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Show that the sequence $(x_n)$ defined as $x_{n+1}=\frac{1}{2}(x_n+x_{n-1}), x_0=2, x_1=7$ is convergent. My attempt is to show that $(x_n)$ is Cauchy. So, i would like to have a feedback on my proof, please. There also might be a problem in a limit calculation... I have also one question in the end of the proof.

First of all we got $\forall p\ge2$ that:

$|x_p-x_{p-1}|=|\frac{1}{2}x_{p-1}-\frac{1}{2}x_{p-3}|=...=(\frac{1}{2})^{p-1}|x_2-x_0|=5(\frac{1}{2})^{p-1}$

Then, by definition of a Cauchy sequence, we have to show that $\forall \varepsilon>0 \ \exists N \ \forall m,n\ge N$:

$|x_m-x_n|<\varepsilon$.

But,

$|x_m-x_n|\le|x_m-x_{m-1}|+|x_{m-1}-x_{m-2}|+...+|x_{n+1}-x_n|\le 5\sum_{k=n}^{m-1}(\frac{1}{2})^k=5(\frac{1}{2^{n-1}}-\frac{1}{2^{m-1}})\le 5\frac{1}{2^{n-1}}$

and so $(x_n)$ is a Cauchy sequence.

To calculate it's limit we can consider the following:

$x_n=(x_n-x_{n-1})+(x_{n-1}-x_{n-2})+...+(x_1-x_0)+x_0 =5\underbrace{\sum_{k=0}^{n-1}(\frac{1}{2})^k}_{\text{geometric series}}+2=5\Big(\frac{1-(1/2)^n}{1-1/2}\Big)+2$

So $\lim_{n \to \infty}x_n=12$... (Which might be false...)

Now i have some questions:

(i) First of all, my limit is clearly false as if i check first 5 values it is clear that the sequence might converge to $\sim5,3$. So, I would like to understand where i commited an error.

(ii) Then, if my proof is correct, i doubt if it holds for real... If we take $m=n-1$ and $\varepsilon=1$, we have a counter example, probably. So, in $|x_p-x_{p-1}|\le5\Big(\frac{1}{2}\Big)^{p-1}$, the value $5$ makes me doubting...

If possible, i would like an elaborate answer and feedback for these questions and this proof, please.

Daniil
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  • Quick question: Have you learned about solving second order linear difference equations with constant coefficients yet? If you set $x_n=\lambda ^n$, plug in and solve for $\lambda$ you can find a particular solution satisyfing your DE – Matthew H. Feb 13 '21 at 22:08
  • How did you got “ $|x_p-x_{p-1}|=|\frac{1}{2}x_{p-1}-\frac{1}{2}x_{p-2}|$”? I suppose you meant $|\frac{1}{2}x_{p-1}-\frac{1}{2}x_{p-3}|$ – Gio Feb 13 '21 at 22:09
  • Each term is obtained by taking the average of the two before, so visually the sequence looks like the partial sums of a convergent alternating series. One can show by induction that the sequence is bounded between $2$ and $7$, for example, so the limit of $12$ is definitely wrong – pancini Feb 13 '21 at 22:09
  • It should be $(-{1 \over 2})^k$ and not $({1 \over 2})^k$ which affects the computation of the limit. Everything else looks fine to me. – Zarrax Feb 13 '21 at 22:11
  • @Gio oh yes, thank you! I didn’t remark it – Daniil Feb 13 '21 at 22:18
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    @Gio: I think archuser got it by simplifying the RHS of $x_{p} - x_{p-1} = 1/2(x_{p-1}+x_{p-2}) - x_{p-1}$. – Rob Arthan Feb 13 '21 at 22:18
  • In your limit calculation, you're summing up straight differences. This is not the same thing as you were doing previously with proving Cauchy. There you're adding absolute values of differences. – user3733558 Feb 13 '21 at 22:33
  • @Zarrax I've just edited my question. Normally it had to be $|x_p-x_{p-1}|=\frac{1}{2}|x_{p-1}-x_{p-3}|$ so i don't have any negatif sign appearing , or probably i am blind... But, i am agree if you simplify and write that $|x_p-x_{p-1}|=-\frac{1}{2}|x_{p-1}-x_{p-2}|$, there is $(-\frac{1}{2})^k$ then. But, why my simplification would be false or where did i commit an error? – Daniil Feb 13 '21 at 22:35
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    See as well here https://math.stackexchange.com/q/2559687 with a "Cauchy" explanation. – Jean Marie Feb 13 '21 at 22:40
  • @archuser, you still need to plug $-1/2$ in the summation from $k=0$ to $n-1$. I think that is the fundamental mistake in what you did – rmdmc89 Feb 13 '21 at 22:44
  • @rmdmc89 could you elaborate, please? I still don’t see why I have to plug the sign -... – Daniil Feb 13 '21 at 23:02
  • @archuser You have $x_p - x_{p-1} = -{1 \over 2}(x_{p-1} - x_{p-2})$, so the geometric series is $5\sum_{k=0}^{n-1}(-{1 \over 2})^k$ and not $5\sum_{k=0}^{n-1}({1 \over 2})^k$ – Zarrax Feb 13 '21 at 23:19
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    See https://math.stackexchange.com/questions/2559687/show-that-the-sequence-a-1-1-a-2-2-a-n2-a-n1a-n-2-converges-b?noredirect=1&lq=1 – Kavi Rama Murthy Feb 13 '21 at 23:28
  • Note that your sequence is defined as the mean of the the two terms before, and the mean of two numbers must lie between the two numbers. – Some Guy Feb 14 '21 at 04:27

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