Proving a sequence converges when combinations of consecutive terms converge

Solution 1:

Denote $\mu_n=\frac12(2x_{n+1}-x_n-L)$. We know that $\mu_n\to 0$ when $n\to\infty$. Rearranging a bit for convenience $$ x_{n+1}-L=\frac12 (x_n-L)+\mu_n\qquad\Leftrightarrow\qquad y_{n+1}=\frac12 y_n+\mu_n $$ where $y_n=x_n-L$. We need to prove that $y_n\to 0$ when $n\to 0$. By repeating the recursion we can get $$ y_{n+m}=\left(\frac12\right)^my_n+\sum_{k=0}^{m-1}\left(\frac12\right)^{m-1-k}\mu_{n+k} $$ which gives the estimation $$ |y_{n+m}|\le |y_n|\left(\frac12\right)^m+\sup_{k\ge n}|\mu_k| \cdot\underbrace{\sum_{k=0}^{m-1}\left(\frac12\right)^{m-1-k}}_{\le 2}. $$ So taking $n$ large enough to make the second term $\le\epsilon/2$ and then taking $m$ large enough to make the first term $\le\epsilon/2$ we prove that $y_n\to 0$.

Solution 2:

Here is a suggested start (you can fill in details I'm sure).

Note that $$\lim_{n\to \infty}4x_{n+2}-x_n=3L$$ and progressing inductively you get $$\lim_{n\to \infty}2^rx_{n+r}-x_n=(2^r-1)L$$