Series behavior using the Ratio Test My professor gave us some food for thought today. A classmate asked a question and the professor didn't answer it but instead asked us to think about it.
We must find a series $A_n$ with $A_n \geq 0$ such that the $\lim{\frac{A_{n+1}}{A_n}}$doesn't exist but the $\limsup \frac{A_{n+1}}{A_n} < 1$.
Would it suffice to find a series that alternates between addition and subtraction of a small number below 1, such as:
$.5 +(-0.25)^n = A_n$ as $n\rightarrow\infty$?
Thank you guys for the help!
 A: Let $A_0=1$. If $n$ is even, let $A_{n+1}=\frac{A_n}{2}$. If $n$ is odd, let $A_{n+1}=\frac{A_n}{4}$. 
A: Let $q_1, q_2, \dots $ be an enumeration of the rationals in $(0,1/2).$ Set $A_1 = 1$ and $A_{n+1} = q_nA_n, n= 1,2,\dots$
Then $A_{n+1}/A_n = q_n$ has every point in $[0,1/2]$ as a subsequential limit, and there are no others.
A: Forget about the sequence $A_n$, you're just interested in the sequence $A_{n+1}/A_n$. You're looking for a sequence $r_n$ such that $\lim r_n$ doesn't exist but $\limsup r_n$ does. After that you can construct an appropriate $A_n$ from $A_{n+1}/A_n = r_n$.
Now just visualize it. In order for $\lim r_n$ not to exist, $r_n$ has to oscillate somehow, but its $\limsup$ has to be smaller than $1$. One easy to way to keep the $\limsup$ smaller than $1$ is to have every $r_n$ less than $1$. How about:
$$(r_n) = 0.1, 0.9, 0.1, 0.9, 0.1, 0.9, 0.1, 0.9 ...$$
Now just let $A_0$ be whatever you like and define the rest recursively by $A_{n+1} = r_n A_n$. This recurrence is not difficult to solve if you want a more direct formula for $A_n$.
A: Try 
$$
A_n = 2^{-n} \prod_{k=0}^n \sin(1+k) $$
Then 
$\lim_{n \to \infty} \frac{A_{n+1}}{A_n} $ does not exist, yet
$$\lim\sup \left| \frac{A_{n+1}}{A_n} \right| \leq \frac{1}{2}
$$
and the series converges.
