With $y_n$ a sequence of real numbers, prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges Let $y_n$ be a sequence of real numbers. Prove that if $y_n=x_{n-1}+2x_{n}$ converges then $x_n$ also converges.
Let us suppose that $y_{n}$ goes to a limit $L$. Then for all $\varepsilon >0$, for sufficiently large $n$, $|y_n -L| < \varepsilon$ .
But $|y_n - L| < 2|x_n-\frac{L}{3}|+|x_n - \frac{L}{3}|$
But then how to proceed? 
 A: We have
$$
 \begin{aligned}
 x_2 &= \frac 12 y_2 - \frac 12  x_1  \\
 x_3 &= \frac 12  y_3 - \frac 12 x_2 = \frac 12  y_3 + \frac 14  y_2 - \frac 14  x_1 \\
 x_4 &= \frac 12  y_4 - \frac 12 x_3 = \frac 12  y_4 + \frac 14  y_3 - \frac 18  y_2 + \frac 18 x_1 \\
 \ldots 
\end{aligned}
$$
and generally for $n\ge 2$
$$ \begin{aligned}
 x_n &=
  \frac 12 y_n - \frac 14 y_{n-1} + \frac 18  y_{n-2}
 - \ldots + \frac{(-1)^n}{2^{n-1}} y_2 - \frac{(-1)^n}{2^{n-1}} x_1 \\
&= \sum_{k=2}^{n} \frac{(-1)^{n-k}}{2^{n-k+1}}y_{k} - \frac{(-1)^n}{2^{n-1}} x_1 \\
&= L \sum_{k=2}^{n} \frac{(-1)^{n-k}}{2^{n-k+1}}
 + \sum_{k=2}^{n} \frac{(-1)^{n-k}}{2^{n-k+1}} \bigl(y_{k} - L \bigr)
 - \frac{(-1)^n}{2^{n-1}} x_1 \\
 &=:A_n + B_n + C_n 
\end{aligned}
$$
Now 
$$
 A_n = L \sum_{j=0}^{n-2} \frac{(-1)^{j}}{2^{j+1}} \to \frac L3
$$
and
$$
 C_n =  - \frac{(-1)^n}{2^{n-1}} x_1 \to 0
$$
for $n \to \infty$.
For all $\varepsilon > 0$ then is a  $N$ such that
$\lvert y_n - L \rvert < \varepsilon$ for $n \ge N$.
Then
$$
 B_n = \left( \sum_{k=2}^{N-1} + \sum_{k=N}^{n} \right) \frac{(-1)^{n-k}}{2^{n-k+1}} \bigl(y_{k} - L \bigr)
$$
The first (finite) sum converges to zero,
and the absolute value of the second sum can be estimated above by
$$
 \varepsilon \sum_{k=0}^\infty \frac{1}{2^{j+1}}
= \varepsilon \, .
$$
It follows that that $B_n \to 0$ and therefore
$$
 \lim_{n \to \infty} x_n =  \lim_{n \to \infty} A_n
+ \lim_{n \to \infty} B_n + \lim_{n \to \infty} C_n = \frac L3 + 0 + 0
= \frac L3\, .
$$
A: WLOG, by linearity, we assume that $\lim_{n}y_n =0$.
First we will show that $\limsup_n x_n<\infty$ and $\liminf_n x_n<\infty$, by contradiction.
Set $k_n$ the first time $|x_{k_n}|>M$. Then, 
$$|y_{k_n}|=|x_{k_n-1}+2x_{k_n}  |>|2x_{k_n}  |-|x_{k_n-1}|>M$$
Now taking $k_n\rightarrow \infty$ we obtain a contradiction.
Also, it is easy to rule out that $\limsup_n x_n<0$ and $\liminf_n x_n>0$.
Set $A=\max ( \limsup_n x_n,-\liminf_n x_n)$. Assume that $A>0$, and, by symmetry, WLOG take $0\neq s_1=\limsup_n x_n>-\liminf_n x_n$. And, choose $k_n$ such that $x_{k_n}\rightarrow s_1$. 
So,
$$y_{k_n}-2x_{k_n}=x_{k_n-1}  $$ 
Taking $k_n\rightarrow \infty$ we find $\lim_nx_{k_n-1}=-2s_1 $. Which gives $$-2s_1\geq \liminf_n x_n \Rightarrow 0<2s_1\leq -\liminf_n x_n$$ a contradiction, hence $A=0$. Therefore,
$$\limsup_n x_n=-\liminf_n x_n=0$$ 
$\blacksquare$
