Let's say I have a nonnegative, double-indexed sequence (like a table extending to infinity in two directions) $(a_{ij})_{i,j\in \mathbb{N}}$, and assume $\sum_{j}a_{ij}=1/2^{i}$.
By theorem 8.3, Rudin's PMA, we know $\sum_{j}\sum_{i}a_{ij} = \sum_{i}\sum_{j}a_{ij} = 1$. Rudin's theorem 3.55 also says, if series $\sum s_{n}$ converges absolutely, then any rearrangement of it converges to the same limit.
Now lets do the "diagonal trick" and convert the double-indexed sequence $(a_{ij})$ into just a normal sequence $(a_{n})$. Is there a way we can show that $\sum a_{n}$ converges to 1?
I’m actually reviewing measure theory with Donald Cohn’s book. And on page 14, line 8, the author finds the cover of sets $A_{n}$, i.e., $\{I_{n,i}\}_{i=1}^{\infty}$. He does the diagonal trick for indices $n, i$ and turns this sequence of covers $(\{I_{n,i}\}_{i=1}^{\infty})_{n}$ for $(A_{n})$ into a cover for $\cup A_{n}$ with just a single index, i.e., a cover $\{I_{m}\}$. Then he sums up volumes of intervals in this cover. It seems he assumes the sum of the doubly indexed sequence is the same as the sum of the sequence.