If $ (x_{n})_{n \in \mathbb{N}} $ is a sequence in $ \mathbb{R} $ and $ \displaystyle \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x $, then is it necessarily true that $$ \lim_{n \to \infty} x_{n} = x? $$
Could anyone kindly offer suggestions on how to approach this problem? Thanks!
Let $ \epsilon > 0 $. Then there exists an $ N \in \mathbb{N} $ sufficiently large so that $$ \forall k \in \mathbb{N}: \quad x - \epsilon < 2 x_{N + k} - x_{N + k - 1} < x + \epsilon. $$ This means that \begin{equation*} \begin{array}{ccccc} x - \epsilon & < & 2 x_{N + 1} - x_{N} & < & x + \epsilon, \\ 2 (x - \epsilon) & < & 4 x_{N + 2} - 2 x_{N + 1} & < & 2 (x + \epsilon), \\ 4 (x - \epsilon) & < & 8 x_{N + 3} - 4 x_{N + 2} & < & 4 (x + \epsilon), \\ 8 (x - \epsilon) & < & 16 x_{N + 4} - 8 x_{N + 3} & < & 8 (x + \epsilon), \\ \vdots & & \vdots & & \vdots \end{array} \end{equation*} Summing the first $ m $ lines of this system of inequalities yields $$ (2^{m} - 1) (x - \epsilon) < 2^{m} x_{N + m} - x_{N} < (2^{m} - 1) (x + \epsilon). $$ Dividing this throughout by $ 2^{m} $, we obtain $$ \left( 1 - \frac{1}{2^{m}} \right) (x - \epsilon) < x_{N + m} - \frac{x_{N}}{2^{m}} < \left( 1 - \frac{1}{2^{m}} \right) (x + \epsilon). $$ Letting $ m \to \infty $, we then get $$ x - \epsilon \leq \liminf_{m \to \infty} x_{N + m} = \liminf_{n \to \infty} x_{n} \quad \text{and} \quad \limsup_{n \to \infty} x_{n} = \limsup_{m \to \infty} x_{N + m} \leq x + \epsilon. $$ However, $ \epsilon $ is arbitrary, so $$ \liminf_{n \to \infty} x_{n} = \limsup_{n \to \infty} x_{n} = x, $$ which implies that $ \displaystyle \lim_{n \to \infty} x_{n} = x $.
Define $y_n=x_n-x$. You will find $\lim_{n\rightarrow\infty}(2y_{n+1}-y_n)=0$
Response to Andre's comment
@AndreNicolas I agree the above line is not a rigorous proof. So here is the whole proof.
$\lim_{n\rightarrow\infty}(2y_{n+1}-y_n)=0$, so $\forall \epsilon>0$, $\exists N$ s.t. $\forall n>N, |2y_{n+1}-y_n|<\epsilon$
Define $A=|y_N|$, so, $|y_{N+1}|<(A+\epsilon)/2$
Then, it is not hard to prove through mathematical induction that
$$ \forall \text{positive integer} k, |y_{N+k}|<\frac{A}{2^k}+\sum_{i=1}^k \frac{\epsilon}{2^i}< \frac{A}{2^k} + \epsilon $$
Now define $M=\lceil \log_2(A/\epsilon)\rceil$, then $\forall n>N+M, |y_n|<2\epsilon$
Thus $\lim_{n\rightarrow\infty}y_n=0. \quad \blacksquare $
Yes, it has to be true. Take $(a_n), (b_n)$, such that $a_n = 2^n \cdot x_n$ and $b_n = 2^n$. Now calculate:
$$ \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_{n}} = \lim_{n\to \infty} \frac{2^{n+1} x_{n+1} - 2^n x_n}{2^{n+1} -2^n} = \lim_{n \to \infty}2 x_{n+1} - x_n = x $$
Notice $x_n = \frac{a_n}{b_n} \wedge b_n \to \infty^+$. Above limit exists, so Stolz–Cesàro theorem ends the proof.
$$\lim_{n \to \infty} x_n = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_{n}} = x$$
Q.E.D.
The assertion is true even in the more general setting of a locally convex topological vector space (l.c.t.v.s.).
Theorem: Let $ X $ be an l.c.t.v.s. with base field $ \mathbb{F} $ ($ = \mathbb{R} $ or $ \mathbb{C} $). Let $ x \in X $ and suppose that $ (x_{n})_{n \in \mathbb{N}} $ is a sequence in $ X $ satisfying $ \displaystyle \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x $. Then $ \displaystyle \lim_{n \to \infty} x_{n} = x $.
Proof
Suppose that the topology of $ X $ is generated by the family $ (p_{i})_{i \in I} $ of semi-norms. Define a new sequence $ (y_{n})_{n \in \mathbb{N}} $ in $ X $ by $$ \forall n \in \mathbb{N}: \quad y_{n} \stackrel{\text{def}}{=} x_{n} - x. $$ As $$ \forall n \in \mathbb{N}: \quad 2 x_{n + 1} - x_{n} - x = 2 y_{n + 1} - y_{n}, $$ we obtain $ \displaystyle \lim_{n \to \infty} (2 y_{n + 1} - y_{n}) = 0_{X} $. Now, fix an arbitrary $ i \in I $ and an arbitrary $ \epsilon > 0 $. Then there exists an $ N \in \mathbb{N} $ sufficiently large so that $$ \forall k \in \mathbb{N}: \quad {p_{i}}(2 y_{N + k} - y_{N + k - 1}) < \frac{\epsilon}{2}. $$ In particular, $$ \forall k \in \mathbb{N}: \quad {p_{i}}(2^{k} y_{N + k} - 2^{k - 1} y_{N + k - 1}) < 2^{k - 1} \cdot \frac{\epsilon}{2}. \quad (\spadesuit) $$ It follows readily that \begin{align*} \forall m \in \mathbb{N}: \quad {p_{i}}(2^{m} y_{N + m} - y_{N}) & = {p_{i}} \left( \sum_{k = 1}^{m} (2^{k} y_{N + k} - 2^{k - 1} y_{N + k - 1}) \right) \quad (\text{Telescoping sum.}) \\ & \leq \sum_{k = 1}^{m} {p_{i}}(2^{k} y_{N + k} - 2^{k - 1} y_{N + k - 1}) \quad (\text{Triangle Inequality.}) \\ & < \sum_{k = 1}^{m} 2^{k - 1} \cdot \frac{\epsilon}{2} \quad (\text{By $ \spadesuit $.}) \\ & = (2^{m} - 1) \frac{\epsilon}{2}. \end{align*} Hence, $$ \forall m \in \mathbb{N}: \quad {p_{i}} \left( y_{N + m} - \frac{1}{2^{m}} y_{N} \right) < \left( 1 - \frac{1}{2^{m}} \right) \frac{\epsilon}{2}. $$ By another application of the Triangle Inequality, $$ \forall m \in \mathbb{N}: \quad {p_{i}}(y_{N + m}) - {p_{i}} \left( \frac{1}{2^{m}} y_{N} \right) \leq {p_{i}} \left( y_{N + m} - \frac{1}{2^{m}} y_{N} \right). $$ This implies that $$ \forall m \in \mathbb{N}: \quad {p_{i}}(y_{N + m}) < {p_{i}} \left( \frac{1}{2^{m}} y_{N} \right) + \left( 1 - \frac{1}{2^{m}} \right) \frac{\epsilon}{2}. $$ Next, pick an $ m \in \mathbb{N} $ large enough so that $ {p_{i}} \left( \dfrac{1}{2^{n}} y_{N} \right) < \dfrac{\epsilon}{2} $ for all $ n \in \mathbb{N}_{\geq m} $. Hence, $$ \forall n \in \mathbb{N}_{\geq m}: \quad {p_{i}}(y_{N + n}) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon, $$ or equivalently, $$ \forall n \in \mathbb{N}_{\geq N + m}: \quad {p_{i}}(y_{n}) < \epsilon. $$ As $ \epsilon $ is arbitrary, we see that $ \displaystyle \lim_{n \to \infty} {p_{i}}(y_{n}) = 0 $. Finally, as $ i \in I $ is arbitrary, we conclude that $ \displaystyle \lim_{n \to \infty} y_{n} = 0_{X} $, thus yielding $ \displaystyle \lim_{n \to \infty} x_{n} = x $. $ \quad \blacksquare $
Question: Is the theorem true if $ X $ is just a general topological vector space? In other words, if $ X $ is not locally convex, then does the conclusion of the theorem still hold?
The beautiful idea of @Tacet: -- maybe we can generalize it? Say we have two sequences $(x_n)$, and a sequence of numbers $(c_n)$ $>0$ such that $$(c_n+1) x_{n+1}- c_n x_n$$ has limit $x$. Write this as
$$\frac{a_{n+1} x_{n+1} - a_n x_n}{a_{n+1} - a_n} \to x$$ where $c_{n} + 1 = \frac{a_{n+1} }{a_{n+1} - a_n}$ and so $$\frac{a_{n+1}}{a_n} = 1+ \frac{1}{c_n}$$ With $a_1=1$ we get $$a_n = \prod_{k=1}^{n-1} (1+ \frac{1}{c_k})$$
In order to apply Cesaro-Stolz, we want $a_n \to \infty$ and this is equivalent to $\sum \frac{1}{c_n} \to \infty$.
( works if $(c_n)$ constant, or bounded above, or bounded above by an arithmetic progression)
Note: the idea of Cesaro-Stolz is the following : we can strengthen a given drink past $p$ proof by adding a sufficient amount of drinks of strenght $\ge p$.
Let $\lim$ $(2x_{n+1}-x_n)=x.$ Then for every $\epsilon>0,$ there exists $N\in\Bbb{N}$ such that when $n\geq N,$ $|(2x_{n+1}-x_n)-x|<\epsilon.$
Fix $\epsilon>0.$ Then $$|x_n-x|=|x_n+(x_n-x_n)-x|=|(2x_n-x_n)-x|<\epsilon.$$
...Although I'm not exactly sure about that last step. My issue is that this is a recursive sequence. So intuitively, as $n\to\infty,$ we get that $x_{n+1}=x_n.$ I'm not exactly sure how to deal with this one facet rigorously.
