Sum of a Geometric Series

You need the sum of a geometric series, the formula that adds terms where each one is the previous term multiplied by a constant ratio. Here it is: for a finite geometric series with first term a, common ratio r, and n terms, the sum S_n = a(1, r^n) / (1, r) when r ≠ 1. For an infinite geometric series where |r| < 1, the sum S = a / (1, r). This walks you through both formulas, how to use them from sigma notation, and what breaks when r falls outside the convergence range.

The finite geometric series formula comes from a simple trick. Write the sum S_n = a + ar + ar² + … + ar^(n, 1). Multiply every term by r: r S_n = ar + ar² + … + ar^n. Subtract r S_n from S_n: S_n, r S_n = a, ar^n, so S_n(1, r) = a(1, r^n). Divide by (1, r) (r ≠ 1) and you get S_n = a(1, r^n) / (1, r). OpenStax Precalculus 2e, section 11.4 states this formula exactly.

You can also write the formula as S_n = a(r^n, 1) / (r, 1). Both are the same; use the one that keeps the denominator positive. If r = 1, every term equals a and the sum is n·a, no formula needed.

A geometric series in sigma notation looks like Σ_{k=1}^{n} a·r^(k, 1). The index variable (k, i, j, or n) is a dummy variable: rename it without changing the sum. The lower bound tells you the first term's exponent. In that form, the first term a appears when k = 1, and r is the base of the exponential. The number of terms n = (upper bound), (lower bound) + 1.

Watch the Starting Index

If the sum is Σ_{k=0}^{n, 1} a·r^k, the exponent matches the index, and n terms still appear. The general term is a_k = a·r^(k, 1) when the series starts at k = 1, or a·r^k when it starts at k = 0. An index shift changes the bounds and the summand together. For example, Σ_{k=0}^{2} 3·2^k = 3 + 6 + 12 = 21, which is the same as Σ_{k=1}^{3} 3·2^(k, 1).

Example 1: Σ_{k=1}^{5} 3·2^(k, 1). Here a = 3, r = 2, n = 5. Using S_n = 3(1-2^5) / (1-2) = 3(1-32) / (, 1) = 3(, 31)/(, 1) = 93. OpenStax gives 93 as the answer.

Example 2: Σ_{k=1}^{4} 2·3^(k, 1). a = 2, r = 3, n = 4. S_n = 2(1-3^4) / (1-3) = 2(1-81) / (, 2) = 2(, 80)/(, 2) = 80.

Example 3: Σ_{k=1}^{3} 5·4^(k, 1). a = 5, r = 4, n = 3. S_n = 5(1-4^3) / (1-4) = 5(1-64) / (, 3) = 5(, 63)/(, 3) = 105.

An infinite geometric series converges to a finite sum only when the absolute value of the common ratio is less than 1. The sum is S = a / (1, r). OpenStax Precalculus 2e, section 11.4 gives this formula and the convergence condition.

Example: Σ_{k=1}^{∞} 3·(1/2)^(k, 1). a = 3, r = 1/2. Since |1/2| < 1, the sum is 3 / (1-1/2) = 3 / (1/2) = 6.

Example: Σ_{k=1}^{∞} 4·(1/3)^(k, 1). a = 4, r = 1/3, |r| < 1, sum = 4 / (1-1/3) = 4 / (2/3) = 6.

If |r| ≥ 1, the infinite geometric series does not converge. The partial sums grow without bound (or oscillate) and there is no finite sum.

Geometric series appear in savings and loans. Deposit $100 at the end of each year into an account earning 5% annual interest. The total after n years is 100 + 100(1.05) + 100(1.05)^2 + ... + 100(1.05)^(n, 1), a finite geometric series with a = 100, r = 1.05, and n terms.

Repeating decimals are infinite geometric series. The decimal 0.3333... = 0.3 + 0.03 + 0.003 + ... = 3/10 + 3/100 + 3/1000 + ... with a = 3/10, r = 1/10. The sum is (3/10) / (1-1/10) = (3/10) / (9/10) = 1/3.