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Two recursive sequences are given: $$x_{n+1}=(1-\frac{2}{n})x_n-\frac{y_n}{n}+\frac{4}{n}$$ $$y_{n+1}=(1-\frac{1}{n})y_n-\frac{x_n}{n}+\frac{3}{n}$$where $x_1=0$ and $y_1=0$. Find the following limit: $\lim_{\underset{n \to \infty}{}}(x_n+y_n)$.

I tried to denote $a_n=x_n+y_n$ and get the equation only in terms of a_n, but after taking the limit, I saw that the left hand side was the same as the right hand side. I don't know which approach I should take.

perenqi
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    Can you show $(1,2)$ is a fixed point? – D S Feb 18 '24 at 10:50
  • Yes, I can show it. – perenqi Feb 18 '24 at 11:12
  • What about $\lim_{n \to \infty} x_n$ and $\lim_{n \to \infty} y_n$? – D S Feb 18 '24 at 11:31
  • I tried to find these limits, but I couldn't even get the recursive formulas separately for $x_n$ and $y_n$. It looks like it's impossible. $x_n$ must be in the equation for $y_n$ and vice versa. – perenqi Feb 18 '24 at 11:42
  • Try to prove there are no other fixed points. Then show that $\forall n > N$ for some $N$, $x_n$ is bounded above and below by $1$ and $x_{n-1}$. Do something similar for $y_n$. – D S Feb 18 '24 at 11:51
  • Are you asked to have a precise value for the limit or just show that the limit exists and is finite ? – Jean Marie Feb 18 '24 at 15:07
  • I asked to have a precise value for the limit. – perenqi Feb 18 '24 at 15:17

1 Answers1

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Let us write the problem in matrix form $$\pmatrix{x_{n+1}\\y_{n+1}}=\underbrace{\pmatrix{1-2/n&-1/n\\-1/n&1-1/n}}_ {:=M}\pmatrix{x_{n}\\y_{n}}+\pmatrix{4/n\\3/n} \tag{1}$$ Make a change of variable $\color{red}{(z_n,t_n)=(x_n -1,y_n - 2)}$, the equation $(1)$ can be simplified as $$\pmatrix{z_{n+1}\\t_{n+1}}=M\cdot\pmatrix{z_{n}\\t_{n}}\tag{2}$$ Diagonalize the matrix $M= U^{-1}D_nU$ where $$U= \pmatrix{\frac{1+\sqrt{5}}{2}&1 \\\frac{1 -\sqrt{5}}{2}&1}$$ $$D_n=\pmatrix{1-\frac{3+\sqrt{5}}{2n}&0 \\0&1-\frac{3-\sqrt{5}}{2n}}$$ Then $$(2) \iff \pmatrix{z_{n+1}\\t_{n+1}}=U^{-1}D_nU\cdot\pmatrix{z_{n}\\t_{n}} \iff U\pmatrix{z_{n+1}\\t_{n+1}} = D_n\cdot U\pmatrix{z_{n}\\t_{n}}\tag{3}$$ Denote $\color{red}{V_n = \pmatrix{V_{n,1}\\V_{n,2}} = U\pmatrix{z_{n}\\t_{n}}} \in \mathbb{R}^2$, then $$(3) \iff V_{n+1} = D_n\cdot V_n$$ $$\implies V_n = \left(\prod_{i=1}^{n-1}D_i \right)V_1 = \color{red}{ \pmatrix{\prod_{i=1}^{n-1}\left(1-\frac{3+\sqrt{5}}{2i}\right)&0 \\0&\prod_{i=1}^{n-1}\left(1-\frac{3-\sqrt{5}}{2i}\right)}} \cdot V_1 \tag{4}$$

We notice that, for all $a>0$: $$\begin{align} \prod_{i=1}^{n-1}\left(1-\frac{a}{i}\right)&=\exp\left(\sum_{i=1}^{n-1}\ln \left( 1-\frac{a}{i}\right) \right) \xrightarrow{n\to+\infty} 0 \tag{5} \end{align}$$ because the series $\sum_{i=1}^{n-1}\ln \left( 1-\frac{a}{i}\right)$ diverges to $-\infty$ (We have : $\ln \left( 1-\frac{a}{i}\right) \sim -\frac{a}{i}$ for $i$ large and $\sum_{i=1}^{n-1} \frac{a}{n} \xrightarrow{n\to+\infty} +\infty$).

From $(4)(5)$, we deduce: $$V_n \xrightarrow{n\to+\infty} \pmatrix{0\\0}\implies \pmatrix{z_{n}\\t_{n}} = U^{-1}V_n \xrightarrow{n\to+\infty} \pmatrix{0\\0} \implies \color{red}{\pmatrix{x_{n}\\y_{n}} \xrightarrow{n\to+\infty} \pmatrix{1\\2} } $$

We conclude that $a_n = x_n +y_n \xrightarrow{n\to+\infty} 3$.

NN2
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  • Thank you so much! – perenqi Feb 18 '24 at 16:34
  • @perenqi You're welcome! – NN2 Feb 18 '24 at 16:46
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    [+1] Nice answer. – Jean Marie Feb 18 '24 at 18:45
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    @NN2 An afterthought : the infinite product you have considered is "close" to the famous expression of $\Gamma$ function as an infinite product : $\frac{1}{\Gamma(z)}=ze^{\gamma z} \prod_{k=1}^{\infty}(1+\tfrac{z}{n})e^{- \frac{z}{n}}$ ($\gamma =$ Euler-Mascheroni constant). I wonder if a slightly different derivation could be possible taking into account this formula. – Jean Marie Feb 19 '24 at 00:17