Are infinite linear combinations that converge in a Banach space uniquely represented?

Question

If $X$ is a Banach space, and $(x_n)$ is a sequence of linearly independent elements of X, then is the following true? If $\sum_{n=1}^\infty a_n x_n = 0_X,$ then $a_n = 0$ for every $n$.

Thank you.

Answer 1

By the usual definition of linearly independent, this fails to be true.

For example: define a sequence in $X = \ell^2$ as follows: for $n,k \in \Bbb N$, $$ x_1(k) = \frac{1}{k}\\ x_n(k) = \begin{cases} 1 & k = n-1\\ 0 & \text{otherwise} \end{cases} \quad \text{for } n \geq 2 $$ Confirm that this sequence is linearly independent. Nevertheless, we have $$ x_1 - \sum_{n=2}^\infty \left(\frac 1{n-1}\right)x_n = 0 $$