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I have the following sequence: $a_{n+2}=0.5(a_n+a_{n+1})$, $a_1=2, a_2=5$, and I need to prove that $\lim\limits_{n\to \infty}a_n$ exists, and find it.

I don't know ho to prove that the limit exists, since the sequence neither decreasing nor increasing.

In addition, I don't know how to find it, since the equation I am getting is $L=0.5(L+L)$, whitch is true for all $L$.

Daniel
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    You can prove convergence by proving that it is a Cauchy sequence - $ | a_{n +2} - a_{n +1} | = \dfrac{1}{2} | (a_{n+1} - a_n) |$ i.e. absolute difference gets halved as $n$ increases. – ab123 Nov 20 '20 at 08:00
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    Yes that is another standard way to deal with such problems (provided that the underlying space has this Cauchy property, which in this case is true). –  Nov 20 '20 at 08:26

3 Answers3

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$a_1 = 2$, $a_2 = 5$

$$2a_3 = a_1 + a_2$$ $$2a_4 = a_2 + a_3$$ $$2a_5 = a_3 + a_4$$ $$\cdots$$ $$2a_{n - 1} = a_{n- 2} + a_{n - 3}$$ $$2a_n = a_{n- 1} + a_{n - 2}$$

Add all equations, you will get

$$2a_n + a_{n - 1} = a_1 + 2a_2$$

Since you can show that the sequence is Cauchy and hence converges as I explained in the comment above, $a_n \to L$ and $a_{n - 1}\to L$ as $n \to \infty$

Hence, $$2L + L = a_1 + 2a_2$$ or $$L = \dfrac{2 + 2(5)}{3} = 4$$

ab123
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As mentioned in comments, the sequence may be shown to be Cauchy and hence convergent.

This is a second-order homogeneous linear recurrence. The characteristic equation is $$2\lambda^2-\lambda-1=(2\lambda+1)(\lambda-1)=0$$ with roots $1$ and $-\frac12$. Therefore the general equation is $a_n=p+q(-1/2)^n$ where $p$ and $q$ are determined from the initial conditions; here $p=q=4$. Since $(-1/2)^n$ tends to zero as $n\to\infty$, the limit is $p=4$.

Parcly Taxel
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  • I don't really understand how to arrive at the characteristic equation (more precisely, in general) or how the roots give a general solution for this $a_{n}$, but if you were to give more detail I would happily upvote. – Derek Luna Nov 20 '20 at 08:30
  • @DerekLuna Read up Wikipedia... – Parcly Taxel Nov 20 '20 at 08:31
  • Ok well you're answer is practically useless for nearly anyone that would be asking this question in the first place. If your goal is to be helpful you should probably explain exactly how your results are working here for the specific problem at hand. – Derek Luna Nov 20 '20 at 08:33
  • @DerekLuna I do not have any results per se. I am merely applying well-known techniques for solving such problems. – Parcly Taxel Nov 20 '20 at 08:40
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    @DerekLuna I agree with Parcly Taxel. It's like you asked to explain quadratic formula. This is linear recurrence 101 and everybody dealing with them should know how to solve – Raffaele Nov 20 '20 at 09:43
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Here I leave a sketch. I may return and provide the details if someone asks for it when I am free.

Let $x_n = a_{2n+1}$ for each $n \ge 1$, and $y_n = a_{2n}$ for each $n \ge 1.$

Show that $(x_n)_{n \ge 1}$ and $(y_n)_{n \ge 1}$ are monotone and bounded and hence conclude that each of them converge to a finite limit. Let $l, l'$ be the limits of these two sequences respectively.

Now, use $a_{n+2} = \frac{a_n + a_{n+1}}{2}$ to show that $l = l'.$

Conclude from here that $(a_n)_{n \ge 1}$ converges.

Also you can find the value of the limits $l , l'$ and hence find the point where $(a_n)_{n \ge 1}$ converges.