[Math] Proof of Double Sum Manipulation in a Proof using Cauchy Product for Series

Question

In a proof that uses Cauchy product for series (baby rudin, 3.50), it's used the following manipulation:$$\sum _{k=0}^n \sum _{i=0}^k a_{i}b_{k-i}=a_0b_0+(a_0b_1+a_1b_0)+(a_0b_2+a_1b_1+a_2 b_0) +…+(a_0b_n+a_1b_{n-1}+…+a_nb_0)\\=a_0B_n+a_1B_{n-1}+…+a_nB_0,$$
where $B_j=\sum _{i=0}^j b_i$. Even if the formula is intuitive and the argument is convincing, I'm struggling to find a proof that avoids the dots-dots. How would you prove the identity $$\sum _{k=0}^n \sum _{i=0}^k a_{i}b_{k-i}=\sum _{k=0}^n a_k\sum _{i=0}^{n-k}b_i$$
by directly manipulating the sum?

Answer 1

$$\sum_{k=0}^n\sum_{i=0}^ka_ib_{k-i}=\sum_{i=0}^n\sum_{k=i}^na_ib_{k-i}=\sum_{i=0}^n\sum_{k=0}^{n-i}a_ib_k$$

The first equality is just reversing the order of summation, and the second is an index shift, dropping the index $k$ by $i$. Now just interchange the names $i$ and $k$ in the final expression, and you’re done.