Sum of Series Calculator
Sum a series without adding term by term. Choose geometric (first term a, ratio r, n terms), arithmetic (first term a, difference d, n terms), or infinite geometric (converges when |r| < 1) and get the sum, the last term, and the formula applied.
Example: with Series type Geometric — each term × r · First term (a) 3 · Common ratio r (geometric) 2 · Common difference d (arithmetic) 5 · Number of terms (n) 8 → Sum of the series: 765.
- Last term384 (term #8)
- The series3 + 6 + 12 + 24 + … + 384
- Formula appliedS = a(rⁿ − 1)/(r − 1) with a = 3, r = 2, n = 8
Computed by the calculator below using its default values. Change any input to see your own numbers.
Geometric: S = a(rⁿ − 1)/(r − 1). Arithmetic: S = n/2 × (2a + (n − 1)d). Infinite geometric: S = a/(1 − r), only when |r| < 1.
Why series formulas beat adding by hand
A series is a sequence with plus signs: 3 + 6 + 12 + 24 + ... The formulas exist because the sums telescope. For a geometric series, multiply the whole sum by r and subtract — every middle term cancels, leaving S = a(rⁿ − 1)/(r − 1). For an arithmetic series, pair the first term with the last, the second with the second-to-last, and so on: each pair adds to the same value, giving Gauss's classroom trick S = n/2 × (first + last).
Infinite geometric series are the surprising case: when |r| < 1 the terms shrink fast enough that infinitely many of them still total a finite number. The formula S = a/(1 − r) is what the finite formula becomes as rⁿ fades to zero — it is why 5 + 2.5 + 1.25 + ... lands exactly on 10, and why 0.999... equals 1 (a = 0.9, r = 0.1 gives 0.9/0.9).
How it’s calculated
Geometric: S = a(rⁿ − 1)/(r − 1) for r ≠ 1; S = a·n when r = 1; nth term = a·rⁿ⁻¹. Arithmetic: S = n/2 × (2a + (n − 1)d); nth term = a + (n − 1)d. Infinite geometric: S = a/(1 − r), valid only for |r| < 1. Sums above 10¹⁵ display in scientific notation.
Computed in double-precision floating point — for huge n and |r| > 1 the result is accurate to about 15 significant digits, not exact to the last digit.
The three formulas at a glance
| Series | Sum formula | Requirement |
|---|---|---|
| Arithmetic, n terms | S = n/2 × (2a + (n − 1)d) | any a, d |
| Geometric, n terms | S = a(rⁿ − 1)/(r − 1) | r ≠ 1 |
| Geometric, r = 1 | S = a × n | every term equals a |
| Infinite geometric | S = a/(1 − r) | |r| < 1, else diverges |
Standard results from any algebra or precalculus text; each follows from the telescoping/pairing arguments above.
Common mistakes
- Using the infinite formula when |r| ≥ 1 — the series 3 + 6 + 12 + ... diverges; a/(1 − r) would report a meaningless negative number.
- Off-by-one on n: the last term uses (n − 1) — the 8th term of 3, 6, 12, ... is 3 × 2⁷ = 384, not 3 × 2⁸.
- Mixing up d and r — arithmetic series add a constant, geometric series multiply by one.
- Plugging r = 1 into the geometric formula and dividing by zero; when every term is equal, the sum is simply a × n.
Frequently asked questions
What is the sum of a geometric series?
S = a(rⁿ − 1)/(r − 1) for n terms with first term a and ratio r ≠ 1. For 3 + 6 + 12 + ... with 8 terms: 3 × (2⁸ − 1)/(2 − 1) = 3 × 255 = 765.
What is the arithmetic series formula?
S = n/2 × (2a + (n − 1)d), or equivalently n/2 × (first + last). For 2 + 5 + 8 + ... with 10 terms: 10/2 × (2 + 29) = 155.
When does an infinite geometric series have a finite sum?
Only when the ratio is strictly between −1 and 1. Then S = a/(1 − r): the series 5 + 2.5 + 1.25 + ... sums to exactly 10. At |r| ≥ 1 the running total grows (or oscillates) without settling.
What is the difference between a sequence and a series?
A sequence is the ordered list of terms (3, 6, 12, 24); a series is their sum (3 + 6 + 12 + 24 = 45). The formulas here answer the series question.
Does the geometric formula work with a negative ratio?
Yes — terms simply alternate sign. With a = 8, r = −1/2: 8 − 4 + 2 − 1 + ... converges (infinitely) to 8/(1 − (−1/2)) = 5.33.