The transformed series
Assume aₙ≥0 and the series of aₙ diverges. The lesson proves that the series with terms aₙ/(ₙ) also diverges. Nonnegativity is part of the hypothesis.
Math Mastery with Amitesh · YouTube · 7:51
Prove a divergence result for nonnegative series by comparison and a necessary condition for convergence.
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Assume aₙ≥0 and the series of aₙ diverges. The lesson proves that the series with terms aₙ/(ₙ) also diverges. Nonnegativity is part of the hypothesis.
If aₙ≤1 eventually, then ₙ≤2 and aₙ/(ₙ)≥aₙ/2. Comparison with the divergent original series forces divergence.
If the transformed series converged, its terms would tend to zero. Writing bₙ=aₙ/(ₙ) gives aₙ=bₙ/(ₙ), so aₙ would tend to zero and the eventual comparison would apply, contradicting convergence.
Assume aₙ≥0 and the series of aₙ diverges. The lesson proves that the series with terms aₙ/(ₙ) also diverges. Nonnegativity is part of the hypothesis.
If aₙ≤1 eventually, then ₙ≤2 and aₙ/(ₙ)≥aₙ/2. Comparison with the divergent original series forces divergence.
If the transformed series converged, its terms would tend to zero. Writing bₙ=aₙ/(ₙ) gives aₙ=bₙ/(ₙ), so aₙ would tend to zero and the eventual comparison would apply, contradicting convergence.
Assume aₙ≥0 and the series of aₙ diverges. The lesson proves that the series with terms aₙ/(ₙ) also diverges. Nonnegativity is part of the hypothesis. If aₙ≤1 eventually, then ₙ≤2 and aₙ/(ₙ)≥aₙ/2. Comparison with the divergent original series forces divergence. If the transformed series converged, its terms would tend to zero. Writing bₙ=aₙ/(ₙ) gives aₙ=bₙ/(ₙ), so aₙ would tend to zero and the eventual comparison would apply, contradicting convergence.