$\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1 \Rightarrow \exists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c$

Question

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

Answer 1

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!

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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 $\implies a_n \to L$

Answer 2

Thanks to MisterRiemann for the counterexample of $a_n = n$.

$$\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim\limits_{n \to \infty} \frac{n}{{n+1}} = 1$$ but: $$\lim\limits_{n \to \infty} {a_n} =\lim\limits_{n \to \infty} n = \infty \Rightarrow \nexists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c $$

Answer 3

The conclusion is wrong even for bounded sequences.

An example is $a_ n = 2 + \sin(\log(n))$ which is not convergent (it oscillates between $1$ and $3$). But $$ \left\lvert \frac{a_{n+1}}{a_n} - 1 \right\rvert = \left\lvert \frac{\sin(\log(n+1))- \sin(\log(n))}{2 + \sin(\log(n))} \right\rvert \\ \le \left\lvert \sin(\log(n+1))- \sin(\log(n)) \right\rvert = \left\lvert \frac{\cos(\log(x_n))}{x_n}\right\rvert $$ for some $x_n \in (n, n+1)$, using the mean value theorem. It follows that $$ \left\lvert \frac{a_{n+1}}{a_n} - 1 \right\rvert \le \frac{1}{n} \to 0 \, , $$ i.e. $\frac{a_{n+1}}{a_n} \to 1$.

Roughly speaking, $(a_n)$ oscillates, but with decreasing frequency, so that the ratio of successive sequence elements approaches one.