Proving the Sum Rule

Real Analysis · Axiom Academy

An - N proof that , built on the trick and the triangle inequality. Suppose and . Prove, directly from the - N definition of a limit, that the sum sequence converges to the sum of the limits: The idea: keep each error under , so the total stays under Each sequence is forced within of its limit (top two bands). Because the errors can only add , the worst the combined error can do is — so the sum lands within of A+B (bottom band). Nice work. You proved the Sum Rule the way every limit theorem is proven — by forcing the error below an arbitrary . The moving parts: The trick: two error terms add, so split the budget — demand from each so the total stays under . Taking : a single threshold past which both convergence guarantees are active at once. The triangle inequality: regroup as (a_n - A) + (b_n - B) , then splits the combined error into the two pieces you control. The payoff: — exactly the bound the definition asks for. This same -splitting strategy reappears throughout analysis — in the proofs of the difference rule, the scalar-multiple rule, and continuity of sums. Master it once and the rest follow the same template.

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