Geometric Series Sum Calculator
Enter the first term a, the common ratio r and the number of terms n to get the nth term, the partial sum Sₙ and the sum to infinity (when |r| < 1) — with the working shown.
- nth term aₙ
- 512
- Sum to infinity
- Diverges
Calculator
Worked solution
nth term: aₙ = a·rⁿ⁻¹ = 1·(2)^9 = 512
Partial sum: Sₙ = a·(1 − rⁿ)/(1 − r) = 1·(1 − (2)^10)/(1 − 2) = 1,023
Sum to infinity: with |r| = 2 ≥ 1 the series diverges, so S∞ is undefined.
About this calculator
This calculator works with a geometric series — a sequence where each term is the previous one multiplied by a fixed common ratio r, starting from a first term a. Enter a, r and the number of terms n to instantly see the nth term aₙ, the partial sum of the first n terms Sₙ, and the sum to infinity S∞ when the series converges.
How to read your results
The result card shows the partial sum Sₙ of the first n terms at the top. Below it you will find the nth (last) term aₙ = a·rⁿ⁻¹ and the sum to infinity. The sum to infinity is a finite number only when |r| < 1, because that is the condition for the terms to shrink toward zero fast enough for the total to settle; if |r| ≥ 1 the series diverges and the calculator says so instead of printing a misleading number. The worked solution underneath substitutes your values into each closed form so you can follow every step.
How it's calculated
The nth term is computed with the closed form aₙ = a·rⁿ⁻¹. The partial sum uses Sₙ = a·(1 − rⁿ)/(1 − r) for r ≠ 1, falling back to Sₙ = n·a when r = 1 to avoid dividing by zero. The sum to infinity is S∞ = a/(1 − r) and is reported only when |r| < 1, the standard convergence condition for a geometric series; otherwise the calculator states that the series diverges rather than printing a value. These formulas and the convergence rule are documented by Wolfram MathWorld and Wikipedia.
Worked example
Enter a = 1, r = 2, n = 10 (the series 1 + 2 + 4 + … + 512).
The nth term is a₁₀ = 1·2⁹ = 512 and the partial sum is S₁₀ = 1·(1 − 2¹⁰)/(1 − 2) = 1023. Because |r| = 2 ≥ 1, the series diverges, so there is no sum to infinity. Switch to r = 0.5 and the sum to infinity becomes a/(1 − r) = 1/0.5 = 2.
Frequently asked questions
When does a geometric series have a sum to infinity?
A geometric series converges to a finite sum to infinity only when the common ratio satisfies |r| < 1. In that case S∞ = a/(1 − r). When |r| ≥ 1 — including r = 1 and r = −1 — the terms do not shrink toward zero, the partial sums keep growing or oscillating, and there is no finite total, so the series is said to diverge.
What is the difference between the nth term and the partial sum?
The nth term aₙ = a·rⁿ⁻¹ is the value of a single term — the nth entry in the sequence. The partial sum Sₙ adds up the first n terms together: Sₙ = a·(1 − rⁿ)/(1 − r) for r ≠ 1, or simply n·a when r = 1. So aₙ is one number from the list, while Sₙ is the running total of the whole list up to that point.
Can the first term or common ratio be negative?
Yes. The first term a and the common ratio r can be any finite numbers, including negatives and fractions. A negative ratio makes the terms alternate in sign (for example 1, −2, 4, −8…). The closed-form formulas handle every case; the only requirement for a finite sum to infinity is still |r| < 1.
Why does the formula change when r = 1?
The general partial-sum formula Sₙ = a·(1 − rⁿ)/(1 − r) divides by (1 − r), which is zero when r = 1. In that case every term equals a, so the sum is simply n copies of a: Sₙ = n·a. The calculator detects r = 1 and uses this special case automatically.
Sources
Reviewed by the YouCalc Team · Last reviewed
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