Suppose that $\{a_n\}_{n\ge1}$ and $\{b_n\}_{n\ge1}$ are two positive strictly decreasing summable sequences. Suppose further that there exists $k>0$ and $K>0$ such that $kb_n\le a_n\le Kb_n$ when $n>N$ (this is denoted $a_n=\Theta(b_n)$ as $n\to\infty$).
Does this imply that $a_n/b_n\to c$ with $c>0$ as $n\to\infty$?
I think that the answer should be positive, but I am looking for a nice argument. Since $\{b_n\}_{n\ge1}$ is positive $$ k\le a_n/b_n\le K $$ for $n>N$. It seems that there are three possibilities for the limit of $a_n/b_n$: (i) $a_n/b_n\to0$ as $n\to\infty$, (ii) $a_n/b_n\to c$ with $c>0$ and (iii) $a_n/b_n\to\infty$ as $n\to\infty$. Since $k\le a_n/b_n\le K$ for $n>N$, the only possibility is that $a_n/b_n\to c$ with $c>0$ as $n\to\infty$. However, this seem to be based on the assumption that $\{a_n/b_n\}_{n\ge1}$ is a monotone sequence. Is this is true? Is $\{a_n/b_n\}_{n\ge1}$ a monotone sequence? It seems that the answer is negative (see here) and then my argument fails.
Any help is much appreciated!
Consider if $b_n = a_n\cdot 2^{(-1)^n}$, which is to say $b_n$ alternates between being $2a_n$ and $\frac12a_n$. We have $\frac13b_n< a_n<3b_n$, but $\frac{a_n}{b_n}$ is an oscillating divergent sequence, and not monotone as you assumed.
As long as $a_n$ decreases quickly enough (say $a_n = 5^{-n}$), then $b_n$ is also strictly decreasing, and as they are both dominated by, say, $2^{-n}$ in that case, they are also summable.