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Here is the problem: Σ 1 / (2 n minus 1) (2 n plus 1), n equals 1 to ∞.

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Decide whether Σ 1 / (2 n minus 1) (2 n plus 1) (n from 1 to ∞) converges or diverges.

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Write the series. Σ a(n) for n equals 1, 2, 3, … with a(n) equals 1 / (2 n minus 1) (2 n plus 1). First terms: 0.333333, 0.0666667, 0.0285714, 0.015873, ….

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nth-term test. lim a(n) equals 0, so the divergence test is inconclusive; the terms do shrink, so a finer test is needed.

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Limit comparison with a p-series. Positive terms: the terms shrink roughly like 1/n squared (estimated from a(2N)/a(N)), so compare with Σ 1/n squared: lim a(n) times n squared equals 1 over 4. The limit is a positive finite number, so the series behaves exactly like the p-series Σ 1/n squared, which converges because p greater than 1.

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Estimate the sum numerically. Partial sums S N were computed for several N and extrapolated to N → ∞ (Richardson extrapolation; alternating terms are averaged first). Two independent extrapolations agree to better than 10 to the power minus 10, giving S approximately 0.5. Any closed form shown is a numerical identification, not a proof.

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So the answer is The series converges (limit comparison with Σ 1/n squared); sum approximately 0.5 (matches 1 over 2 to 10+ digits).

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A series converges when its partial sums approach a finite number. The terms must tend to 0 for that to happen, but that alone is not enough (the harmonic series diverges), so a comparison, ratio, root, alternating or integral test is used to decide.

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Partial sums (supporting evidence, not a proof): S₁₀ equals 0.47619048, S₁₀₀ equals 0.49751244, S₁₀₀₀ equals 0.49975012, S₁₀₀₀₀ equals 0.499975.