Weak* subseries convergent implies norm convergent?

151 Views Asked by At

Suppose $X$ is a Banach space and $X^{*}$ is separable. Suppose that $\sum x_{n}^{*}$ is a series in $X^{*}$ which has the property that every subseries $\sum x_{n_{k}}^{*}$ converges weak*. Show that $\sum x_{n}^{*}$ converges in norm. [Hint: Every $x^{**}\in X^{**}$ is the weak* limit of a sequence in $X$.]

I know the hint basically follows from Goldstine's theorem that $B_{X}$ is weak* dense in $B_{X^{**}}$ and from the fact that the weak* topology on $B_{X^{**}}$ is metrizable by the assumption that $X^{*}$ is separable. But I can't figure out how to use the hint.