Originally Posted by

**slevvio** 2. Explain why $\displaystyle \sum_{r,s =0}^{n} a_r b_s \rightarrow AB $ [done]

3. Now show that $\displaystyle \sum_{n=0}^{\infty} c_n = AB $.

HINT: [You might want to do this by estimating $\displaystyle | \sum_{r=0}^{2n} c_r -\sum_{r,s=0}^{n} a_r b_s | $. Here you will have to use that $\displaystyle \sum_{n=0}^{\infty} a_n $ and $\displaystyle \sum_{n=0}^{\infty} b_n $ are absolutely convergent.]

I dont understand where the 2n is coming from or how to even use it. I think I am trying to prove the statement

$\displaystyle \forall \epsilon > 0 , \exists N \in \mathbb{N} $ such that $\displaystyle \forall n \ge N, | \sum_{r=0}^{n} c_r - AB | < \epsilon $.