Let $T:\mathcal{l}^2\to\mathcal{l}^2$ be an operator with
$$(Tu)_n:= \frac{1}{2}u_n + \frac{1}{n^2}u_n.$$
I've found that the operator is not compact, since
$$(Tu)_n = \left(\frac{1}{2}+\frac{1}{n^2}\right)u_n$$
consists a sequence which converges not to zero. But how can I show whether the operator is continuous?
Suppose that $u^{(k)} \to u$ and $Tu^{(k)} \to v$ in $\ell^2$. Then, for all $n\geq 1$, \begin{align} |(Tu)_n-v_n|&\leq |(Tu)_n - (Tu^{(k)})_n|+|(Tu^{(k)})_n-v_n|\\ &= \left(\frac{1}{2}+\frac{1}{n^2}\right)|u_n-u^{(k)}_n|+|(Tu^{(k)})_n-v_n|\\ &\leq \frac{3}{2}\Vert u-u^{(k)} \Vert_2 + \Vert Tu^{(k)}-v \Vert_2 \to 0, \end{align} as $k\to\infty$, so $(Tu)_n = v_n$ for each $n\geq 1$, and thus $Tu=v$. The Closed graph theorem does the rest.
Alternatively, observe that $$ \Vert Tu \Vert_2^2 = \sum_{n=1}^\infty \left(\frac{1}{2}+\frac{1}{n^2}\right)^2 |u_n|^2 \leq \frac{9}{4} \sum_{n=1}^\infty |u_n|^2 = \frac{9}{4} \Vert u\Vert_2^2. $$