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There is a sequence $x_n$ defined recursively:

$$x_{n+2} = \frac{x_n + x_{n+1}}{2}, n\geqslant1 $$

And $x_1=a$, $x_2=b$.

I need to find $lim_{n\to\infty} x_n$.

First, I decieded to get rid of recurrent definition, and I got the following:

$$x_n=a+\sum_{k=1}^{n}\frac{b-a}{2^k} = a+(b-a)\sum_{k=1}^{n}\frac{1}{2^k}$$

And then I started finding the limit:

$$lim_{n\to\infty} x_n = a + (b-a) (lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{2^k})$$

As long as $\sum_{k=1}^{inf}\frac{1}{2^k}$ is 1, it equals to $a+(b - a)$ which is $b$. But this answer is marked as wrong.

What is my mistake?

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    $\begingroup$ The closed form of $x_n$ is $x_n =(\dfrac{a+b}{3})(1-(-0.5)^{n-2})+\dfrac{b}{3}(1-(-0.5)^{n-3})$ $\endgroup$ – Aryaman Jal Dec 2 '17 at 12:37
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Consider substitution $x_n = \xi^n$. Recurrent definition says now that $$2\xi^2 = 1 + \xi.$$

Arbitrary solution to second order recurrent equation is a linear combination of $\xi_{1,2}$. But we have two initial conditions $x_1 =a$ and $x_2 = b$.

PS. Concerning your solution. I think it doesn't satisfy the recurrent equation. Probably you should check it.

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rearrange $x_{n+2} = \frac{x_n + x_{n+1}}{2}$ into $x_{n}=\frac12x_{n-1}+\frac12x_{n-2}$, you can see that this is a linear homogenous recurrence relation of degree $2$. in this case let's solve it by saying $x_n=r^n$ thus $x_{n}=\frac12x_{n-1}+\frac12x_{n-2}\implies r^n=\frac12r^{n-1}+\frac12r^{n-2}\implies r^2=\frac12r^1+\frac12\implies r_{1,2}=\begin{cases}1\\-\frac12\end{cases}$

with this we can get to the equation $x_n=c_1\cdot1^n+c_2\cdot\left(-\frac12\right)^n=c_1+c_2\cdot\left(-\frac12\right)^n$ and with the initial condition we have $x_1=c_1-\frac12c_2=a$ and $x_2=c_1+\frac14c_2=b$ solve this:$$c_1=a+\frac12c_2\implies c_1+\frac14c_2=a+\frac12c_2+\frac14c_2=a+\frac34c_2=b\implies c_2=\frac43\left(b-a\right)\\\implies c_1=a+\frac12\cdot\frac43\left(b-a\right)=a+\frac23\left(b-a\right)$$


Edit

Now that I think about it you don't even need to find $c_2$, because $\lim_{n\to\infty}(-1/2)^n=0$ you have $\lim_{n\to\infty}x_n=c_1$. So instead of what I did you should solve for $c_1$ first and then you can stop.

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