# Show that $\sum_n\sum_{k\ell=n}\dfrac{u_kv_k}{n^s} = \left(\sum_n\dfrac{u_n}{n^s}\right)\left(\sum_n \dfrac{v_n}{n^s}\right)$

How to show that $$\sum_{n\in\Bbb N^*}\sum_{k\ell=n}\dfrac{u_kv_\ell}{n^s} = \left(\sum_{n\in\Bbb N^*}\dfrac{u_n}{n^s}\right)\left(\sum_{n\in\Bbb N^*}\dfrac{v_n}{n^s}\right)$$ when $\sum_{n\in\Bbb N^*}\dfrac{u_n}{n^s}$ and $\sum_{n\in\Bbb N^*}\dfrac{v_n}{n^s}$ converge absolutely ?

Define $I=\mathbb N^* \times \mathbb N^*$ and $I_n=\{(k,l)\in I, kl=n\}$. $(I_n)_n$ is a partition of $I$.
For $i=(k,l)\in I$ define $w_i = \frac{u_kv_l}{k^sl^s}$. The family $(w_i)_{i\in I}$is summable:
$$\sum_{i\in I} |w_i| = \sum_{n,m \in\Bbb N^*} \frac{|u_n||v_m|}{(nm)^s}=(\sum_{n\in\Bbb N^*}\dfrac{|u_n|}{n^s})(\sum_{m\in\Bbb N^*}\dfrac{|v_m|}{m^s})<\infty$$
The rearrangement theorem you've been taught applies : $$(\sum_{n\in\Bbb N^*}\dfrac{u_n}{n^s})(\sum_{m\in\Bbb N^*}\dfrac{v_m}{m^s}) = \sum_{n,m \in\Bbb N^*} \frac{u_nv_m}{(nm)^s} = \sum_{i\in I}w_i = \sum_{n \in\Bbb N^*} \sum_{i\in I_n} w_i =\sum_{n \in\Bbb N^*} \sum_{kl=n}\frac{u_kv_l}{k^sl^s} =\sum_{n \in\Bbb N^*} \sum_{kl=n}\frac{u_kv_l}{n^s}$$