Question_
Suppose that $a_n>0$ and $b_n>0$ for $n\ge N$ where $N$ is an integer. Given that $\lim_{n \to \infty} {a_n \over b_n}=\infty$ and $\sum a_n$ converges, can anything be said about $\sum b_n$ ?
I've found the case of $\sum b_n$ converging while $\sum a_n$ converges:
$$a_n={1\over n^2}, b_n={1\over n^3}$$
(Easy:D) I've also tried to find the case of $\sum b_n$ diverges, but unfortunately, I still have no idea. Nevertheless, I believe that there's no case for $\sum b_n$ diverging. The reason is as follows. When we use the definition of the limits:
$$\forall M>0, \exists N \space s.t. \space \forall n\ge N \implies a_n/b_n >M$$
So, $a_n\gt Mb_n$. Due to this,
$$\sum_{n=N}^\infty a_n \gt M\sum_{n=N}^\infty b_n$$
Thus, $\sum_{n=N}^\infty b_n$ converges. I thought the part $n=1 \to N$ does not affect the convergence of $\sum_{n=N}^\infty b_n$ since the number of terms is finite.
I'm asking the question to check whether the approach I've made is correct or have some contradictory points. Also, I'm wondering if there exists a sequence $b_n$ that $\sum b_n$ diverges. Thanks for giving your ideas!
Best Answer
HINT:
For large $n$ we have $a_n/b_n>1$, hence $a_n>b_n$.