Asymptotic behaviour of the difference of consecutive values of null sequence

Question

What is the asymptotic behaviour of the difference $$ c_j - c_{j+1} $$ for $j\rightarrow \infty$ if $(c_j)_{j\in\mathbb{N}}$ is a null sequence?

Answer 1

Yes, because the limits exist for $\lim_{n \to \infty}c_n = 0$ and so $\lim_{n \to \infty}c_{n+1} = 0$ then $\lim_{n \to \infty}c_n - c_{n+1} = \lim_{n \to \infty}c_n - \lim_{n \to \infty}c_{n+1}= 0-0=0$.

Or by definition: Let $\epsilon > 0$ then there exist $N \in \mathbb{N}$ such that for all $n > N$ we have $|c_n| < \frac{\epsilon}{2} $ and $|c_{n+1}| < \frac{\epsilon}{2} $.

We want to show $|c_n - c_{n+1}| < \epsilon$ and indeed by the triangle inequality:

$|c_n - c_{n+1}| \leq |c_n| + |c_{n+1}| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

Answer 2

Oh..., it was obvious - now I have it.

The asymptotic behaviour is of course as $c_j$ since $(c_j - c_{j+1})/c_j = 1-c_{j+1}/c_j\rightarrow 1.$