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Given $x_1 = a$ and $x_2 = b$ find the values of $a, b \in \Bbb R, n\in\Bbb N$ for which the following recurrences converge (diverge): $$ \begin{align*} x_{n+2} &= 2x_{n+1} - x_n \tag1\\ x_{n+2} &= 4x_{n+1} - 3x_n\tag2\\ x_{n+2} &= -2x_{n+1} - x_n\tag3\\ x_{n+2} &= x_{n+1} + 2x_n\tag4\\ \end{align*} $$


$(1)$: $$ \begin{align} x_{n+2} &= 2x_{n+1} - x_n \iff \\ x_{n+2} - x_{n+1} &= x_{n+1} - x_n = \\ &= x_{n+1} - x_{n}\\ &= x_{n} - x_{n-1}\\ &\cdots\\ &= x_{2} - x_{1}\\ &= b - a \end{align} $$ Now taking the limit: $$ \lim_{n\to\infty}(x_{n+1} - x_{n}) = \lim_{n\to\infty}(a-b)=a-b $$ For this recurrence to be convergent $a$ must be equal to $b$, otherwise it doesn't satisfy Cauchy criteria, thus: $$ a = b \implies \exists \lim_{n\to\infty}x_n \\ a \ne b \implies \exists! \lim_{n\to\infty}x_n $$


$(2)$: $$ \begin{align} x_{n+2} &= 4x_{n+1} - 3x_n \iff\\ x_{n+2} - x_{n+1} &= 3(x_{n+1} - x_n) \\ &= 3^2(x_{n} - x_{n-1})\\ &\cdots \\ &= 3^{n-1}(x_{2} - x_{1})\\ &= 3^{n-1}(b - a) \end{align} $$

This case is similar to $(1)$: $$ b = a \implies \exists \lim_{n\to\infty}x_n\\ b \ne a \implies \exists! \lim_{n\to\infty}x_n $$


$(3)$: $$ \begin{align} x_{n+2} &= -2x_{n+1} - x_n \iff \\ x_{n+2} + x_{n+1} &= -(x_{n+1} + x_n) \\ &= (x_{n} + x_{n-1}) \\ &= -(x_{n-1} + x_{n-2}) \\ &\cdots\\ &= (-1)^{n-1}(x_2 + x_1) \\ &= (-1)^{n-1}(b + a) \\ \end{align} $$

In this case convergence is only possible if: $$ a + b = 0 \implies \exists \lim_{n\to\infty}x_n\\ a + b \ne 0 \implies \exists! \lim_{n\to\infty}x_n $$


$(4)$. I'm stuck with this one, not sure what transformations to apply. I could probably use generating functions or solve through a characteristic polynomial, but that is heavy machinery for such a simple case. I would also like to not apply any solutions involving matrices since i'm not very familiar with linear algebra yet.

Could you please verify the first three cases and either suggest a solution or give a hint for the last one? Thank you!

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  • $\begingroup$ how $\lim_{n\to\infty}|x_{n+1} -x_n|=0 $ will ensure that limit of sequence will exist ? $\endgroup$ – neelkanth Feb 6 at 17:02
  • $\begingroup$ @neelkanth well, it might not, i'm not sure. However if we choose some $C\in\Bbb R^+$ and $\lambda \in (0,1)$ such that $|x_{n+1} - x_n| = |a - b| \le C\lambda^n$ then $x_n$ converges. And this seems to only be possible for $a-b = 0$ $\endgroup$ – roman Feb 6 at 17:25
  • $\begingroup$ You could have tried characteristic polynomial method indeed, e.g. very similar question here $\endgroup$ – rtybase Feb 25 at 21:02
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The first two points are correct (but you need to justify $a= b \implies (x_n) \text{ is convergent}$).

For the third point $x_{n+1}+x_n=0$ for all $n$ does not ensure that $(x_n)$ is convergent. In fact it leads to $x_{n+1}=-x_n$ so $$x_n=(-1)^{n-1} a = (-1)^{n-2} b$$ so the condition is $a+b=0 \text{ and } a=b=0$.

For the last point in you want to applies a similar method the best way is to define $$y_{n}=x_{n+1}+\lambda x_n$$ where $\lambda$ is a parameter we will choose later.

In this case we have $$y_{n+1}=x_{n+2}+\lambda x_{n+1}=x_{n+1}+2x_n+\lambda x_{n+1}\\=(1+\lambda)\left(x_{n+1}+\frac{2}{1+\lambda} x_n \right).$$

So by choosing $\lambda$ such that $2/(1+\lambda)=\lambda$ i.e $\lambda=1$ or $\lambda=-2$ we obtain

  • $\lambda=1$

$$(x_{n+2}+x_{n+1})=2(x_{n+1}+x_n)$$

  • $\lambda=-2$

$$(x_{n+2}-2x_{n+1})=-(x_{n+1}-2x_n)$$

so you obtain a necessary condition for convergence $$b+a=0$$ $$b-2a=0$$ and so the sequence is convergent iff $a=b=0$.


Note that by considering a generic recurrence you can show that the good choice of $\lambda$ are always the opposite of the roots of the characteristic polynomial.

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  • $\begingroup$ Nice approach for the last case. Thank you! $\endgroup$ – roman Feb 25 at 18:52

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