Complex sequence $(z_i)$ such that $\sum_i z_i^k$ converges and $\sum_i \vert z_i \vert^k$ diverges for all $k$

Question

How to find a complex sequence $(z_i)$ such that $$\sum_i z_i^k \text{ converges for all } k \in \mathbb N$$ but $$\sum_i \vert z_i \vert^k \text{ diverges for all } k \in \mathbb N ?$$

Answer 1

Let $\alpha=(1+\sqrt{5})/2$. It has the Diophantine property: $|e^{i k\alpha}-1| \geq \frac{c}{k^2}$ for integer $k\neq 0$. Then $$ a_n = r_n e^{i\alpha n}= \frac{1}{\log(n+1)} e^{i\alpha n}$$ will have the desired property. First, $$ \sum_n |a_n|^k \geq \sum_n r_n^k= \sum_n \frac{1}{(\log(n+1))^k} = + \infty$$ Second, consider $$S_n=\sum_{p=1}^{n} e^{ipk\alpha}= e^{ik\alpha}\frac{e^{ink\alpha} -1}{e^{ik\alpha}-1} $$ for which $|S_n|\leq 2 k^2/c$. Then by an Abel partial summation $$ \sum_{n= 1}^{N} a_n^k = \sum_{n=1}^{N} (S_{n}-S_{n-1}) r_n^k = \sum_{n=1}^{N} S_{n} (r_{n}^k -r_{n+1}^k)+S_{N} r_{N+1}^k $$ and $$\sum_n|S_n(r_{n}^k-r_{n+1}^k) | \leq \frac{2k^2}{c} \sum_n (r^k_{n}-r^k_{n+1}) = \frac{2k^2}{c} r_1^k $$ It follows that $\sum_n a_n^k$ is convergent.

As is evident from the proof you may use any irrational rotation number $\alpha$ and any decreasing sequence $r_n$ going to zero sufficiently slowly (so that $\sum_n r_n^k=+\infty$ for all $k$).

Answer 2

Consider the example \begin{align} \sum^\infty_{n=2} \frac{(-1)^{nk}}{\log^k n} \end{align} for each fixed $k$. By the alternating series test, the series converges. Now, consider the absolute series \begin{align} \sum^\infty_{n=2} \frac{1}{\log^k n}. \end{align} By the Cauchy consendation test, it suffices to check the following series \begin{align} \sum^\infty_{m=1} 2^m \frac{1}{\log^k(2^m)} = \sum^\infty_{m=1} \frac{2^{m}}{m^k\log^k 2}=\infty. \end{align}

Thus, the series diverges.

Edit: The above example works when $k$ is odd, but doesn't when $k$ is even.

However, we can consider the following modification \begin{align} \sum^\infty_{n=2} \left(\frac{e^{in}}{\log n}\right)^k. \ \ \ (*) \end{align} For a fixed $k$, observe the partial sum of the $e^{ink}$ is bounded since \begin{align} \left|\sum^N_{n=2} e^{ink}\right| = \left|e^{i2k}\left( \frac{1-e^{i(N-1)k}}{1-e^{ik}} \right)\right| \leq \frac{2}{|1-e^{ik}|}<\infty \end{align} since $k$ is an integer. Thus, citing theorem 3.42 in Rudin's Principles of Mathematical Analysis, we have that the series $(\ast)$ converges. But by Cauchy's condensation test, the absolute series diverges.

Comment: I hope this one will do the trick and sorry for my earlier mistakes.