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$\{a_n\}$ is a sequence of real numbers.$\space\sum_{n=1}^{\infty} a_{2n}$ and $\sum_{n=1}^{\infty} a_{2n-1}$ are both conditionally convergent. Is there such $\sum_{n=1}^{\infty} a_{n}$ that is divergent?

I understand that $\sum_{n=1}^{\infty} a_{2n}$ and $\sum_{n=1}^{\infty} a_{2n-1}$ being conditionally convergent means that $\sum_{n=1}^{\infty} (a_{2n}+a_{2n-1})$ is also conditionally convergent, and that $\sum_{n=1}^{\infty} (a_{2n}+a_{2n-1})$ is also a rearrangement of $\sum_{n=1}^{\infty} a_{n}$. I also know that due to the Riemann series theorem there should be such rearrangements that are both conditionally convergent and divergent. I just don't know if specifically $\sum_{n=1}^{\infty} a_{n}$ can be divergent. How could I find such series? Or show that there isn't any?

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  • $\begingroup$ Consider the partial sums with an odd number of terms and the partial sums with an even number of terms. $\endgroup$ – saulspatz Jun 24 '20 at 15:15
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No. Asserting that the series $\sum_{n=1}^\infty a_n$ converges is equivalent to asserting that the series$$0+a_2+0+a_4+0+a_6+\cdots\tag1$$converges. And asserting that the series $\sum_{n=1}^\infty a_{2n-1}$ converges is equivalent to asserting that the series$$a_1+0+a_3+0+a_5+0+\cdots\tag2$$converges. But the sum of the series $(1)$ and $(2)$ is the series$$a_1+a_2+a_3+a_4+\cdots,$$which therefore converges.

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