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If $\sum a_k$ converges absolutely, then $|a_k|<\frac{1}{k}$ for all sufficiently large k.

I'm trying to give a proof or a counterexample about the above statement, but I'm not really sure where to start. I think it's false, but don't know how to go about finding a counterexample. Any help would be much appreciated.

Thanks

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    Yep I did, thanks2012-12-30

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HINT: The idea is to have infinitely many terms $a_k$ such that $a_k\ge\frac1k$, but to spread them out very thinly. Try letting

$a_k=\begin{cases} \frac1k,&\text{if }k\text{ is a perfect square}\\\\ 0,&\text{otherwise}\;. \end{cases}$

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    @user51897: Absolute convergence is a special case of [unconditional convergence](http://en.wikipedia.org/wiki/Unconditional_convergence).2012-12-31