Let $k_n,a_n$ be sequences with $k_n>0, a_n>0$. I want to take a condition on the divergence of one series involving the $k_n,a_n$, and the non-divergence of the sequence $a_n$, and conclude the divergence of another series.
Explicitly, take the following assumptions
- $\sum\limits_{n=1}^{\infty}\log(1+k_na_n) = \infty$
- For some $\delta>0$ there are infinitely many $n$ such that $a_n\geq \delta$.
Then is it the case that $\sum\limits_{n=1}^{\infty}\log(1+k_na_n)a_n = \infty$? (Note the new $a_n$ term)
I don't know whether this is true or not, and I can't think of any counter examples.
Attempt so far:
First, for any $\epsilon$ we can make the two sums $$A^{\epsilon} = \sum\limits_{a_n\geq \epsilon}\log(1+k_na_n)$$ and $$B^{\epsilon} =\sum\limits_{a_n < \epsilon}\log(1+k_na_n)$$ By assumption, for each $\epsilon$ at least one of $A^{\epsilon}, B^{\epsilon}$ must diverge. If $A^{\epsilon}$ diverges for at least one $\epsilon$, then the result follows since $$\sum_{n=1}^{\infty}\log(1+k_na_n)a_n \geq \sum_{a_n\geq \epsilon}\log(1+k_na_n)a_n \geq \sum_{a_n\geq \epsilon}\log(1+k_na_n)\epsilon = \epsilon A^{\epsilon}$$
The difficulty for me comes when I assume $A^{\epsilon} < \infty$ for all $\epsilon>0$. One idea is to show the two assumptions mean the sequence $\log(1+k_na_n)a_n$ cannot converge.