An element not in any finite dimensional subspace An exercise from Real Analysis by Lang

Let $\{x_n\}$ be a sequence of linear independent elements of norm 1 in a Banach space. Find an element in closure of the space generated by all $x_n$ but not in any subspace generated by finite number of $x_n$.

I choose $\sum_{n=1}^{\infty} 3^{-n}x_n$ but I am not sure how to prove the last statement.
 A: Your construction does not necessarily work. Consider the Banach space $\ell^1(\mathbb{N})$ (here I use the convention $0 \notin \mathbb{N}$) of absolutely summable (real or complex, doesn't matter) sequences.
Let $e_k$ denote the $k^{\text{th}}$ "standard basis" (not a basis in the sense of linear algebra, but a Schauder basis) element, i.e. $e_k(k) = 1$ and $e_k(n) = 0$ for $n \neq k$. Then choose $x_1 = e_1$ and
$$x_n = \frac{1}{4}e_n - \frac{3}{4}e_{n-1} $$
for $n \geqslant 2$. It's easy to see that $\lVert x_n\rVert = 1$ for all $n$ and that the $x_n$ are linearly independent. But we have
$$\sum_{n = 1}^{\infty} 3^{-n}x_n = \frac{1}{3}e_1 + \frac{1}{4}\sum_{n = 2}^{\infty}3^{-n}(e_n - 3e_{n-1}) = \frac{1}{3}e_1 + \frac{1}{4}\sum_{n = 2}^{\infty} 3^{-n}e_n - \frac{1}{4}\sum_{n = 1}^{\infty} 3^{-n}e_n = \frac{1}{4}x_1\,.$$
The problem here (and the problem to consider generally) is that for $n \geqslant 2$ the vector $x_n$ is too close to the space spanned by $x_1, \dotsc, x_{n-1}$, this way we can get back to a space spanned by finitely many (in this case one) of the $x_n$. What "too close" means depends on the coefficient sequence. If we look at the general construction
$$\sum_{n = 1}^{\infty} c_nx_n$$
with $(c_n) \in \ell^1$, then a return to $\operatorname{span} \{ x_1, \dotsc, x_{n-1}\}$ is impossible if we have
$$\lvert c_n\rvert \cdot \operatorname{dist}(x_n, \operatorname{span} \{x_1, \dotsc, x_{n-1}\}) > \sum_{k = n+1}^{\infty} \lvert c_k\rvert\,. \tag{$\ast$}$$
You can always choose a sequence $(c_n)$ such that $(\ast)$ holds for all $n \geqslant 2$ (or just infinitely many $n$). But it may be simpler to instead show that you can find a sequence $(y_n)$ of unit vectors with $y_n \in \operatorname{span} \{x_1, \dotsc, x_n\}$ and $\operatorname{dist}(y_n, \operatorname{span} \{x_1, \dotsc, x_{n-1}\}) > \frac{2}{3}$ for all $n$ (Riesz lemma), and then you can take e.g.
$$\sum_{n = 1}^{\infty} 3^{-n}y_n\,.$$
A: Your construction $x = \sum_{n=1}^\infty \frac{x_n}{3^n}$ doesn't work in general. Consider the canonical vectors $(e_n)_n$ in $\ell^2$ and let $$x_1 = \sqrt{\frac{71}{72}}e_2- 3\sum_{n=3}^\infty \frac{e_n}{3^n}$$ and $x_n = e_n$ for $n \ge 2$. Then $(x_n)_n$ are linearly independent and have norm one but $$\sum_{n=1}^\infty \frac{x_n}{3^n} = \left(\frac13\sqrt{\frac{71}{72}} + \frac19\right)e_2 \in \operatorname{span}\{x_1,x_2\}.$$
We can give a nonconstructive proof. We always have
$$\bigcup_{n\in\Bbb{N}} \operatorname{span}\{x_1,\ldots, x_n\} \subseteq \overline{\operatorname{span}\{x_n : n\in\Bbb{N}\}}$$
but the equality does not hold. Namely, RHS is a Banach space and LHS is a countable union of nowhere dense closed sets and hence cannot be a Banach space by the Baire category theorem. Therefore we can pick $x \in \overline{\operatorname{span}\{x_n : n\in\Bbb{N}\}}$ which is not contained in any $\operatorname{span}\{x_1,\ldots, x_n\}$.
