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Sequences and seriesAlgebraCollegeabout 2:45

Evaluate the sum of 1/(2n-1)(2n+1) from n=1 to infinity: animated lesson

A narrated, step-by-step sequences and series video lesson with captions. Final answer: The series converges (limit comparison with Σ 1/n^2); sum ≈ 0.5 (matches 1/2 to 10+ digits).

Video with a synthetic English voice and captions. Pick another language for subtitles in that language, or open the interactive version.

Video narrated by a synthetic voice (Piper TTS, en_US-ljspeech-high voice (LJ Speech dataset, public domain)); the steps come from our maths engine and the answer was checked independently. Download MP4 (1.0 MB) · Captions (WebVTT)

Steps on screen

  1. 1. The problem
    Worked example
    ∑n=1∞1(2n−1)(2n+1)\sum_{n=1}^{\infty} \frac{1}{\left(2n-1\right) \left(2n+1\right)}
  2. 2. Step by step
    1. ∑n=1∞1(2n−1)(2n+1)\sum_{n=1}^{\infty} \frac{1}{\left(2n-1\right) \left(2n+1\right)}
      Write the series
    2. lim⁡n→∞1(2n−1)(2n+1)=0\lim_{n\to\infty} \frac{1}{\left(2n-1\right) \left(2n+1\right)} = 0
      nth-term test
    3. lim⁡n→∞an1/n2=14\lim_{n\to\infty} \frac{a_n}{1/n^{2}} = \frac{1}{4}
      Limit comparison with a p-series
    4. Estimate the sum numerically
  3. 3. The answer
    ∑n=1∞1(2n−1)(2n+1) converges≈0.5\sum_{n=1}^{\infty} \frac{1}{\left(2n-1\right) \left(2n+1\right)} \ \text{converges} \approx 0.5

    limit comparison with Σ 1/n^2 ⇒ converges, S ≈ 0.5 (matches 1/2 to 10+ digits)

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

Transcript

The problem. Here is the problem: Σ 1 / (2 n minus 1) (2 n plus 1), n equals 1 to ∞. Decide whether Σ 1 / (2 n minus 1) (2 n plus 1) (n from 1 to ∞) converges or diverges.

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

The answer. 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). 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.

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

About sequences and series

Evaluate finite sums and arithmetic and geometric series.

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Evaluate the sum of 1/(2n-1)(2n+1) from n=1 to infinity