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Prove or disprove: If (aₙ) is a sequence in R and 0 ES [aₙ] then there exists a subsequence (aₙ) of (aₙ) such that the series ∑[infinity]ⱼ₌₁ aₙⱼ an converges absolutely.

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Final answer:

To prove the statement, we need to show that given a sequence (aₙ) in R such that 0 ≤ aₙ, there exists a subsequence (aₙ) of (aₙ) such that the series ∑[infinity]ⱼ₌₁ aₙⱼ an converges absolutely.

Step-by-step explanation:

To prove the statement, we need to show that given a sequence (aₙ) in ℝ such that 0 ≤ aₙ, there exists a subsequence (aₙⱼ) such that the series ∑[infinity]ⱼ₌₁ aₙⱼ converges absolutely. Since the terms of the sequence (aₙ) are non-negative, we can consider a subsequence (aₙⱼ) that consists of terms that are non-decreasing. By the Monotone Convergence Theorem, this subsequence will converge to a limit or go to infinity. If (aₙⱼ) goes to infinity, then the series ∑[infinity]ⱼ₌₁ aₙⱼ diverges. Therefore, we consider the case where (aₙⱼ) converges to a limit L. Since |aₙⱼ - L| ≤ aₙⱼ for all j, the series ∑[infinity]ⱼ₌₁ |aₙⱼ - L| is bounded by the series ∑[infinity]ⱼ₌₁ aₙⱼ, and therefore converges. Thus, we have proven that if (aₙ) is a sequence in ℝ and 0 ≤ aₙ, then there exists a subsequence (aₙⱼ) such that the series ∑[infinity]ⱼ₌₁ aₙⱼ converges absolutely.

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