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 : $$(\sum_{n\in\Bbb N^*}\dfrac{u_n}{n^s})(\sum_{m\in\Bbb N^*}\dfrac{v_m}{m^s}) = \sum_{n,m \in\Bbb N^*} \frac{u_nv_m}{(nm)^s} = \sum_{i\in I}w_i = \sum_{n \in\Bbb N^*} \sum_{i\in I_n} w_i =\sum_{n \in\Bbb N^*} \sum_{kl=n}\frac{u_kv_l}{k^sl^s} =\sum_{n \in\Bbb N^*} \sum_{kl=n}\frac{u_kv_l}{n^s}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.