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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?

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Note that since $\sum_n a_n$ converges, we have $\vert a_n \vert < 1$ for $n>N$. From this $N$ on, we have $\vert a_n \vert^k < \vert a_n \vert$ – user17762 Jan 27 '13 at 22:58
I believe not. For k=1, the sum converges and $a_n=o(\frac{1}{n})$. For $k>1$ odd, this is $o(\frac{1}{n^k})$ and converges. – CBenni Jan 27 '13 at 22:59
@Marvis I guess that can be made into an answer. – CBenni Jan 27 '13 at 23:00
@CBenni: Where does $a_n=o\left(\cfrac{1}{n}\right)$ come from? – xavierm02 Jan 27 '13 at 23:02
@xavierm02 you can show that the sum over every sequence that is asymptotically bigger or equal to 1/n and with a finite number of sign switches diverges and converges otherwise. My comment was effectively the same as Marvis' anyways – CBenni Jan 27 '13 at 23:05

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.

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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.

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interesting. We know where to begin to look. – ncmathsadist Jan 27 '13 at 23:03

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