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Sum of Series Calculator – Arithmetic & Geometric

Quick answer

Series sums with formulas: Σk for n=100 is 5050, Σk² = n(n+1)(2n+1)/6 gives 14 at n=3, and geometric 1+½+¼+… = 2.

  • Steps shown.

Figures are estimates from this calculator’s standard formula — adjust the inputs in the tool below for your exact number.

How the Sum of Series Calculator Works

Σk = n(n+1)/2 | Σk² = n(n+1)(2n+1)/6 | Sₙ(geo) = a₁(rⁿ−1)/(r−1) | S∞ = a₁/(1−r)

Series come in families, each with a closed form. The arithmetic series 1+2+…+n sums to n(n+1)/2, so the first 100 integers total 100×101/2 = 5,050. Squares follow Σk² = n(n+1)(2n+1)/6: for n = 3 that is 3×4×7/6 = 14 (indeed 1+4+9 = 14). Geometric series a₁ + a₁r + a₁r² + … sum to a₁(rⁿ−1)/(r−1); for a₁ = 3, r = 2, n = 4 the total is 3×15 = 45. When |r| < 1 the infinite sum converges: 1 + ½ + ¼ + … = 1/(1−½) = 2. The calculator identifies which family your series belongs to, applies the matching formula with full substitution, and cross-checks by direct addition for small n. Partial sums are tabulated so the growth pattern (quadratic for Σk², exponential for geometric r > 1) is visible.

  1. 1

    Enter the series type and terms

    Arithmetic, geometric, or a raw list of terms

  2. 2

    Apply the closed form

    The matching formula substitution is displayed

  3. 3

    Cross-check by addition

    Small n values are summed directly as verification

  4. 4

    Test convergence

    Geometric series report S∞ when |r| < 1

Use Cases

Gauss-Style Problems

Sum 1..100 = 5,050 and 1..365 = 66,795 instantly

Compound Totals

3+6+12+24 = 45 via the geometric form

Proof Practice

Verify Σk² = 14 for n = 3 before trusting the general case

Tips

  • 1

    Σk = n(n+1)/2 and Σk² = n(n+1)(2n+1)/6 cover integer series

  • 2

    Geometric sums need r ≠ 1; r = 1 means n equal terms

  • 3

    Infinite geometric sums exist only for |r| < 1

Common Mistakes

  • Using n(n+1)/2 for squares — Σk² has its own (2n+1)/6 factor

  • Dividing by (1−r) when r > 1 for a finite sum; that form needs (r−1)

  • Summing a divergent series like 1+2+4+… to a finite "answer"

  • Off-by-one: summing to n−1 after miscounting terms

FAQs

What is 1+2+…+100?

n(n+1)/2 = 100×101/2 = 5,050.

What is 1²+2²+3²?

3×4×7/6 = 14 — formula Σk² = n(n+1)(2n+1)/6.

What is 1+½+¼+…?

2 — an infinite geometric series with r = ½: a₁/(1−r) = 2.

What is 3+6+12+24?

Geometric, a₁ = 3, r = 2, n = 4: 3(2⁴−1)/(2−1) = 45.

Does every infinite series converge?

No — geometric series need |r| < 1; 1+2+4+8+… grows without bound.