# Does $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ imply that $a_n$ convergent?

Does the condition $$\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$$ imply that $$a_n$$ convergent?

When I say convergent I mean to $$L\in \mathbb{R}$$ or to $$\infty$$ or to $$-\infty$$

I have a gut feeling this statement is false, just can't find a series which contradicts it.

Thanks a lot

• @Plopperzz It goes to infinity. – Mark Feb 21 at 21:38
• It goes to infinity.. read the first line in my question – Shaq Feb 21 at 21:38
• Here is a counter-example: math.stackexchange.com/a/3015738/42969. – Martin R Feb 21 at 21:43
• Ok great Martin this a fine contradiction! I am happy my gut feeling didn't disappoint me – Shaq Feb 21 at 21:48

A counter-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.

(This example is taken from $\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1 \Rightarrow \exists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c$. I copied it here because the other question excludes sequences converging to $$\pm \infty$$, so it is not an exact duplicate.)

A counterexample: $$a_n = \exp(\sin\sqrt{n})$$. The ratio $$\frac{a_{n+1}}{a_n}$$ is $$\frac{a_{n+1}}{a_n}=\exp(\sin(\sqrt{n+1})-\sin(\sqrt{n}))\approx \exp\frac{\cos(\sqrt{n})}{2\sqrt{n}}\approx 1+\frac{\cos(\sqrt{n})}{2\sqrt{n}}\to 1$$ but $$a_n$$ oscillates between $$e^1$$ and $$e^{-1}$$, not converging to anything or diverging to $$\infty$$.

Hint

Think to a sequence $$(a_n)$$ such that $$\lim\limits_{n \to \infty} (a_{n+1}-a_n)=0$$, but that oscillates infinitely between $$1$$ and $$2$$.

The limit points of $$(a_n)$$ is the segment $$[1,2]$$ and $$\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}=1$$.

Some examples here and there.