Suppose the sequence $\{b_n\}$ is positive and monotonically decreasing and that $\lim_{n \to \infty} b_n = 0$ but the series $\sum_{n=1}^\infty b_n = \infty$ diverges. Suppose the series $\sum_{n=1}^\infty a_n$ converges.
Show that
$$\liminf \frac{1}{n} \sum_{i=1}^n \frac{a_i}{b_i} \leq 0$$
I tried summation by parts:
$$\frac{1}{n} \sum_{i=1}^n \frac{a_i}{b_i} = \frac{1}{nb_n}\sum_{i=1}^n a_i + \frac{1}{n}\sum_{i=1}^{n-1}\left(\sum_{j=1}^ia_i\right) \left(\frac{1}{b_i}- \frac{1}{b_{i+1}} \right)$$
I know that $\sum_{i=1}^n a_i$ is bounded but can't proceed since I'm not even sure how $nb_n$ behaves since $b_n \to 0$ as $n \to \infty$.
Apply, instead, the summation by parts formula
$$\sum_{i=1}^n a_i = \sum_{i=1}^n \frac{a_i}{b_i}b_i = S_nb_n + \sum_{i=1}^{n-1}S_i(b_i - b_{i+1}), $$
where $\displaystyle S_n = \sum_{i=1}^n \frac{a_i}{b_i}$.
Now we can prove the assertion by contradiction. Suppose that
$$\liminf_{n \to \infty} \frac{1}{n} \sum_{i=1}^n \frac{a_i}{b_i} = \liminf_{n \to \infty} \frac{S_n}{n} = r > 0.$$
Then there exists a positive integer $N$ such that for all $n \geqslant N$ we have $S_n > nr/2$. Hence,
$$\sum_{i=1}^n a_i = S_nb_n + \sum_{i=1}^{N-1}S_i(b_i - b_{i+1}) + \sum_{i=N}^{n-1}S_i(b_i - b_{i+1}) \\ \geqslant \frac{rn}{2}b_n + \sum_{i=1}^{N-1}S_i(b_i - b_{i+1}) + \frac{r}{2}\sum_{i=N}^{n-1}i(b_i - b_{i+1}) \\ = \frac{rn}{2}b_n + \sum_{i=1}^{N-1}S_i(b_i - b_{i+1}) + \frac{r}{2}\sum_{i=N}^{n-1}(ib_i - (i+1)b_{i+1}) + \frac{r}{2}\sum_{i=N}^{n-1}b_{i+1} \\ = \sum_{i=1}^{N-1}S_i(b_i - b_{i+1}) + \frac{rN}{2}b_N + \frac{r}{2}\sum_{i=N}^{n-1}b_{i+1}. $$
Note that $N$ is fixed and $\sum_{i=N}^\infty b_{i+1} = +\infty$. This contradicts the convergence of $\sum a_n$.
Therefore,
$$\liminf_{n \to \infty} \frac{1}{n} \sum_{i=1}^n \frac{a_i}{b_i} \leqslant 0.$$