# How to show that the following sequence is convergent?

I am trying to solve this question from a math book. Suppose $$a_n$$ is convergent, then show whether the following sequence is convergent or divergent: $$\begin{pmatrix}\cfrac{a_n}{a_{n+1}} \end{pmatrix}_n$$

my initial thoughts were that the series should converge to $$1$$. But then looking at the d'Alembert's ratio, where it's the reciprocal of this series, I am starting to think that I am wrong with the limit being $$1$$.

What is ít that I am doing wrong here. Any suggestion would be appreciated.

• Look at $a_n=\frac 1 {n!}$. Jun 17, 2021 at 9:48

If $$(a_n)_n$$ converges to a nonzero real, then the sequence you've written also converges and converges to $$1$$. This is due to a more general result: Suppose $$(a_n)_n$$ and $$(b_n)_n$$ are convergent sequences such that $$\lim_n a_n \neq 0$$. Then $$\lim_{n \to \infty} \frac{b_n}{a_n} = \frac{\lim_n b_n}{\lim_n a_n}.$$ ($$a_n$$ is eventually nonzero and so the limit makes sense.)
However, if $$a_n \to 0$$, then the situation is more delicate. For example, if $$a_n = 1/n$$, then the ratio converges to $$1$$. If $$a_n = 1/n!$$, then the ratio does not converge (or it diverges to $$\infty$$).
Indeed, fix any $$\alpha \in (1, \infty)$$. Define $$(a_n)_n$$ as $$a_n = \alpha^{-n}$$. Then, $$a_n \to 0$$ but $$a_n/a_{n + 1} = \alpha$$ for all $$n$$.
You can also get oscillatory behaviour by considering a sequence as $$1, 1, -\frac12, -\frac12, \frac13, \frac13, \ldots.$$