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Given a sequence $\{a_n\}_{n\in \mathbb N}$ of real numbers such that $$\begin{align} \lim_{n\to\infty}\frac{a_na_{n+1} - a_{n-1}a_{n+2}}{a_{n+1}^2 - a_na_{n+2}} &= p + q && (1)\\[1mm] \lim_{n\to\infty} \frac{a_n^2 - a_{n-1}a_{n+1}}{a_{n+1}^2 - a_na_{n+2}} &= pq && (2)\end{align}$$ where $|p| < |q|$, prove that $$\lim_{n\to\infty} \frac{a_n}{a_{n+1}} = p$$

Attempts:

Idea #1: Let us denote $b_n = \dfrac{a_n}{a_{n+1}}$. If we divide both numerators and denominators of $(1)$ and $(2)$ by $a_na_{n+2}$ and $a_{n+1}a_n$, respectively, we have $$ \begin{align} \lim_{n\to\infty}\frac{\dfrac{a_{n+1}}{a_{n+2}} - \dfrac{a_{n-1}}{a_{n}}}{\dfrac{a_{n+1}}{a_n}\dfrac{a_{n+1}}{a_{n+2}} - 1} &= p+q && (1') \\[1mm] \lim_{n\to\infty} \frac{\dfrac{a_n}{a_{a+1}} - \dfrac{a_{n-1}}{a_n}}{\dfrac{a_{n+1}}{a_n} - \dfrac{a_{n+2}}{a_{n+1}}} &= pq && (2') \end{align}$$ which now can be written as $$\begin{align} \lim_{n\to\infty}\frac{b_{n+1}b_n - b_nb_{n-1}}{b_{n+1} - b_n} &= p+q && (1'') \\[1mm] \lim_{n\to\infty} \frac{b_n - b_{n-1}}{\dfrac{1}{b_{n+1}} - \dfrac{1}{b_{n}}} &= -pq && (2'') \end{align}$$

Now, every numerator and denominator contains the difference of consecutive terms of some sequence (reminds me of Cesaro-Stolz, but that cannot be applied here).

Idea #2: The given conditions can be written as $$\begin{align} \lim_{n\to\infty}\frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} &= p + q && (1')\\[2mm] \lim_{n\to\infty} \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} &= pq && (2')\end{align}$$ Now, maybe a bit of Linear Algebra could be incorporated somehow.

If we set (B. Grossman) $$\begin{align} x_n = \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}}, \quad y_n = \frac{\begin{vmatrix}a_n & a_{n-1} \\ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \\ a_{n+2} & a_{n+1}\end{vmatrix}} \end{align}$$ Then, we have $$\pmatrix{a_{n+1} & a_{n+2}\\a_n & a_{n+1}} \pmatrix{x_n\\y_n} = \pmatrix{a_n\\a_{n-1}}$$ This gives $$\frac{a_{n+1}x_n + a_{n+2}y_n}{a_nx_n + a_{n+1}y_n} = \frac{a_n}{a_{n-1}}$$ Dividing the numerator and the denominator by $a_{n+1}$, we get $$\frac{x_n + \dfrac{a_{n+2}}{a_{n+1}}y_n}{\dfrac{a_n}{a_{n+1}}x_n + y_n} = \frac{a_n}{a_{n-1}}$$ Noting that $x_n \to p+q$, $y_n \to pq$ and assuming that $\dfrac{a_n}{a_{n-1}} \to A$, and sending $n$ to infinity, we get $$\frac{p+q + Apq}{\dfrac{1}{A}(p+q) + pq} = A$$ which simplifies to $$0=0$$

I think I did something wrong somewhere.

Any help is appreciated.

VIVID
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    If $B$ is the expression in the first limit and $C$ is the expression in the second limit, I would compute $-B+\sqrt{B^2-4C}$. – plop May 10 '21 at 15:06
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    With Cramer's rule, we can identify the quotients as the entries of the solution to a system of equations: if we define \begin{align} x_n = \frac{\begin{vmatrix}a_n & a_{n-1} \ a_{n+2} & a_{n+1}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \ a_{n+2} & a_{n+1}\end{vmatrix}}, \quad y_n = \frac{\begin{vmatrix}a_n & a_{n-1} \ a_{n+1} & a_{n}\end{vmatrix}}{\begin{vmatrix}a_{n+1} & a_{n} \ a_{n+2} & a_{n+1}\end{vmatrix}}, \end{align} Then we have $$ \pmatrix{a_{n+1} & a_{n+2}\a_n & a_{n+1}} \pmatrix{x_n\y_n} = \pmatrix{a_n\a_{n-1}} $$ – Ben Grossmann May 10 '21 at 15:10
  • @BenGrossmann Yes, I also had something roughly similar in mind. Could this help? – VIVID May 10 '21 at 15:12
  • @BenGrossmann Added more thoughts with your hint. I think I made a mistake though since the answer has been different than the desired one. – VIVID May 10 '21 at 15:33
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    @VIVID I think that I have a sign error for $y_n$, that might explain your incorrect result – Ben Grossmann May 10 '21 at 15:53
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    @VIVID Actually, even with the correct substitution you end up getting $0=0$. One way to proceed is as follows. suppose that $a_n/a_{n-1} \to A$ and that $A \neq 0$. Divide by $a_{n-1}$ to get the equation $$\pmatrix{a_{n+1}/a_{n-1} & a_{n+2}/a_{n-1}\a_n/a_{n-1} & a_{n+1}/a_{n-1}}\pmatrix{x_n\-y_n} = \pmatrix{a_n/a_{n-1}\1}.$$ Letting $n \to \infty$ in the above equation yields $$ \pmatrix{A^2 & A^3\ A & A^2} \pmatrix{p+q\-pq} = \pmatrix{A\1} \implies\ A(p+q) - A^2 pq = 1 \implies\ (pq)A^2 - (p+q)A + 1 = 0 \implies\ (pA - 1)(qA - 1) = 0 \implies\ A = 1/p \text{ or } A = 1/q. $$ – Ben Grossmann May 10 '21 at 16:18
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    That said, I see no way to differentiate between $1/p$ and $1/q$ with this method, and I don't know how to justify the assumption that $a_n/a_{n-1}$ converges. – Ben Grossmann May 10 '21 at 16:19
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    Another idea is to use the substitution $b_n = p^n a_n$. This leads to the equations \begin{align} \lim_{n\to\infty}\frac{(b_nb_{n+1} - b_{n-1}b_{n+2})}{(b_{n+1}^2 - b_nb_{n+2})} &= 1 + q/p\ \lim_{n \to \infty}\frac{b_n^2 - b_{n-1}b_{n+1}}{b_{n+1}^2 - b_nb_{n+2}} &= q/p, \end{align} and we've reduced the problem to showing that $b_n/b_{n+1} \to 1$. – Ben Grossmann May 10 '21 at 16:22
  • @BenGrossmann Thank you. I will try to use your hints. – VIVID May 10 '21 at 16:50
  • Can you tell us where the problems comes from? – Martin R May 11 '21 at 07:03
  • Previous identical question found with SearchOnMath – Martin R May 11 '21 at 07:13
  • @MartinR Thank you. I was wondering why Approach0 was not working for some time now... – VIVID May 11 '21 at 07:22
  • @VIVID: https://chat.stackexchange.com/transcript/message/57889293#57889293 – Martin R May 11 '21 at 07:23
  • @MartinR I didn't understand why you used problems (in plural). This problem comes from an old book written by Sadovnichiy and Podkoldzin. – VIVID May 11 '21 at 07:23
  • That was a typo. – It is always a good idea to add the source of a problem to the question. – Martin R May 11 '21 at 07:25
  • @MartinR Ok. Got you, and thank you. – VIVID May 11 '21 at 07:27

2 Answers2

1

The following proof is not complete, hoped this is helpful to you.

Denote $b_n=\frac{a_n}{a_{n+1}}$ , then $$\lim\limits_{n\to\infty}\frac{b_n(b_{n+1}-b_{n-1})}{b_{n+1}-b_n}=p+q,\quad \lim\limits_{n\to\infty}\frac{b_n b_{n+1}(b_{n}-b_{n-1})}{b_{n+1}-b_n}=pq.$$

If you assume that $\lim\limits_{n\to\infty} b_n=x$ exists, since $$p+q=\lim\limits_{n\to\infty}\frac{b_n(b_{n+1}-b_{n-1})}{b_{n+1}-b_n}=\lim\limits_{n\to\infty}[\frac{b_n(b_{n}-b_{n-1})}{b_{n+1}-b_n}+b_n],$$ hence $\lim\limits_{n\to\infty}\frac{b_n(b_{n}-b_{n-1}\ )}{b_{n+1}-b_n}=y$ also exists. Moreover, $$x+y=p+q,\quad xy=pq\implies \left\{\begin{array}{l} x=p \\ y=q \end{array}\right.,\ or \left\{\begin{array}{l} x=q \\ y=p \end{array}\right..$$

If you can check that $\lim\limits_{n\to\infty} |b_n|\le \lim\limits_{n\to\infty}|\frac{b_n(b_{n}-b_{n-1}\ )}{b_{n+1}-b_n}|$, then $$\lim\limits_{n\to\infty} b_n=x=p.$$

Liufeng Yang
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HINT: Call $$ X_n=\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n+1}^2-a_na_{n+2}}\\ Y_n=\frac{a_{n}^2-a_{n-1}a_{n+1}}{a_{n+1}^2-a_na_{n+2}} $$ then $$ \lim_nX_n=p+q\\ \lim_nY_n=pq $$ thus $$ \lim_n\frac{X_n}{Y_n}=\frac{p+q}{pq} $$ but $$ \frac{X_n}{Y_n} =\frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n}^2-a_{n-1}a_{n+1}}\;. $$ Then you get $$ \frac{X_n}{Y_nY_{n-1}}= \frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{n-1}^2-a_{n-2}a_{n}} $$ and inductively $$ \frac{X_n}{Y_nY_{n-1}\cdots Y_2}= \frac{a_na_{n+1}-a_{n-1}a_{n+2}}{a_{1}^2-a_{0}a_{2}} $$ from which $$ \frac{X_n}{Y_nY_{n-1}\cdots Y_2}(a_{1}^2-a_{0}a_{2})+a_{n-1}a_{n+2}=a_na_{n+1} $$ divide by $a_{n+1}^2$ and work on LHS.

Joe
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