If $(a_n)_{n\geq 1}$ and $(b_n)_{n\geq 1}$ are decreasing, non-negative sequences of real numbers and $\sum_{n=1}^{\infty} a_n$ converges and $\sum_{n=1}^{\infty} b_n$ diverges, then is it always the case that there exists some positive integer $N$ such that for all $n\geq N$ we have $b_n \geq a_n$?
I was originally trying to prove that this result is true as a lemma to prove that if $(a_n)_{n \geq 1}$ was positive and decreasing and $\sum_{n=1}^{\infty} a_n$ converges that $\lim_{n \to \infty} na_n=0$.
I originally wanted to prove the first statement as the second statement follows after taking $b_n = {1\over{n\ln(n)}}$. I have since realized that there is another solution to the original question which doesn't involve this lemma, but I still am curious to know whether the lemma is true.
I have figured out that the lemma is not true if you take away the requirement that both $a_n$ and $b_n$ are decreasing, but I believe with this restriction, the result should be true.
As discussed in the comments:
To build a counterexample, start with your favorite convergent series $\sum a_n$. Now, let $b_n=a_1$ for enough terms to sum over $1$. (you can let $b_k=a_1-\epsilon+\epsilon \times 10^{-k!}$ if you want something strictly decreasing). Say that takes $N$ terms. Now let $b_n=a_{N+1}-\epsilon$ for enough terms to sum over $1$ (similar note). And repeat.
As you see, the idea is to let $b_n$ decrease but only in relative big gaps, presumably very , very far apart.