Does there exist a sequence of real numbers $\{a_n\}$ such that $\sum_na_n^k$ converges for $k=1$ but diverges for every other odd positive integer?

Question

Does there exist a sequence of real numbers $\{a_n\}$ such that $\sum_na_n^k$ converges for $k=1$ but diverges for every other odd positive integer?

Answer 1

Yes. In fact, a stronger fact holds:

For any set $C$ (finite or infinite) of odd positive integers, there is a sequence of real numbers $\{a_n\}$ such that for $k$ odd, $$ \sum_n a_n^k $$ converges iff $k\in C$.

Whether this is possible was asked by Polya as problem 4142 in the American Mathematical Monthly. It was solved by N.J. Fine, the solution appeared in 1946 (pp 283-284), and can be found here.

Answer 2

Here is a partial answer. If $a_n \ge 0$ for all $n$, no such series exists. Since $\sum_n a_n < \infty$, $a_n \to 0$. For $n$ sufficiently large, $a_n < 1$, so $a_n^k < a_n < 1$. This shows that $\sum_n a_n^k$ converges for any $k > 1$. In fact this applies if the series is absolutely convergent.

You would need a mighty strange conditionally convergent series to meet your criterion.