A}$$

Consequently, $n! x_n$ diverges to $\infty$ at a linear speed.

• Great solution. Now only to consider if $x_n=1$ can happen for some $n$ – Jakobian May 26 '18 at 23:13
• Yeah, great solution and nice method (I'll use it again surely)! Surely we should avoid the case $x_n=1$ for some $n$ as the recurrence would be definitely zero. +1, answer accepted. – Alberto Andrenucci May 27 '18 at 13:34

Assume $x_n>0$. Let's suppose $\lim\limits_{n\to \infty} 2^nx_n = g$. Using Stolz theorem: $$\lim\limits_{n\to \infty} 2^nx_n = \lim\limits_{n\to \infty} \frac{2^n}{\frac{1}{x_n}} = \lim\limits_{n\to \infty} \frac{2^{n+1}-2^n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}} = \lim\limits_{n\to \infty} 2^nx_n\cdot \frac{x_{n+1}}{x_n-x_{n+1}}$$ $$= g \lim\limits_{n\to \infty} \frac{|x_n-x_n^3|}{(n+x_n^2)x_n-|x_n-x_n^3|} = g \lim\limits_{n\to \infty} \frac{1-x_n^2}{n-1+2x_n^2} = 0$$ It shows that if the limit exists, it must be either $0$ or $\infty$. That can lead to assume, the sequence is monotone. It can be shown it indeed is from some point on. We know the sequence $x_n$ is decreasing to $0$, so we will assume that $n$ is large enough, so that $x_n \leq 1$ $$2^{n+1}x_{n+1}\leq 2^nx_n \iff 2x_{n+1}\leq x_n \iff 2|x_n-x_n^3|\leq(n+x_n^2)x_n$$ $$\iff 2x_n-2x_n^3\leq(n+x_n^2)x_n \iff -3x_n^3\leq (n-2)x_n$$ Which is obvious for $n\geq2$. For big enough $n$ we have:

$$\frac{(n+1)!x_{n+1}}{n!x_n}=\frac{(n+1)(1-x_n^2)}{n+x_n^2}>1$$ Let's suppose the limit is finite, and equals $h$. I also use Stolz theorem here.

$$h=\lim\limits_{n\to\infty} n!x_n = \lim\limits_{n\to\infty} (n+1)!\cdot\frac{x_n-x_n^3}{n+x_n^2} = \lim\limits_{n\to\infty} \frac{(n+2)!(x_{n+1}-x_{n+1}^3)-(n+1)!(x_n-x_n^3)}{(n+1)+x_{n+1}^2-(n+x_n^2)}$$ $$= \lim\limits_{n\to\infty} [(n+2)!x_{n+1}-(n+1)!x_n] = h\lim\limits_{n\to\infty} (n+1)[(n+2)\frac{1-x_n^2}{n+x_n^2}-1]$$ $$= h\lim\limits_{n\to\infty} (n+1)\cdot\frac{2-nx_n^2-3x_n^2}{n+x_n^2} = 2h$$ Hence $h=0$, which can't be.

• Perfect, thanks! Do you have any idea for the other question i added? – Alberto Andrenucci May 26 '18 at 21:23
• I've added some more, what i think might be solution. But you need to check if i didn't make any mistakes. – Jakobian May 26 '18 at 22:23