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?
On
Check that it's the limit of the following finite rank operators and invoke the theorem that a uniform limit of finite rank operators on a Banach Space is a compact operator.
$$T_n(x_1,x_2....)=(x_1,\frac{x_2}2...\frac{x_n}n,0,0,0,0)$$
You can verify this by checking that $T_n$ is a Cauchy sequence in $L(\mathcal l_1)$
Define $$T_j(x):=\left(x_1,\frac{x_2}2,\dots,\frac{x_j}j,0,\dots,0\right).$$ It's a compact operator (because it's finite ranked) and $$T(x)-T_j(x)=\left(0,\dots,0,\frac{x_{j+1}}{j+1},\dots\right),$$ hence $$\lVert T(x)-T_j(x)\rVert_{\ell^1}=\sum_{k=j+1}^{+\infty}\frac{|x_k|}{k}\leq \frac 1{j+1}\sum_{k=j+1}^{+\infty}|x_k|\leq \frac 1{j+1}\lVert x\rVert_{\ell^1},$$ which proves that $\lVert T-T_j\rVert\leq \frac 1{j+1}$.
To conclude, notice that a norm limit of compact operators is compact.