Let \(\alpha\in(0,1)\). Prove that every real sequence satisfying \[ x_{n+1}=\alpha x_n+(1-\alpha)x_{n-1} \] has a limit, and express the limit in terms of \(\alpha,x_0,x_1\).
Solution
1. Define\[ d_n=x_n-x_{n-1} \qquad (n\geq1). \]Then\[ d_{n+1}=(\alpha-1)d_n. \]
Using the recurrence, \[ \begin{aligned} d_{n+1} &= x_{n+1}-x_n\\ &= \alpha x_n+(1-\alpha)x_{n-1}-x_n\\ &= -(1-\alpha)(x_n-x_{n-1})\\ &= (\alpha-1)d_n. \end{aligned} \]
2. For every \(n\geq1\),\[ d_n=(\alpha-1)^{n-1}(x_1-x_0). \]
Step 1 is a first-order geometric recurrence with initial value \[ d_1=x_1-x_0. \] Iteration gives the formula.
3. For every \(n\geq1\),\[ x_n = x_0 + (x_1-x_0) \sum_{j=0}^{n-1}(\alpha-1)^j. \]
Telescope the first differences: \[ x_n-x_0 = \sum_{k=1}^n(x_k-x_{k-1}) = \sum_{k=1}^n d_k. \] Substitute step 2 and reindex.
4. The sequence \((x_n)\) converges and\[ \boxed{ \lim_{n\to\infty}x_n = \frac{x_1+(1-\alpha)x_0}{2-\alpha}. } \]
Because \(0<\alpha<1\), \[ |\alpha-1|=1-\alpha<1. \] Thus the geometric series in step 3 converges to \[ \frac{1}{1-(\alpha-1)} = \frac{1}{2-\alpha}. \] Therefore \[ \begin{aligned} \lim_{n\to\infty}x_n &= x_0+\frac{x_1-x_0}{2-\alpha}\\ &= \frac{x_1+(1-\alpha)x_0}{2-\alpha}. \end{aligned} \]
5. Q.E.D.
Step 4 proves convergence and gives the requested limit.