Assume that $\sum_{n=1}^{\infty} c_nv^{(n)}=0 $ where $\left \{v^{(n)} \right \}_{n\in\mathbb{N}}$ is an idependent set in a Banach space $X$. Must it be true that $c_n=0$ for all $n$?
I thought about this question while trying to solve another related problem. My intuition says that it's correct, but when I try to prove it, I cannot find a way to use the independency of the vectors (because having a linear combination with a very small norm has nothing to do with having a linear combination that equals $0$). But maybe I'm wrong when I try to mix linear independence and series convergence like that.
I'd like to hear your thoughts of this.
Assuming that here "independent" means that no vector is a finite linear combination of the others, the answer is no:
Say $v_1,v_2,\dots$ are orthonormal vectors in a Hilbert space. Let $v_0=\sum_{n=1}^\infty v_n/n$. Then $v_0,v_1,\dots$ are independent (you can easily verify this using orthonormality of $v_1,\dots$), although $$-v_0+v_1+v_2/2+\dots=0.$$