L. Kronecker's theorem for sequences and series: $\lim_{n\to\infty}\frac{1}{b_n}\sum_{k=1}^na_kb_k=0$ Assume $\sum a_i$ is a convergent series and $b_1,b_2,\dots$ is a divergent monotonically increasing sequence. How can we see that $$\lim_{n\to\infty}\frac{1}{b_n}\sum_{k=1}^na_kb_k=0$$
Attempt: We have $$|\frac{1}{b_n}\sum_{k=1}^na_kb_k|=|\sum_{k=1}^n\frac{a_kb_k}{b_n}|\leq \sum_{k=1}^n|\frac{a_kb_k}{b_n}|\leq\sum_{k=1}^n|a_k|$$ since $b_k\leq b_n$ for $k\leq n$.
This is not even enough to claim that the sequence converges. How can we see that its limit is zero?
 A: Let $R_n = \sum_{k=n}^\infty a_k$. Then
$$\sum_{k=1}^n a_k b_k = \sum_{k=1}^n (R_k - R_{k+1})b_k = \sum_{k=1}^n R_k b_k - \sum_{k=2}^{n+1} R_kb_{k-1} = R_1b_1 - R_{n+1}b_n + \sum_{k=2}^n R_k(b_k - b_{k-1}).$$
So, since $b_k$ is monotonically increasing,
$$\left\lvert \frac{1}{b_n}\sum_{k=1}^n a_kb_k\right\rvert \leqslant \frac{\lvert R_1b_1\rvert}{\lvert b_n\rvert} + \lvert R_{n+1}\rvert + R\frac{b_m - b_1}{\lvert b_n\rvert} + S_m\frac{b_n-b_m}{\lvert b_n\rvert},$$
where $R = \sup \{\lvert R_k\rvert : k \in \mathbb{N}\}$ and $S_m = \sup \{ \lvert R_k\rvert : k \geqslant m\}$.
Now since $R_n \to 0$, we also have $S_n \to 0$, thus we can make the second and fourth terms small by choosing $m$ large enough, and then the first and third by choosing $n > m$ large enough.
More precisely, let $m_0$ such that $b_{m_0} \geqslant \lvert b_1\rvert$. Then, for $m \geqslant m_0$ we have $\frac{b_m-b_1}{b_n} \leqslant \frac{2 b_m}{b_n}$ and $0\leqslant \frac{b_n-b_m}{b_n} \leqslant 1$ for all $n \geqslant m$, so
$$\left\lvert \frac{1}{b_n}\sum_{k=1}^n a_kb_k\right\rvert \leqslant R\frac{\lvert b_1\rvert + 2 b_m}{b_n} + 2 S_m \leqslant R\frac{3b_m}{b_n} + 2S_m$$
for all $n \geqslant m \geqslant m_0$. Choose $m$ so that $S_m < \frac{\varepsilon}{4}$, and $n\geqslant m$ so large that $\frac{3b_m}{b_n} < \frac{\varepsilon}{4R}$.
A: This is a (long) comment rather than an answer, to indicate why Daniel's answer needs the complications it introduces. It may help clarify why it proceeds as it does.
If the $a_i$ and $b_i$ are positive, you are almost there. Usually in analysis, you split sums you want to analyze into two pieces, one near zero and one near infinity, and you estimate both pieces separately.
Given $n_0<n$, we have that 
 $$ S_n:=\frac1{b_n}\sum_{k=1}^n a_k b_k = \left(\sum_{k=1}^{n_0}\frac{a_k b_k}{b_n}\right)+\left(\sum_{k=n_0+1}^{n}\frac{a_k b_k}{b_n}\right). $$
Now, to prove the result, fix $\epsilon>0$. We need to show that there is $N$ such that if $n\ge N$, then $S_n<\epsilon$. Begin by picking $n_0$ large enough that for all $n>n_0$, we have $\sum_{k=n_0+1}^{n} a_k <\epsilon/2$, which is possible since the series of the $a_i$ converges. 
Now, with $n_0$ fixed as above, pick $N$ large enough that for any $n\ge N$, we have
 $$ \frac{b_{n_0}}{b_n}<\frac{\epsilon}{2(\sum_{k=1}^{n_0} a_k)}. $$
This is possible since $n_0$ is fixed and the $b_i$ increase unboundedly. 
Now, we see that for $n\ge N$ we have
 $$ \sum_{k=1}^{n_0}\frac{a_k b_k}{b_n}\le \sum_{k=1}^{n_0}\frac{a_k b_{n_0}}{b_n}<\frac{\epsilon}2, $$
while 
 $$ \sum_{k=n_0+1}^{n}\frac{a_k b_k}{b_n}\le \sum_{k=n_0+1}^{n}a_k<\frac{\epsilon}2, $$
and we conclude that $S_n<\epsilon$, as wanted.

If the $a_i$ and $b_i$ are only assumed nonnegative, the same argument works with two small changes: First, make sure that $n_0$ is large enough that $b_{n_0}>0$ and, second, replace $$ \frac{\epsilon}{2(\sum_{k=1}^{n_0} a_k)} $$ with $$ \frac{\epsilon}{2(1+\sum_{k=1}^{n_0} a_k)}, $$ to ensure that the denominator is nonzero.
If the $a_i$ are not assumed nonnegative, more work is needed, and it is here that the simple approach is found wanting. 
Now we need to show that $|S_n|<\epsilon$. We have that 
 $$ |S_n|= \left|\left(\sum_{k=1}^{n_0}\frac{a_k b_k}{b_n}\right)+\left(\sum_{k=n_0+1}^{n}\frac{a_k b_k}{b_n}\right)\right|\le\left|\sum_{k=1}^{n_0}\frac{a_k b_k}{b_n}\right|+\left|\sum_{k=n_0+1}^{n}\frac{a_k b_k}{b_n}\right|.$$
First, $$ \left|\sum_{k=1}^{n_0}\frac{a_k b_k}{b_n}\right|\le\sum_{k=1}^{n_0}\frac{|a_k| |b_k|}{b_n} $$ can be estimated essentially as before. Now we can require that $n_0$ is large enough that $b_{n_0}>0$ and $b_{n_0}=\max\{|b_i|:i\le n_0\}$. Also, we let $N>n_0$ be large enough that if $n\ge N$, then 
 $$ \frac {b_{n_0}}{b_n}<\frac{\epsilon}{2(1+\sum_{k=1}^{n_0} |a_k|)}. $$
The real problem comes with the second sum, $$ \left|\sum_{k=n_0+1}^{n}\frac{a_k b_k}{b_n}\right|, $$ which before was the easy one to estimate. Now we cannot use the same argument, since the $a_i$ are not necessarily positive, so we cannot appeal to monotonicity in the same way. Of course, we cannot assume the series of $a_i$ converges absolutely, so we cannot replace the $a_i$ with $|a_i|$ to proceed as before. Really, a different idea is needed here. We need to appeal to the monotonicity of the $b_i$, and to use that the sum $\sum_i a_i$ converges, but keeping the terms the way we have here, in quotients $a_ib_i/b_n$ is not going to work any longer. 
At this point it may be natural to try to "separate" the two expressions, and the obvious attempt to do this is via summation by parts, which ends up leading to Daniel's argument.
