= Solution
The complex <Hahn-Banach theorem> says that every bounded complex-<linear functional> on a <linear subspace> of a normed complex space extends to the whole space with the same <norm>.
For the separation assertion choose $x_0\notin M$ and $d=\operatorname{dist}(x_0,M)>0$. On $M+\mathbb Cx_0$ define $g_0(m+\lambda x_0)=\lambda d$. The inequality $\|m+\lambda x_0\|\geq|\lambda|d$ makes its <norm> at most one; approximating the distance $d$ shows the <norm> is exactly one. Hahn-Banach gives an extension $g$ of <norm> one annihilating $M$.
Each <geometric sequence> satisfies $\|x_k\|_1=(1-|\beta_k|)^{-1}\leq(1-\delta)^{-1}$. Let $M$ be their <closed linear span>. If $M\ne\ell^1$, the separation just proved gives a nonzero bounded functional annihilating all $x_k$. Represent it by $b\in\ell^\infty$ and form
$$
F(z)=\sum_{j=1}^\infty b_jz^{j-1},\qquad|z|<1.
$$
The series converges locally uniformly and defines a <holomorphic function>. It vanishes at every $\beta_k$. These distinct points lie in the <compact> disk $|z|\leq\delta<1$, so have an accumulation point inside the domain. The <identity theorem> gives $F=0$, hence every <Taylor coefficient> $b_j=0$, contradicting the functional's nonzero <norm>. Therefore
$$
\boxed{\overline{\operatorname{span}\{x_k:k\geq1\}}=\ell^1.}
$$
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