Let $x_0, x_1$ be positive numbers and define $$x_{n+2}=x_{n+1}+\frac{1}{x_n + \sqrt{n}}$$

Prove that the sequence $y_n = x_n / \sqrt{n} \to 1 \\ \\ \\$.

My attempt:

I think it would suffice to consider the inequalities $x_n < \sqrt{n}$ and $x_n > \sqrt{n}$. From the given recursion, we would then extract $$x_{n+2} - x_{n+1} = \frac{1}{x_n + \sqrt{n}} > \frac{1}{2\sqrt{n}}$$ (flip the inequality in the other case) and compute $y_n$ from this. But the issue here is that I can't find the necessary limit in both cases.

  • 1
    $\begingroup$ Just an observation so far: The process itself does not necessarily move closer to $\sqrt{n}$ in either relative or absolute terms. Take $x_0 = 2$ and $x_1=0.15$. $\;x_n$ starts below $\sqrt{n}$ but exceeds it before approaching it from above, $\dfrac{x_n}{\sqrt{n}}$ does not begin consistently approaching $1$ until $n=11$, and $x_n - \sqrt{n}$ does not begin consistently approaching $0$ until $n=46$. $\endgroup$ – Neil W Apr 29 '14 at 12:56

According to the general form of the Stolz–Cesàro theorem in the linked wikipedia page,

$$\limsup_{n\to \infty}\frac{x_n}{\sqrt{n}}\le \limsup_{n\to \infty}\frac{x_n-x_{n-1}}{\sqrt{n}-\sqrt{n-1}}=2\cdot\limsup_{n\to \infty}\sqrt{n}(x_n-x_{n-1}),\tag{1}$$ and $$\liminf_{n\to \infty}\frac{x_n}{\sqrt{n}}\ge \liminf_{n\to \infty}\frac{x_n-x_{n-1}}{\sqrt{n}-\sqrt{n-1}}=2\cdot\liminf_{n\to \infty}\sqrt{n}(x_n-x_{n-1}).\tag{2}$$ By the definition of $(x_n)$,

$$\limsup_{n\to \infty}\sqrt{n}(x_n-x_{n-1})=\limsup_{n\to \infty}\frac{\sqrt{n}}{x_{n-2}+\sqrt{n-2}}=\frac{1}{\liminf\limits_{n\to \infty}\frac{x_n}{\sqrt{n}} +1},\tag{3}$$ and

$$\liminf_{n\to \infty}\sqrt{n}(x_n-x_{n-1})=\liminf_{n\to \infty}\frac{\sqrt{n}}{x_{n-2}+\sqrt{n-2}}=\frac{1}{\limsup\limits_{n\to \infty}\frac{x_n}{\sqrt{n}} +1}.\tag{4}$$ From $(1)$ and $(3)$ we know that $$\limsup_{n\to \infty}\frac{x_n}{\sqrt{n}}\left(\liminf_{n\to \infty}\frac{x_n}{\sqrt{n}} +1\right) \le 2,\tag{5}$$ and from $(2)$ and $(4)$ we know that $$\liminf_{n\to \infty}\frac{x_n}{\sqrt{n}}\left(\limsup_{n\to \infty}\frac{x_n}{\sqrt{n}} +1\right) \ge 2.\tag{6}$$ Comparing $(5)$ with $(6)$, we can conclude that $$\limsup_{n\to \infty}\frac{x_n}{\sqrt{n}}=\liminf_{n\to \infty}\frac{x_n}{\sqrt{n}}=1.$$

  • $\begingroup$ Nice. Should the middle expression in (4) be the analogue of the middle expression of (3)? $\endgroup$ – user21467 Apr 30 '14 at 12:48
  • $\begingroup$ @StevenTaschuk: Yes, of course. I have corrected it. Thank you for reminding me. $\endgroup$ – user104254 Apr 30 '14 at 15:44
  • $\begingroup$ Thank you! It's a great answer. $\endgroup$ – Ayesha Apr 30 '14 at 19:02
  • 1
    $\begingroup$ @Ayesha: You are welcome! Your question is very interesting. $\endgroup$ – user104254 May 1 '14 at 6:12
  • $\begingroup$ How do you know that $$\limsup_{n\to \infty}\frac{x_n-x_{n-1}}{\sqrt{n}-\sqrt{n-1}}=2\cdot\limsup_{n\to \infty}\sqrt{n}(x_n-x_{n-1})$$? $\endgroup$ – user41281 Sep 12 '14 at 17:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.