Can a sequence which decays more slowly still yield a converging series? In Bergman's companion notes to Rudin, he says that "If a sequence of positive terms has convergent sum, so does every sequence of positive terms which decays more rapidly." So given a sequence $\{a_n\}$ of positive terms such that $\sum_n a_n$ converges, if $\{b_n\}$ is such that
$$
\lim_{n\to\infty} \frac{a_n}{b_n} = +\infty,
$$ then $\sum_nb_n$ converges. I can prove this given $\{a_n\}$ and $\{b_n\}$. However, can we find $b_n$ which decays more slowly, i.e.
$$
\lim_{n\to\infty} \frac{b_n}{a_n} = +\infty
$$ such that $\sum_n b_n$ converges?

Similarly, we have the claim "if a sequence of positive terms has divergent sum, then so does every sequence of positive terms which decays more slowly."
The question is the following:


  Show that for every converging series $\sum\limits_na_n$ with positive terms there exists a  converging series $\sum\limits_nb_n$ with positive terms such that $\lim\limits_{n\to\infty}\frac{a_n}{b_n}=0$.


A hands-on approach is as follows: for every $n$, consider $$A_n=\sum_{k\geqslant n}a_k,$$ then, by hypothesis, $A_n\to0$ hence, for each $k\geqslant0$, there exists some finite $\nu(k)$ such that $$A_{\nu(k)}\leqslant2^{-k}.$$ Assume without loss of generality that the sequence $(\nu(k))$ is nondecreasing and define $(b_n)$ by
$$\color{red}{\forall k\geqslant0,\quad\forall n\in[\nu(k),\nu(k+1))},\quad \color{red}{b_n=k\,a_n}.$$
Then $\frac{a_n}{b_n}\to0$ when $n\to\infty$ and $$\sum_{n\geqslant\nu(0)}b_n=\sum_{k\geqslant0}k\sum_{n=\nu(k)}^{\nu(k+1)-1}a_n\leqslant\sum_{k\geqslant0}kA_{\nu(k)}\leqslant\sum_{k\geqslant0}k2^{-k},$$ which is finite, hence the series $\sum\limits_nb_n$ converges, as desired.
There are a couple of things... slightly off... in your understanding.

The theorem does not say, as you imply, that if $$\sum_{n=1}^\infty a_n$$ exists and if $$\lim_{n\to\infty}\frac{b_n}{a_n} = \infty,$$ then the sum $$\sum_{n=1}^\infty b_n$$ exists. I advise you to re-check what your notes say.