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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?

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marked as duplicate by punctured dusk, Delta-u, Arnaud Mortier, José Carlos Santos, Eric Wofsey Aug 5 '18 at 19:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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 $$ \lim_{n \to \infty} x_n = \lim_{n \to \infty} A_n + \lim_{n \to \infty} B_n + \lim_{n \to \infty} C_n = \frac L3 + 0 + 0 = \frac L3\, . $$

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  • $\begingroup$ really nice answer. +1 the ingenuity of the argument beats me. $\endgroup$ – Paramanand Singh Jun 13 '16 at 9:38
  • $\begingroup$ @ParamanandSingh: Thank you! $\endgroup$ – Martin R Jun 13 '16 at 10:12
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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,

$$\limsup_n x_n=-\liminf_n x_n=0$$

$\blacksquare$

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