Let ∑∞n=1an be an absolutely converging series. By definition, this means ∑∞n=1|an| converges. We want to show that ∑∞n=1a2n converges absolutely as well.
Here is what I tried:
Let's say ∑∞n=1|a2n| converges. Let ϵ>0 be given and there exists N∈N such that
|a2m+1|+|a2m+2|+⋯+|a2n|<ϵ for all n>m≥N. By the triangle inequality,
|a2m+1+a2m+2+⋯+a2n|≤|a2m+1|+|a2m+2|+⋯+|a2n|
so the Cauchy Criterion guarantees that ∑∞n=1a2n also converges.
Here is my question:
Is this correct, or do I need to prove anything else?
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