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The ultimate and twice the penultimate

Sopantyadvayamantyam

This sutra helps sum telescoping fraction series of the form 1/(n(n+1)). Each fraction splits into partial fractions that cancel in sequence, leaving only the first and last terms. The sum of the first k terms equals k/(k+1).

How It Works

  1. 1.Recognize each term as 1/(n(n+1)) which equals 1/n − 1/(n+1).
  2. 2.Write out the partial fractions: 1/1 − 1/2 + 1/2 − 1/3 + 1/3 − ...
  3. 3.Most terms cancel (telescope), leaving 1 − 1/(k+1).
  4. 4.Simplify to k/(k+1), where k is the number of terms.
  5. 5.The 'ultimate' (last factor) and 'penultimate' (second-to-last) guide the result.

Examples

1/(1×2) + 1/(2×3) + 1/(3×4)
Step 1 / 4

Sum the series: 1/(1×2) + 1/(2×3) + 1/(3×4).

Number of terms3
Series1/(1×2) + 1/(2×3) + 1/(3×4)

Try It Yourself

1/(1×2) + 1/(2×3) + 1/(3×4) + 1/(4×5) = ?/5

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