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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.2011-12-19
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    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.2011-12-19
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    that was supposed to read $N(n+1) \ge k > N(n)$....2011-12-19
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    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}$2011-12-19
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    @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 $\infty$.2011-12-19
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    Related questions: [79204](http://math.stackexchange.com/questions/79204/) and [82998](http://math.stackexchange.com/questions/82998/).2011-12-19
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    @Srivatsan: A I see there can't be a "smallest" divergent series. My problem would be the other way round - looking for a "greatest" convergent series to test against. I guess a similiar argument can be made there. Need to think about it all; thanks for the other comments as well.2011-12-19

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