Proving a sequence converges when combinations of consecutive terms converge Problem: Let $\{x_n\}$ be a sequence of real numbers such that $$\lim_{n\to\infty} 2x_{n+1}-x_n=L \in \mathbf{R}.$$ Prove that $x_n \to L$ as $n\to\infty$.
I can see that if $\{x_n\}$ converges to a limit, then that limit must be $L$. I am having trouble proving the convergence of the sequence. I first tried seeing if I could make $|x_n-L|$ small by playing around with the triangle inequalities, but that didn't pan out. I also tried to prove that $\{x_n\}$ is Cauchy or monotone and bounded, but I couldn't prove the sequence was Cauchy, and I found a counter-example for monotone and bounded: $x_n=(-1)^n/n$.
Any help is appreciated.
 A: 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 $$\lim_{n\to \infty}2^rx_{n+r}-x_n=(2^r-1)L$$
A: 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 $y_n\to 0$.
A: Let me sketch a proof using $\mathbf{\liminf}$ and $\mathbf{\limsup}$:  
Set $y_n = 2x_{n+1}-x_n$, then $x_{n+1} = 1/2(y_n+x_n)$. Since $(y_n)$ coverges to $L$, then it is bounded by $M>0$. Choose $K = \max\{M, x_1\}$. By induction, we can show that $|x_n|\le K, \forall n\in \mathbb{N}$. Therefore
\begin{align}
\liminf_{n\to \infty} x_{n+1} &= \frac{1}{2}L + \frac{1}{2}\liminf_{n\to \infty}x_n \\
\limsup_{n\to \infty} x_{n+1} &= \frac{1}{2}L + \frac{1}{2}\limsup_{n\to \infty}x_n 
\end{align}
Since $(x_n)$ is bounded, then $\mathbf{\liminf}$ and $\mathbf{\limsup}$ of $(x_n)$ must be finite.  It follows from the previous two equations that
$$\liminf_{n\to \infty}x_n = \limsup_{n\to \infty} x_n= L.$$
A: Another way to prove that $\lim_\limits{n\to\infty}x_n=L$ is to apply Stolz-Cesàro theorem (∙/∞ case) by letting $\;a_n=x_0+\sum_\limits{i=0}^{n-1}2^iy_i\;$ and $\;b_n=2^n\;$ where $\;y_n=2x_{n+1}-x_n\;.$
Since $\;\dfrac{a_{n+1}\!-\!a_n}{b_{n+1}\!-\!b_n}\!=\!y_n\;$ and $\;\dfrac{a_n}{b_n}\!=\!x_n\;,\;$ we get the required result.
