Solution: Let us prove that the given sequence of functions converges to $f\equiv 0$ pointwise. Let $\varepsilon > 0$ be given. For any $x \in (0,1]$, by the Archimedean property, choose $n_0 \in \mathbb{N} $ such that $\frac{2}{n_0} \leq x$. Thus, for any $n \geq n_0$, we have \begin{align*} \frac{2}{n} \leq \frac{2}{n_0} \leq x \leq 1 & \implies f_n(x) = 0 \\ & \implies \left\vert f_n(x) - f(x) \right\vert = \left\vert f_n(x) \right\vert = 0 < \varepsilon . \end{align*}

We now prove that the convergence is not uniform. If possible, let us assume that the convergence is uniform. Then the limit must by the same as the pointwise limit. That is, \[ \forall\ \varepsilon >0,\ \exists\ n_0 \in \mathbb{N} \text{ s.t. } \forall\ x \in [0,1] \text{ and } \forall\ n \geq n_0 \ \left\vert f_n(x) \right\vert < \varepsilon . \] For $\varepsilon = \frac{1}{2}$, we can find $n_0 \in \mathbb{N} $ such that for every $x \in [0,1]$ and $n \geq n_0$ we have $f_n(x) < \frac{1}{2}$. Take $x = \frac{1}{n_0}$. Then \[ f_{n_0}(x) = n_0^2 \times \frac{1}{n_0} = n_0 < \frac{1}{2}, \] is absurd. Therefore, the convergence is not uniform.