How to show that $y_K< 4?$

Question

$X = C[0,1]$. $$x_n(t) = \begin{cases} nt, & \text{for $0 \leq t \leq \frac{1}{n}$ } \\ 2-nt, & \text{for $\frac{1}{n} \leq t \leq \frac{2}{n}$ } \\ 0, & \text{for $\frac{2}{n} \leq t \leq 1$} \\ \end{cases}$$

Prove that $x_n$ tend to zero weak but not strongly ?

My attempt : Proof is given in the book Functional analysis by Peter D . Lax courant institute

Here is an outline of the proof

let $l$ be a bounded linear function we claim that $l(x_n)=0$ when $n \to \infty$ Now suppose that $l(x_n)\neq0$ when $n \to \infty$, then there would be infinitely many $n$ such that $|l(x_n)|> \delta >0$ say $l(x_n) > \delta.$ choose a subsequence ${n_k}, n_{k+1} >2n_k$ for which $l(x_n) > \delta$ hold. It is not hard to show that for all $t$ in $[0,1] $

$y_K= \sum_{1}^{K}x_{n_k}(t) < 4\tag1$

which implies that $|y_K|< 4$ for all $K$.From $(1)$ it follow that $l(y_K)=\sum_{1}^{K}l(x_n) > K\delta$

Since this holds for all $K$ , and since |$y_K| <4$ for all $K$ , the boundedness of $l$ is contradicted. Since $|x_n|=\max_n(t)=1$ $\{x_n\}$ does not tend to zero strongly

My confusion : why $y_K= \sum_{1}^{K}x_{n_k}(t) < 4?$ why does the author put series in $y_K$ ?

My thinking : Here $x_n(t) $ is a sequence but not a series.

Suppose $ y_K$ is series then

$y_K= \sum_{1}^{K}x_{n_k}(t) = x_{n_1}(t) + x_{n_2}(t)+...........+x_{n_{k+1}}(t)= \underbrace{ 1+1+.....+1}_{K \text{times}}= K >4$