Why $ \overline{\operatorname{span}}\left\{f_k\right\}_{k=1}^{\infty}=X $ and not $\operatorname{span}\left\{f_k\right\}_{k=1}^{\infty}=X$? Let $X$ be a normed vector space. We say that an infinite series $\sum_{k=1}^{\infty} c_k f_k$ is convergent with sum $f \in X$ if
$$
\left\|f-\sum_{k=1}^n c_k f_k\right\| \rightarrow 0 \text { as } n \rightarrow \infty .
$$
If this condition is satisfied, we write
$$
f=\sum_{k=1}^{\infty} c_k f_k .
$$
For a given sequence $\left\{f_k\right\}_{k=1}^{\infty}$ in $X$, we let $\operatorname{span}\left\{f_k\right\}_{k=1}^{\infty}$ denote the vector space consisting of all finite linear combinations of vectors $f_k$, i.e.,
$$\operatorname{span}\left\{f_k\right\}_{k=1}^{\infty}=\left\{\alpha_1 f_1+\alpha_2 f_2+\cdots+\alpha_N f_N \mid N \in \mathbb{N}, \alpha_1, \alpha_2, \ldots, \alpha_N \in \mathbb{C}\right\}$$
The definition of convergence shows that if each $f \in X$ has a representation of this type, then each $f \in X$ can be approximated arbitrarily well in norm by elements in $\operatorname{span}\left\{f_k\right\}_{k=1}^{\infty}$, i.e.,
$$
\overline{\operatorname{span}}\left\{f_k\right\}_{k=1}^{\infty}=X
$$
I got this from a book. My question is why $\overline{\operatorname{span}}$ and not simply $\operatorname{span}$?
 A: Because $\mathrm{span}$ only consists of finite linear combinations - no matter how arbitrarily long you make that sum.
Quick example: $X = \ell^2$. Then, if we define
$$
f_k := (0, ..., \underset{\text{k-th spot}}{1}, 0, 0,...,),
$$
i.e.
$$
(f_k)_n = \delta_{kn}
$$
for all $k, n \in \mathbb{N}$, then indeed
$$
\overline{\mathrm{span}\lbrace f_k \rbrace_{k=1}^\infty} = \ell^2.
$$
This is very easy to check.
But if we do not take the closure, note that every sequence in
$$
\mathrm{span}\lbrace f_k \rbrace_{k=1}^\infty
$$
has infinitely many members that are zero, simply because all those linear combinations are finite. For example,
$$
(\pi, 0, e, 0, ..., 0, ..., 42, 0, 0, ...) \in \mathrm{span}\lbrace f_k \rbrace_{k=1}^\infty.
$$
But clearly, it holds for the reciprocal natural numbers
$$
\ell^2 \ni (1, 2^{-1}, 3^{-1}, ..., 567^{-1}, 568^{-1}, ...) \notin \mathrm{span}\lbrace f_k \rbrace_{k=1}^\infty
$$
because it does not have zero-entries.
Taking the closure means, you also include all $\ell^2$-limits of those linear combinations. This is where we can get rid of all those zeros.
