Convergence of a rearrangement of conditionally convergent series

$$\{a_n\}$$ 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?

• Consider the partial sums with an odd number of terms and the partial sums with an even number of terms. Jun 24, 2020 at 15:15

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.