Inputs
Computation History
| Time | Mode | a | r | S∞ | Converges | ε | n for ε | Note |
|---|
Formula Used
Convergence: An infinite geometric series S = a + ar + ar² + … converges iff |r| < 1.
Sum: S∞ = a / (1 − r), valid only when |r| < 1.
Remainder after n terms: |Rₙ| = |a·rⁿ| / (1 − |r|).
n for given tolerance ε: choose smallest n with |a·rⁿ|/(1−|r|) ≤ ε.
Inverse relations: a = S∞·(1−r); r = 1 − a/S∞ (must satisfy |r| < 1).
How to Use
- Select a mode: compute S∞, recover a, or recover r.
- Enter known values. Leave the unknown field blank.
- Optional: set a tolerance ε to estimate required terms.
- Choose decimals for output rounding and click Compute.
- Review convergence status. If |r| ≥ 1, S∞ does not exist.
- Export the result or history as CSV or PDF.
Example Data
| a | r | Converges? | S∞ |
|---|---|---|---|
| 3 | 0.5 | Yes | 6 |
| 5 | -0.2 | Yes | 4.1666667 |
| 1 | 0.99 | Yes | 100 |
| 2 | 1.1 | No | — |
Values formatted to ~7 decimals for illustration.
FAQs
It exists only when the common ratio satisfies |r| < 1. Otherwise the partial sums do not approach a finite limit.
Negative r is allowed; the series alternates and still converges if |r| < 1. This tool assumes real inputs; complex support isn’t included.
S∞ = a/(1−r). When r is close to 1, the denominator is small, so the sum becomes very large in magnitude even though it still converges if |r| < 1.
Parameters are inconsistent with convergence. Recheck signs, magnitudes, and rounding; otherwise an infinite sum isn’t defined for those values.
We use |Rₙ| = |a·rⁿ|/(1−|r|) ≤ ε, solve for n using logarithms, and round up to the smallest integer meeting the inequality.
Convergence depends on exact inequalities. However, rounding can mask near‑boundary cases; prefer more decimals when r is very close to ±1.