Let's say we have the double infinite series $\sum\limits_{n=1}^\infty \sum\limits_{m=1}^\infty a_n b_m$ which is absolutely convergent. Furthermore, the two series $\sum\limits_{n=1}^\infty a_n$ and $\sum\limits_{n=1}^\infty b_n$ are absolutely convergent.
I know the Cauchy Product is: $\sum\limits_{n=1}^\infty\sum\limits_{i=2}^n a_i b_{n-i} = \left(\sum\limits_{n=1}^\infty a_n\right) \left(\sum\limits_{n=1}^\infty b_n\right)$
1) However, is it also true that the double infinite series equals the Cauchy Product in the given scenario? I.e. that:
$$\sum_{n=1}^\infty \sum_{m=1}^\infty a_n b_m = \sum_{n=1}^\infty\sum_{i=2}^n a_i b_{n-i} = \left(\sum_{n=1}^\infty a_n\right) \left(\sum_{n=1}^\infty b_n\right)$$
2) In particular I am interested in: does the above (abs. convergent) double infinite series "split" into two (abs. convergent) infinite series? If yes, is this also true for general convergence?
3) I am thinking about doing this:
$$\sum_{n=1}^\infty \sum_{m=1}^\infty a_n b_m = \sum_{n=1}^\infty \left(\sum_{m=1}^\infty a_n b_m\right) = \sum_{n=1}^\infty \left(a_n \sum_{m=1}^\infty b_m \right) = \left(\sum_{n=1}^\infty a_n \right) \left(\sum_{m=1}^\infty b_m \right)$$
Is that correct?
Yes. This is true whenever at least one of the series is absolutely convergent. See Wikipedia.
You can interpret it that way. Absolute convergence implies that the order of summation or taking limits does not influence the result. This is not true for general convergence, where that order matters.
Yes. This is an algebraic manipulation where the last step requires absolute convergence (because you pull one limit out of the other).
Edit: You should let all the sums start at $0$ for the formulas to be correct.