If $ \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x $, then is it true that $ \lim_{n \to \infty} x_{n} = x $?

Solution 1:

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 $.

Solution 2:

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 $