TITLE: If $ \lim_{n \to \infty} (2 x_{n + 1} - x_{n}) = x $, then is it true that $ \lim_{n \to \infty} x_{n} = x $?
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: 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.