I've got a lingering question from a midterm in real analysis that I'd really like to have answered. The first time I answered the question, I received a 2/10 for absolutely mangling the definition of convergence. (By my logic, $\frac{1}{n}, \sqrt{n}$ converged.) I re-did the problem for what amounted to karma, and got a 4/10. A formal statement of what I'm trying to prove is, given a sequence $x_n$ of real numbers: $\exists\ a > 0 \land c \in (0, 1) : \forall\ n \geq 1, |x_{n+1} - x_n| \leq ac^n \implies \exists\ L : x_n \to L$
Essentially, we are given a sequence $x_n$ in the real numbers where the difference of consecutive terms $|x_{n+1} - x_n|$ was always less than the appropriate term in some geometric sequence, $ac^n$. We were told to prove that this sequence was pre-convergent, where pre-convergence was defined as: $\forall\ \varepsilon > 0, \exists\ N \in \Bbb N : n \geq N \implies |a_n - a_N| < \varepsilon$ I tackled the problem first by proving that the set of pre-convergent sequences is equivalent to the set of Cauchy sequences (and thus convergent in complete sets). Then, I noted that $\forall\ r \in \Re, 0 \leq |r|$, so I constructed the sequences $z_n = 0$ where the following held: $ z_n \leq |x_{n+1}-x_n| \leq ac^n$ By applying the Squeeze theorem, it was easy to show that the sequence of differences converged to zero, but I have no idea where to go from there...
Can anyone shed some definite light on this?