Prove that $\sum_{j=0}^n \sum_{k=0}^j p_k q_{j-k} r_{n-j} = \sum_{k=0}^n \sum_{j=k}^n p_k q_{j-k} r_{n-j}$

Question

I encounter this problem when proving that $\Bbb R[[X]] := (\Bbb R^\Bbb N,+,\cdot)$ is actually a formal power series ring over $\Bbb R$.

Let $(p_n \mid n \in \Bbb N), (q_n \mid n \in \Bbb N), (r_n \mid n \in \Bbb N)$ be sequences in $\Bbb R$. Prove that $$\sum_{j=0}^n \sum_{k=0}^j p_k q_{j-k} r_{n-j} = \sum_{k=0}^n \sum_{j=k}^n p_k q_{j-k} r_{n-j}$$

I have tried to manipulate the indices $j,k,n$ but to no avail. Please leave me some hints. Thank you so much!

Answer 1

All you have to prove is that the sets of indices the sums run over are the same. In fact, they are both equal to

$$S\{(j, k)| 0\leq j,l\leq n\land k\leq j\}$$

That is, you can prove that in both sums, the index pair $(j,k)$ appears in the sum if and only if $(j, k)\in S$.

Answer 2

You are adding the same numbers in a different order The condition $0 \leq k \leq j \leq n$ is same as $j \geq k $ , $j \leq n$ and $0\leq k$ so the terms on the two sides are the same.