# Convergence of a sequence 4

Suppose there is a sequence $\{ x_n \}$. Let us define another sequence $\{ y_n \}$ such that $$y_n=2x_{n+1} - x_n$$ Please prove that if $\{y_n\}$ converges to $L$, then $\{x_n\}$ is convergent and also converges to $L$.

I have approached this problem like:
If I consider that $\{x_n\}$ converges to $\alpha$, then taking limit, we can show that: $$L=\lim y_n = \lim (2x_{n+1}-x_n)=2\lim x_{n+1}-\lim x_n = 2\alpha-\alpha = \alpha$$

However, I am having problem in proving that limit for $\{x_n\}$ exists.

We have that $x_{n+1} = \frac{1}{2} y_n + \frac{1}{2} x_n$. This can be inductively continued to rewrite: $$x_ {n+1} = \frac{1}{2} y_n + \frac{1}{4} y_{n-1} + \frac{1}{8} y_{n-2} + ... \frac{1}{2^n}y_1 + \frac{1}{2^n} x_1$$ Can you see where to go from here?
Edit: A cleaner way of thinking about it without the big sum although roughly the same underlying strategy. Fix $\epsilon > 0$ and suppose that for $n \geq N$ we have $|y_n - L| < \epsilon$. Now in particular we have for all $n > N$:
$$|x_{n} - L| < \frac{1}{2}|x_{n-1} - L| + \frac{1}{2}|y_{n-1} - L| < \frac{1}{2}|x_{n-1} - L| + \frac{\epsilon}{2}$$
$$|x_{N + k}-L| < \frac{1}{2^k} |x_N - L| + \epsilon \frac{2^k - 1}{2^k}$$