Let $(a_n)_{n\in \mathbb{N}}$ be a sequence of real numbers. I was wondering if the follwing implication is true:
$$\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1 \Rightarrow \exists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c$$
Put into words: If $\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$ then $\lim\limits_{n \to \infty} a_{n}$ converges.
My intuition behind $\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$ is that at some point $a_n$ and $a_{n+1}$ are alsmost equal. If this is the case $(a_n)_{n\in \mathbb{N}}$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:
- Is the statement really true?
- (If so, how could you prove it?)
Cheers,
Pascal
Best Answer
As noticed by MisterRiemann $a_n=n$ is a first counterexample but also $a_n=n^2$ works or $a_n=\log n$ and so on.
Therefore unfortunately your guess is definitely not true!
As a remark, other common myths on limits are:
1) $a_n \to \infty \implies a_{n+1}\ge a_n$
2) $a_n \to L \implies a_n \to L^+ \quad \lor \quad a_n \to L^-$
3) $a_n \to 0^+ \implies a_{n+1}\le a_n$
4) $a_n$ bounded $\implies a_n \to L$