convergente series $\sum_{n=1}^{+\infty}a_n$ such that all the power diverges This question arose from here,

Can we find a convergente series $\sum_{n=1}^{+\infty}a_n$ such that all the series $\sum_{n=1}^{+\infty}a_n^k$ for $k\ge2$ diverges ?


The answer is Yes.
Consider : $$1+\frac{1}{2\ln(2)}+\frac{1}{2\ln(2)}-\frac{1}{\ln(2)}+\cdots+\underbrace{\frac{1}{n\ln(n)}+\cdots+\frac{1}{n\ln(n)}}_{n \quad \text{terms}}-\frac{1}{\ln(n)}+\cdots$$
We have,
$$
\sum_{n=1}^{\frac{N^2+3N-2}{2}}a_n^k = \left\{
    \begin{array}{ll}
        1+\sum_{n=2}^{N}\bigr(\frac{1}{n^{k-1}\ln(n)^k}+\frac{1}{\ln(n)^k}\bigl) & \mbox{if } k \quad\mbox{is even } \\
       1+\sum_{n=2}^{N}\bigr(\frac{1}{n^{k-1}\ln(n)^k}-\frac{1}{\ln(n)^k}\bigl) & \mbox{if } k \quad\mbox{is odd }
    \end{array}
\right.
$$
Then by Cauchy condensation test, the series $\sum_{n=2}^{+\infty}\frac{1}{ln(n)^k}$ diverges for all $k\in\mathbb{N}$. However, the series $\sum_{n=2}^{+\infty}\frac{1}{n^{k-1}\ln(n)^k}$ converges if $k\ge2$
