Show that a subset $K$ of the Hilbert space $L_2[0,1]$ is closed.

Let $K$ be a subset of $L_2[0,1]$ consisting of the functions $f_n(x)=(1+2^{-n})e^{2\pi inx}$, where $n=1,2,3,\ldots$. Show that the subset $K$ of the Hilbert space $L_2[0,1]$ is closed.

If I want show $K$ is closed , I have to show $f_n$ converges to an element in $L_2[0,1]$. But I couldn't find that element. Or is there any other way to conclude the result?

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Note that $\langle f_n, f_m \rangle = 0$ when $n\neq m$. Hence $f_n$ are orthogonal, and $\|f_n-f_m\|^2 = \|f_n\|^2+\|f_m\|^2 \geq 2$. In particular, $f_n$ is not Cauchy. In fact, no subsequence of $f_n$ can be Cauchy.
Now let $g \notin K$. Then we have $\sqrt{2} \leq \|f_n-f_m\| \leq \|g-f_n\|+\|g-f_m\|$ whenever $n\neq m$. Now let $\delta = \inf_k \|g-f_k\|$. If $\delta = 0$, then there must exist a subsequence $f_{n_k}$ such that $\lim_k \|g-f_{n_k}\| = 0$, which contradicts the previous estimate for large $k$. Hence $\delta > 0$, and consequently, $B(g,\delta) \cap K = \emptyset$. Hence $K^C$ is open, and so $K$ is closed.
You don't need to show it approaches a limit; rather what you need is to show that every limit that some subsequence approaches is already a member of the sequence. And if you were to show that the sequence has a limit, that would not prove the set is closed. If the limit were not a member of the sequence, then that would show that this set is not closed. If you can show that any two members of this set are orthogonal to each other, that would imply that any limit point of this set (which would be a limit of some subsequence) would be in the intersection of the one-dimensional subspaces that they span. But the only point in that intersection is $0$. So it's enough to show that no subsequence of this sequence can approach $0$. That implies no such subsequence can approach anything. The set has no limit points; therefore every limit point is in the set; therefore it's closed.