If $\sum_{1}^{\infty} a_n^2$ converges and $a_n$ is complex valued, then how about $\sum_{1}^{\infty} a_n/n$

Question

If $a_n$ is nonnegative-valued, then I can use the CS inequality to show that $\sum_{1}^{\infty} a_n/n$ converges. However, if $a_n$ is complex valued, then I think a sequence $a_n = 1/(\ln{n})^{0.5}$ for odd $n$ and $a_n = i/(\ln{n})^{0.5}$ for even $n$ works as a counterexample. Am I correct?

Answer 1

Yes, that works (although the square root in the denominator is unnecessary, and it's not defined for $n=1$).

$a_n^2$ is real and positive for odd $n$, real and negative for even $n$, and the sum converges by the alternating series test.

$\frac{a_n}{n}$ is real and positive for odd $n$ and purely imaginary for even $n$. So $\Re(\sum_{j=1}^{2k}\frac{a_j}j)=\sum_{j=1}^k\frac{a_{2j}}{2j}$, which doesn't converge; since the real part doesn't converge, the whole thing certainly doesn't.