# Example of series such that every $\sum_{i=1}^{\infty} b_{n_i}$ converges but $\sum_{n=1}^{\infty} |b_n|$ diverges

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?

• 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. – blub Aug 26 '18 at 23:38
• @zzuussee Fixed – CuriousKid7 Aug 26 '18 at 23:39
• 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. – LPenguin Aug 26 '18 at 23:48
• @SangchulLee But how do we know the rearrangement preserves the order of the $n_i$? – CuriousKid7 Aug 26 '18 at 23:49

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.