Convergence of a product series with one divergent factor I'm currently struggling with the following problem: 

Let $\displaystyle \sum_{k=1}^{\infty} a_k$ be a convergent series with $a_k \in \mathbb{R} \setminus \{0\}$. Then is there always a sequence $\{b_k\}$ of real numbers with $\displaystyle \lim_{k \to \infty} b_k = \infty$ such that the series $\displaystyle \sum_{k=1}^{\infty} a_k b_k$ will still converge?

My intuition of course says there is, as one should always be able to find some sequence that increases "much slower" than $a_k$ decreases. But how can I state this vague notion more precisely and actually prove my guess? I thought of choosing $b_k := -\log a_k$ or something, but that won't hold in all possible cases, won't it?
Could you give any hints, please?

With thanks to Thomas and Robert:
Suppose $ \sum\limits_{n=1}^\infty a_n$ converges.  
Choose a positive number $\alpha$ so that 
$$\tag{1}\left|\,\sum\limits_{n=1}^m a_n\,\right|\le \alpha  $$ for all positive integers $m$.
Set $$S_1=\sum_{n=1}^\infty\, \alpha a_n.$$
$S_1$ is convergent, so we may, and do,  choose  $n_1$ so that for all $l\ge m\ge n_1$
$$
\tag{2} \Bigl|\,\sum_{n=m}^l a_i\,\Bigr|\le {1\over 2^2(\alpha+1)} .
$$
Set 
$$S_2=\underbrace{\sum_{n=1}^{n_1-1} \alpha a_n}_{D_1} + \sum_{n=n_1}^{\infty} (\alpha+1) a_n  .$$
Note that  by (1), $ \left| \, \sum\limits_{n=1}^{m} \alpha a_n\,\right| \le \alpha  $ for all $m\le n_1-1$.
Now choose  $n_2>n_1$ so that for all $l\ge m\ge n_2$
$$
\left|\,\sum_{n=m}^l a_i\,\right|\le {1\over 2^3(\alpha+2)} .
$$
Set 
$$S_3=\sum_{n=1}^{n_1-1} \alpha a_n + \underbrace{\sum_{n=n_1}^{n_2-1} (\alpha+1) a_n }_{D_2} 
+ \sum_{n=n_2}^{\infty} (\alpha+2) a_n .$$
Note that, by (2), $\Bigl|\,\sum\limits_{n=n_1}^{m} (\alpha+1) a_n \,\Bigr|\le {1\over 2^2}  $ for all $m\le n_2-1$.
Continuing in the obvious manner, we define integers $ n_3<n_4<\cdots\,$  and sums $$D_k=\sum\limits_{n=n_{k-1}}^{n_k-1} (\alpha+k-1)a_n$$ satisfying 
$$ 
\tag{4}\left|\,\sum_{n= n_{k-1}}^{ m}(\alpha+k-1)a_n\,\right|\le {1\over 2^k}
$$ 
for all $m\le n_k-1$.
Consider the sum
$$
S=D_1+D_2+D_3+\cdots.
$$
We have, by the triangle inequality, that
$$\eqalign{
|D_n+D_{n+1}+\cdots+ D_m|&\le {1\over 2^n} +{1\over 2^{n+1}} +\cdots+{1\over 2^m} \cr &\le {1\over 2^{n-1} }   \cr
&\buildrel{n \rightarrow\infty}\over{\longrightarrow }\ 0,}
$$
for all  for $m\ge n>1$.
From this and (4),  it follows that $S$ converges.
