divergence of a series involving logarithm

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.

• Would the down voter please explain what is wrong with my post? – ttb Jan 11 '17 at 18:32
• Let $a_n = 1$ for even $n$, and $a_n = \frac{1}{n}$ for odd $n$. Let $k_n = 1$ for odd $n$, and $k_n = \frac{1}{n^2}$ for even $n$. – Daniel Fischer Jan 11 '17 at 20:36
• @DanielFischer The power of proof by contradiction... where the experience easily disprove all our efforts. – Simply Beautiful Art Jan 11 '17 at 20:40
• indeed, this does the trick! – ttb Jan 11 '17 at 21:17