Convergent $\{a_n\}$ $\nRightarrow$ convergent $\{\frac{a_{n+1}}{a_n}\}$ Suppose $\{a_n\}$ converges and $\forall n \; a_n \neq 0$. Then $\{\frac{a_{n+1}}{a_n}\}$ converges.
How to disprove this?
 A: $$\{a_n\}:=\left\{1\,,\,\frac12\,,\,-\frac13\,,\,-\frac14\,,\,\frac15\,,\,\frac16\,,\,-\frac17\,,\,-\frac18\,,\,\frac19\,,\,\frac1{10}\,,\ldots\right\}$$
A: $$\{a_n\}:=\left\{\frac{1}{3},\dfrac{1}{2},\frac{1}{3^2},\frac{1}{2^2},\frac{1}{3^3},\frac{1}{2^3},\cdots \right\}$$
Here even $\sum a_n$ converges, $a_n$ is positive,  but $\frac{a_{n+1}}{a_n}$ has a subsequence that $\rightarrow +\infty$
A: $$ a_n=\frac{2+(-1)^n}{2^n}\implies \frac{a_{n+1}}{a_n}=\frac{2-(-1)^n}{4+2\cdot (-1)^n}=\frac32,\frac16,\frac32,\frac16,\frac32,\ldots$$
A: $$(a_m)=(1,-1,\frac{1}{2},\frac{1}{2},-\frac{1}{3},\frac{1}{3},\frac{1}{4},\frac{1}{4},\cdots)$$
then  $$(\frac{a_{m+1}}{a_m}) = (-1,-\frac{1}{2},+1,\frac{-3}{2},-1,\frac{4}{3},+1,\cdots)$$
Essentially, the idea here is that the first sequence is just defined with the denominators decreasing as they are, repeating every term once and changing signs as needed so that the second sequence contains the subsequence $$(-1,+1,-1,+1,\cdots )$$ which clearly doesn't converge. (we don't really care about terms like $-\frac{1}{2}$ or $\frac{4}{3}$ etc).
