Proving a sequence is convergent Let $\
(x_n )_{n \ge 0} 
$ be a convergent sequence . 
Prove that another sequence $\
(y_n )_{n \ge 0} 
\
$ defined as $
x_n  = y_n  + 2y_{n + 1} 
 $ is convergent as well . I tried mixing the $\
\varepsilon ,N(\varepsilon )
$ definition of a limit with the Bolzano-Weierstrass theorem but I didn't get anything useful .
Any ideas ?

Suppose that $x_n$ converges to $L$ and for the sake of argument assume that $y_n$ converges to $A$. Then $L = A + 2 A = 3A$ and $A = L/3$.
Now let us examine the sequence
$x_0 = y_0 + 2y_1$ so $\frac{x_0 - y_0}{2} = y_1$, and $$y_2 = \frac{x_1 - y_1}{2} = \frac{x_1}{2} - \frac{x_0}{4} + \frac{y_0}{4}$$
Then $$y_3 = \frac{x_2 - y_2}{2} = \frac{x_2}{2} - \frac{x_1}{4} + \frac{x_0}{8} - \frac{y_0}{8}$$
And $$y_4 = \frac{x_3}{2} - \frac{x_2}{4} + \frac{x_1}{8} - \frac{x_0}{16} + \frac{y_0}{16}$$
By induction we have:
$$y_{n+1} = \sum_{k=0}^n (-1)^{n-k} \frac{x_k}{2^{n-k+1}} + \frac{y_0}{2^n}$$
Now let $\epsilon > 0$, and suppose $N \in \mathbb{N}$ is chosen so that $|x_n-L| < \epsilon$ when $n>N$.
For $n>N$ write $$M_n = \sum_{k=1}^N (-1)^{n-k} \frac{x_k}{2^{n-k+1}} = \frac{1}{2^{n-N}} \sum_{k=1}^N (-1)^{n-k} \frac{x_k}{2^{N-k+1}}.$$ Pick $n$ large enough so that $M_n < \epsilon$ and $y_0/2^n < \epsilon$.
Then show that $$\left|y_{n+1} - \sum_{k=N}^n(-1)^{n-k}\frac{L}{2^{n-k+1}}\right|$$ is bounded by $4\epsilon.$
