# 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?

#### Solutions Collecting From Web of "Convergence of a product series with one divergent factor"

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.