If $x_1< x_2$ are real numbers, and $x_n = \frac{1}{2}(x_{n-2} + x_{n-1})$ for every $n > 2$, how do I show $\lim x_n = \frac{1}{3}x_1 + \frac{2}{3}x_2$?
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1Did you try writing out explicit formulas for the first few terms? – Umberto P. Mar 29 '17 at 14:55
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I assumed you meant $x_{n-2}$ not $x_{x-2}$, please override edit if wrong. – Arin Chaudhuri Mar 29 '17 at 14:55
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I'm sure this is a duplicate. – marty cohen Mar 29 '17 at 15:50
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Due to the form of the recursion, you can always write $$ x_n = a_k x_{n-k} + b_k x_{n-k-1}, $$ where $a_k$ and $b_k$ are some appropriate coefficients. Now, assume $n$ is very large and solve $a_k$, $b_k$. Using the recursion we get $$ x_n = (a_k/2+b_k) x_{n-k-1} + a_k/2 \cdot x_{n-k-2}, $$ so $a_{k+1} = a_k/2+b_k$ and $b_{k+1} = a_k/2$. Using these we hence obtain the recursion $a_{k+1} = \frac{1}{2}(a_k+a_{k-1})$. With standard techniques (assuming $a_k = u^k$ and solving a quadratic equation that follows), we get the full solution as $a_k = C_1 + C_2(-1/2)^k$. From the initial values $a_1 = 1/2$ and $a_2 = 3/4$ we get the solution that $$ a_k = 2/3 + (-1/2)^k \cdot 1/3 $$ and $$ b_k = a_{k-1}/2 = 1/3 + (-1/2)^{k-1} \cdot 1/6. $$ Now, we assumed $n \rightarrow \infty$, so we can take $k \rightarrow \infty$ as well and get that $a_k \rightarrow 2/3$ and $b_k \rightarrow 1/3$. At the limit the coefficient $a_k$ is the coefficient of $x_2$ and $b_k$ the coefficient of $x_1$, so the claim follows.
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