recurrence relation, all terms of the sequence positive

Finally I resolved the issue of existence and uniqueness. We will regard each $a_n$ as rational function of $a_1$. That is, we will consider the sequence of rational functions

$$ a_0 = a_0(x) = 0, \qquad a_{1} = a_1(x) = x, \qquad a_{n+1} = \frac{n}{a_n} - a_n - a_{n-1}. \tag{*} $$

We would like to address the set of all $x\in\mathbb{R}$ for which $a_n(x) > 0$ for all $n \geq 1$. To this end, write

$$ I_n = \{ x \in \mathbb{R} : a_1(x) > 0, \cdots, a_n(x) > 0 \}, \qquad I_{\infty} = \bigcap_{n=1}^{\infty} I_n. $$

Then the solution set is exactly $I_{\infty}$, which we seek to identify. In this answer, we will prove that $I_{\infty}$ is a singleton, i.e.,

Proposition. There exists a unique $x \in \mathbb{R}$ for which $a_n(x) > 0$ for all $n \geq 1$.

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We know that $I_1 = (0, \infty)$. Now we prove the following statement:

Claim. For any $n \geq 2$, the followings hold:

1. $I_n = (\alpha, \beta)$ for some $0 \leq \alpha < \beta < \infty$. Moreover,
   • If $n$ is even, then $a_{n-1}(\alpha) = 0$ and $a_n(\beta) = 0$,
   • If $n$ is odd, then $a_{n-1}(\beta) = 0$ and $a_n(\alpha) = 0$.
2. $(-1)^k a_{k+1}'(x) \geq (-1)^{k-1}a_k'(x) \geq 1$ for all $1 \leq k \leq n-1$ and $x \in I_n$.

Proof of Claim. We invoke induction.

• (Base case) The claim is easy to verify directly when $n = 2$. Indeed, we know that $a_2(x) = \frac{1}{x}-x$, and so, $a_2(1) = 0$ and $I_2 = (0, 1)$. Also, $-a_2'(x) = \frac{1}{x^2} + 1 \geq 1$ and $a_1'(x) = 1$ proves part (2) of the claim for $n = 2$.

• (Induction step) Assume that the claim holds for a given $n \geq 2$. Then by the induction hypothesis,

  \begin{align*} (-1)^n a_{n+1}'(x) &= \left(\frac{n}{a_n^2} + 1\right) (-1)^{n-1}a_n'(x) - (-1)^{n-2}a_{n-1}'(x) \\ &\geq \frac{n}{a_n^2} (-1)^{n-1}a_n'(x) \\ & > 0. \end{align*}

  Now we consider two case

  $\text{Case 1.} \ $ If $n$ is even, then $a_n(\alpha^+) = +\infty$ and hence $a_{n+1}(\alpha^+) = -\infty$. Also we have $a_{n+1}(\beta^-) = +\infty$. Since $\alpha_{n+1}' > 0$ on $I_n$, there exists a unique zero $\gamma \in I_n$. Therefore $I_{n+1} = (\gamma, \beta)$.

  $\text{Case 2.} \ $ If $n$ is odd, then by a similar argument, we can check that there exists a unique zero $\gamma$ of $a_{n+1}$ in $I_n$ and $I_{n+1} = (\alpha, \gamma)$.

  So the part 1 of Claim for $n+1$ is proved. Finally, on $I_{n+1}$, we have $a_{n+1} > 0$, and so,

  $$ \frac{n}{a_n^2} = 1 + \frac{a_{n+1} + a_{n-1}}{a_n} > 1. $$

  Plugging this back shows that $(-1)^n a_{n+1}' \geq (-1)^{n-1}a_n'$ and therefore the induction step is proved. ////

Now write $I_n = (\alpha_n, \beta_n)$. Then by the intermediate step of the above proof, we can easily check that $(\alpha_n)_{n\geq 2}$ is non-decreasing, $(\beta_n)_{n\geq 2}$ is non-increasing and

$$ \alpha_{n} < \alpha_{n+2} < \beta_{n+2} < \beta_n. $$

So both $(\alpha_n)$ and $(\beta_n)$ converges. Moreover, if $\alpha = \lim \alpha_n$ and $\beta = \lim \beta_n$, then

$$ I_{\infty} = [\alpha, \beta]. $$

Finally, we prove that $\alpha = \beta$. To this end, we prove the following claim.

Claim. There exists $c \in (0, 1)$ such that $a_n(x) \leq c \sqrt{n}$ for all $n\geq 0$ and $x \in I_{\infty}$. In particular,

$$ |a_{n}'(x)| \geq \frac{1}{c^{2(n-1)}}. $$

Proof of Claim. Write $a_n = a_n(x)$ for simplicity. Then

1. $a_n^2 \leq n$ shows that $a_n \leq \sqrt{n}$ for all $n \geq 0$.

2. We have

   $$\frac{n}{a_n} - a_n = a_{n-1} + a_{n+1} \leq \sqrt{n-1} + \sqrt{n+1} \leq 2\sqrt{n}. $$

   Using this, we can easily conclude that $a_n \geq c_1\sqrt{n}$ for $c_1 = \sqrt{2}-1$.

3. Similarly, we find that

   $$\frac{n}{a_n} - a_n = a_{n-1} + a_{n+1} \geq c_1(\sqrt{n-1} + \sqrt{n+1}) \geq c_1 \sqrt{2n}. $$

   Here, we utilized the inequality $\sqrt{t-1}+\sqrt{t+1} \geq \sqrt{2t}$ which holds for all $t \geq 1$. Then by choosing $c$ as $c = \left(\sqrt{\smash[b]{2+c_1^2}} - c_1\right)/\sqrt{2} \approx 0.749$, we find that $a_n \leq c\sqrt{n}$ for all $n\geq 0$.

Finally, as in the proof of the previous claim, we note that

$$ |a_{n+1}'(x)| \geq \frac{n}{a_n^2} |a_n'(x)| \geq \frac{1}{c^2}|a_n'(x)|. $$

Therefore the desired claim follows. ////

Now we are ready to prove the uniqueness. Assume otherwise that $\alpha < \beta$. Then by the mean-value theorem,

$$|a_n(\alpha) - a_n(\beta)| = |a_n'(x)|(\beta - \alpha) \geq \frac{\beta-\alpha}{c^{2(n-1)}} $$

for some $x \in (\alpha, \beta)$. On the other hand, the left-hand side is bounded by

$$ |a_n(\alpha) - a_n(\beta)| \leq a_n(\alpha) + a_n(\beta) \leq 2c\sqrt{n}. $$

This is a contradiction as $n\to\infty$.

I post it as a curiosity. Considering the difference equation

$$ \frac{n}{x_n^2} = 1 + \frac{x_{n+1}+x_{n-1}}{x_n} $$

and considering

$$ \Delta x_n = x_n-x_{n-1} $$

we have

$$ \frac{n}{x_n^2} = 3+\frac{\Delta x_{n+1}-\Delta x_n}{x_n} $$

which is a difference approximation for the DE

$$ \frac{t}{x^2(t)} = 3 + \frac{\ddot x(t)}{x(t)} $$

now calling $x(1) = a = \frac{2 \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}$ and $x(2) = a-\frac 1a$ and integrating we have

In red the plot for $x_n = \sqrt{\frac n3 + \frac{1}{36n}+ \mathcal{O}(\frac{1}{n^3})}$ and in blue the DE solution.

and now considering instead the initial conditions $x(2) = a-\frac 1a, x(3) = \frac{1-3a^2}{a^3-3}$ we obtain

This result points in the direction of good agreement between the DE and the recurrence. Now following with the initial conditions

$$ x(3) = \frac{1-3a^2}{a^3-3}\\ x(4) = \frac{8 a^3-4 a}{3 a^4-4 a^2+1} $$

we obtain