Prove that the operator $T:\ell^1\rightarrow\ell^1$ which maps $x=(x_1,x_2,\dots)$ to $\left(x_1,\frac{x_2}{2},\frac{x_3}{3},\dots\right)$ is compact.
For an arbitrary sequence $x^{(N)}\in\ell^1$ one would have extract a convergent subsequence of $T x^{(N)}$. Maybe via the diagonal argument?