Show $nx_n \to 1$ for sequence defined by $x_{n+1}=\frac{x_n}{1+nx_n^2}$

Question

Let $a>0$. Consider the sequence $x_{n+1}=\frac{x_n}{1+nx_n^2},~x_1=a,~n\in\mathbb{N}^*$. Study the convergence of $(x_n)_{n\ge1}$ and $(nx_n)_{n\ge1}$.

It was easy for me to prove that $(x_n)_n$ is convergent with the limit equal to $0$, but couldn't find a straightforward approach for $(nx_n)_n$. I first proved that $x_n\le\frac{1}{n}$, or equivalently, $nx_n\le1$, for $n\ge2$, using induction. Then I showed that $(nx_n)_n$ is a nondecreasing sequence: $$\frac{(n+1)x_{n+1}}{nx_n}=\frac{n+1}{n+n^2x_n^2}\ge 1$$ Thus $(nx_n)_n$ is nondecreasing and upper bounded, so it is convergent with positive finite limit $l$.

Now consider $a_n=n$ and $b_n=\frac{1}{x_n}$, so $(b_n)_n$ is a positive, increasing sequence with infinite limit. We have that $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\bigg(\frac{1}{x_{n+1}}-\frac{1}{x_n}\bigg)^{-1}=\frac{1}{nx_n}\to\frac{1}{l}$$ By the Stolz–Cesàro theorem, we obtain the identity $l=\frac{1}{l}$, so the only possible value for $l$ is $l=1$.

The hardest part was to prove the convergence of $(nx_n)_n$, which took me a long time. I feel like I am missing an easier solution. Please let me know if there's an elegant proof for $nx_n\to1$. Thanks in advance.

Answer 1

Once you have shown that $nx_n \le 1$ for $n \ge 2$ you can use the general form of the Stolz–Cesàro theorem:

$$ \frac{1}{\liminf_{n \to \infty} nx_n} = \limsup_{n \to \infty} \frac{1}{nx_n} = \limsup_{n \to \infty} \frac{1/x_n}{n} \\ \le \limsup_{n \to \infty} \frac{1/x_{n+1}-1/x_n}{(n+1)-n} = \limsup_{n \to \infty} n x_n \le 1 $$ implies that $\liminf_{n \to \infty} n x_n \ge 1$, so that the limit exists and is equal to one.

Answer 2

Once you have $(nx_n)_n$ converges to a limit $l>0$, suppose $l\neq 1$ : then as $\displaystyle (n+1)x_{n+1}-nx_n=\frac{x_n(1-n^2x_n^2)}{1+nx_n^2} (1)$ you get $\displaystyle (n+1)x_{n+1}-nx_n\sim\frac{l}{n}(1-l^2)$ hence the serie $\sum ((n+1)x_{n+1}-nx_n)$ is divergent hence $(nx_n)_n$ is divergent : contradiction.

Answer 3

Let $a_n = \frac1{nx_n}$ then,

\begin{align} a_{n+1} &= \frac1{n+1} \times \frac1{x_{n+1}}\\ &= \frac1{n+1} \left(\frac{1+nx_n^2}{x_n}\right)\\ &= \frac1{n+1} \left(\frac{n}{nx_n} + nx_n\right)\\ &= \frac{n}{n+1} a_n + \frac1{n+1}\times \frac1{a_n} \end{align}

Let $b_2 = a_2$ and $$b_{n+1} = \frac{n}{n+1} b_n + \frac1{n+1}$$

Using very simple induction, one can easily prove that for $n \ge 2$, $$1 \le a_n \le b_n.$$

It remains to compute the limit of $b_n$. Indeed for $n\ge 3$ \begin{align} b_n = \frac1n \times nb_n &= \frac1n \left(b_2 + \sum_{k=2}^{n-1} \left((k+1)b_{k+1} - kb_k\right)\right)\\ &= \frac{b_2}n + \frac1n \sum_{k=2}^n 1 = 1 + \frac{b_2}{n} \to 1. \end{align}

Can you finish the proof?

Answer 4

Motivation. Let $y_n = \frac{1}{x_n}$. Then the recurrence relation reads as:

$$ y_{n+1} - y_n = \frac{n}{y_n} $$

The continuum analog of the relation is the differential equation

$$ y' = \frac{x}{y}. $$

This can be solved by rearranging the equation in the form

$$ y\,\mathrm{d}y = x\,\mathrm{d}x,\qquad\text{or equivalently,}\qquad \mathrm{d}(y^2) = \mathrm{d}(x^2). $$

The last equality tells that the quantity $\mathrm{d}(y^2)$ is of central importance in solving the differential equation, and we can expect that the same is true for the original recurrence relation.

---

Actual Proof. Note that $\frac{1}{x_{n+1}} = \frac{1}{x_n} + nx_n$. Using this, we can expand $\frac{1}{x_n^2}$ as:

\begin{align*} \frac{1}{x_n^2} &= \frac{1}{x_1^2} + \sum_{k=1}^{n-1} \biggl( \frac{1}{x_{k+1}^2} - \frac{1}{x_k^2} \biggr) \\ &= \frac{1}{x_1^2} + \sum_{k=1}^{n-1} \biggl[ \biggl( \frac{1}{x_{k}} + kx_k \biggr)^2 - \frac{1}{x_k^2} \biggr] \\ &= \frac{1}{x_1^2} + \sum_{k=1}^{n-1} \bigl( 2k + (kx_k)^2 \bigr) \\ &= a^{-2} + n(n-1) + \sum_{k=1}^{n-1} (kx_k)^2 \tag{1}\label{e:int} \end{align*}

Here, $a = x_1$. From this, we observe the following:

1. Using $\eqref{e:int}$ and noting that $x_n > 0$ for all $n$, we get $$ \frac{1}{x_n^2} \geq a^{-2} + n(n-1), \qquad\text{i.e.,}\qquad nx_n \leq \frac{n}{\sqrt{a^{-2} + n(n-1)}}. \tag{2}\label{e:u} $$

2. $\eqref{e:u}$ shows that $C := \sup_{n\geq 1} nx_n$ is bounded. Hence by $\eqref{e:int}$ again, $$ \frac{1}{x_n^2} \leq a^{-2} + n(n-1) + C^2(n-1). \tag{3}\label{e:l} $$

Combining $\eqref{e:u}$ and $\eqref{e:l}$, we get

$$ \frac{n}{\sqrt{a^{-2} + n(n-1) + C^2(n-1)}} \leq nx_n \leq \frac{n}{\sqrt{a^{-2} + n(n-1)}}. $$

Therefore the desired conclusion follows by the squeeze lemma.