# Difference between unconditional and absolute convergence in Banach spaces

One can show that in any finite-dimensional normed vector space absolute convergence is equivalent to unconditional convergence.

It's not hard to show that if we have an orthonormal sequence in Hilbert space $(e_n)_{n \in \mathbb N}, \ \sum_{k = 1}^\infty \alpha_k e_k$ converges absolutely iff $(\alpha_n)_{n \in \mathbb N} \in \ell^1$ but it converges unconditionally iff $(\alpha_n)_{n \in \mathbb N} \in \ell^2$.

So it suffices to take any sequence from $\ell^2 \setminus \ell^1$ to show that unconditional convergence doesn't imply absolute convergence, the most popular one definitely being $(\frac{1}{k})_{k \in \mathbb N}$.

I have a couple of questions about unconditional vs. absolute convergence:

Are there ''easy'' counterexamples in $\ell^p$ spaces for $p \neq 2$? Is there an intuitive way to explain this difference between unconditional and absolute convergence in infinite-dimensional spaces? Why do we need infinite dimensions for this difference to kick in?

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For $p\ne1$, take the same same example you used for $\ell_2$. For $\ell_1$, see this. Also, see this link for a link to the proof of the Dvoretzky-Rogers theorem, which states that in every infinite dimensional Banach space, there is an unconditionally convergent series that is not absolutely convergent. – David Mitra Feb 13 '14 at 9:38