Geometric Series

With sₙ = a₁·(qⁿ − 1)/(q − 1) for q ≠ 1; equivalent is a₁·(1 − qⁿ)/(1 − q). You need only three inputs: first term a₁, ratio q (each term divided by the previous one) and count n. Example: 3 + 6 + 12 + ... with 8 terms: a₁ = 3, q = 2, so s₈ = 3·(2⁸ − 1)/(2 − 1) = 3·255 = 765. Watch n: the exponent holds the number of terms, not the last exponent (the eighth term is 3·2⁷). For q = 1 the formula fails (division by 0); then all terms are equal and simply sₙ = n·a₁.

Exactly when |q| < 1, i.e. the ratio lies strictly between −1 and 1. Then the terms shrink fast enough that the partial sums approach s∞ = a₁/(1 − q); in the finite formula the term qⁿ dies out as n → ∞. Example: 1 + 1/2 + 1/4 + ... has q = 1/2 and s∞ = 1/(1 − 0.5) = 2. For |q| ≥ 1 the series diverges: at q = 2 the sums explode, at q = 1 they grow linearly, at q = −1 they jump forever between a₁ and 0. The most common mistake is thoughtlessly applying s∞ = a₁/(1 − q) to q = 1.05 (an interest factor!); savings plan sums are finite series.

Because 0.999... is nothing other than an infinite geometric series: 0.9 + 0.09 + 0.009 + ... with a₁ = 0.9 and q = 0.1. Since |q| < 1 the limit exists, and it is s∞ = 0.9/(1 − 0.1) = 0.9/0.9 = 1. Exactly, not approximately: 0.999... denotes the LIMIT of the partial sums, not any single partial sum, and this limit is the number 1. Every finite partial sum (0.9; 0.99; 0.999; ...) stays below 1, but the gap 10⁻ⁿ becomes arbitrarily small and is 0 in the limit. Two numbers at distance 0 are the same number. The same argument shows 0.333... = 1/3, which times 3 confirms 0.999... = 1 again.

The sequence is the list of individual terms: aₙ = a₁·qⁿ⁻¹, e.g. 3, 6, 12, 24, ... The series is the sum of these terms: sₙ = a₁ + a₁q + ... + a₁qⁿ⁻¹. Each question has its matching formula: "How large is the 8th term?" requires a₈ = 3·2⁷ = 384; "How large is the sum of the first 8 terms?" requires s₈ = 765. Mix-ups are caught immediately by checking magnitudes: the sum must exceed the largest term. Their limit behaviour also differs: for |q| < 1 the SEQUENCE tends to 0, while the SERIES tends to the positive value a₁/(1 − q).

If you deposit a fixed instalment R at the end of every year, which then earns interest with factor q = 1 + p/100, the final capital after n years is a geometric sum: the first instalment earns interest (n−1) times, the last one not at all, together K = R·(qⁿ − 1)/(q − 1). Example: 1000 € per year, 3 % interest, 10 years: K = 1000·(1.03¹⁰ − 1)/0.03 = 1000·0.3439/0.03 ≈ 11,464 €, i.e. 1464 € of interest on 10,000 € of deposits. The same structure sits inside loans (annuities), pension payouts and drug levels under regular dosing. Careful: here q > 1, the series grows; the s∞ formula is off limits.