Let $\{x_n\}$ be a sequence in a Banach space $X.$ Then the following conditions are equivalent:
(a) $$\sup_{1 \leq n < \infty} \|x_n\| < \infty$$
(b) $$\sum_{i=1}^\infty |\alpha_i| < \infty\quad\text{implies that }\quad \sum_{i=1}^\infty \alpha_i x_i\quad \text{converges}$$
To prove that (b) implies (a) suppose $\sup_{1 \leq n < \infty}\|x_n\| = \infty,$ then, taking indices $n_k$ with $\|x_{n_k}\| > 2^k (k = 1, 2, ...)$ and taking $\alpha_{n_k} = \frac{1}{2^k} (k = 1, 2, ...)$ and $\alpha_n = 0 (n \neq n_1, n_2, ...),$ we obtain that $\sum_{i=1}^\infty |\alpha_i| < \infty,$ but $\|\alpha_{n_k} x_{n_k}\| > 1 (k= 1, 2, ...),$ whence $\sum_{i=1}^\infty \alpha_i x_i$ does not converge.
My question is why then $\sum_{i=1}^\infty \alpha_i x_i$ does not converge??
