Does there exist a series $\sum_{n=1}^{\infty} b_n$ such that $\sum_{i=1}^{\infty} b_{n_i}$ converges for any $n_1 < n_2 < \ldots$ but $\sum_{n=1}^{\infty} |b_n|$ diverges?

Certainly the example would have to be conditionally convergent, but such standard examples as the alternating harmonic series don't seem to work (at least from what I've managed to show).

Any ideas?

  • $\begingroup$ If I don't misunderstand something, shouldn't be the index of the series over the $b_{n_i}$'s be $i$ and not $n$? Also, interesting question, +1. $\endgroup$
    – blub
    Commented Aug 26, 2018 at 23:38
  • $\begingroup$ @zzuussee Fixed $\endgroup$ Commented Aug 26, 2018 at 23:39
  • 2
    $\begingroup$ The sum consisting of all positive terms, $A = \sum_{i; b_{i}\geq 0} b_{i}$ converges. Likewise for negative terms, $B = \sum_{j; b_{j} < 0} b_{j}$. Since $A-B = \sum \left | b_{i} \right |$, $\sum \left | b_{i} \right |$ must then converge. $\endgroup$
    – LPenguin
    Commented Aug 26, 2018 at 23:48
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    $\begingroup$ @SangchulLee But how do we know the rearrangement preserves the order of the $n_i$? $\endgroup$ Commented Aug 26, 2018 at 23:49

1 Answer 1


No. Define $$b_{n_i}=b_n^+=\max\{b_i,0\}.$$ By hypothesis, $\sum_{i=0}^\infty b_{n_i}$ converges. Similarly, we can define $b_n^-$ and see that the sum over $b_n^-$ is convergent by hypothesis.

This means that $\sum_{n=0}^\infty b_n$ is absolutely convergent since both its positive parts and negative parts converge.


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