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?

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I did not check the details, but would approach this as follows: since $\sum a_k$ converges $\forall \varepsilon_n > 0 \exists N(n): |\sum_{k=N}^\infty a_k | \le \varepsilon_n^n$ (Note the power $n$) Now choose $\varepsilon_n \rightarrow 0$ and for $N(n+1) \ge k > N(k)$ multiply $a_k$ by $\varepsilon_n$ and check what kind of estimate results for the new series. – user20266 Dec 19 '11 at 18:20
If the series is convergent, then the tails $\sum_{k=n}^\infty a_k$ should converge to zero as $n\to\infty$, yes? I doubt you can suitably define each $b_k$ in terms of $a_k$, but you should be able to define $b_n$ in terms of the absolute values of these tail sums. – Skatche Dec 19 '11 at 18:24
that was supposed to read $N(n+1) \ge k > N(n)$.... – user20266 Dec 19 '11 at 18:26
If $\{a_n:n\in\mathbb{N}\}\subset \mathbb{R}_{+}$ then we can take $b_n=\left(\sum\limits_{k=n}^{\infty}a_k\right)^{-1/2}$ – Norbert Dec 19 '11 at 18:28
@Thomas: I think you meant multiply $a_k$ by $1/\varepsilon_n$. Also you want to control $\left|\sum_{k=N}^M a_k\right|$ for all $M > N$, not just the sum from $N$ to $\infty$. – Robert Israel Dec 19 '11 at 18:33

1 Answer

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.

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... where $b_k = \alpha$ for $1 \leq k \leq n_1 -1$, $b_k = \alpha+1; n_1 \leq k \leq n_2 - 1$, ... Ah I see, obviously $b_k \to \infty$ but the product's sum converges. Thank you. – Jonathan Bryce Dec 19 '11 at 20:34
"With thanks to Thomas and Robert" -- appreciated! :-) Robert's comments about mine are correct, though, I'm afraid. Was in a hurry..... – user20266 Dec 20 '11 at 17:46