Comparison Test (Series)

Comparison Test

Suppose ∀n∈N we have an,bn∈R satisfy 0≤an≤bn. Then:

  1. If ∑bn converges, then ∑an converges.
  2. If ∑an diverges, then ∑bn diverges (automatically from (1))

Proof
Suppose ∑bn converges. Let's show ∑an satisfies the Cauchy Condition for Series, and thus converges.

Let ε>0. Since ∑bn converges then we can choose N∈N such that n>m≥N has it that:

|bm+1+⋯+bn|<ε

Everything here is non-negative, so we can get rid of the absolute value signs:

bm+1+⋯+bn<ε

Suppose this is true, that we choose our n>m≥N. Then:

|am+1+⋯+an|=am+1+⋯+an≤bm+1+⋯+bn<ε

Thus ∑an satisfies the Cauchy Condition for Series, so then it must converge.
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