Geometric Series Calculator

Sum finite and infinite geometric series with step-by-step working
sum of the first 8 terms of 3, 6, 12, 24, ...
5 + 5/3 + 5/9 + ... to infinity
sum from k=0 to 5 of 2 * 3^k
sum from n=1 to infinity of 4 * (-2/5)^(n-1)

The Geometric Series Formulas

A series is geometric when each term is the previous one times a fixed common ratio rr:

a+ar+ar2+โ‹ฏ=โˆ‘n=1Narnโˆ’1a + ar + ar^2 + \cdots = \sum_{n=1}^{N} ar^{n-1}

Finite sum of the first nn terms, valid for any rโ‰ 1r \neq 1:

Sn=aโ€‰1โˆ’rn1โˆ’r=aโ€‰rnโˆ’1rโˆ’1S_n = a\,\frac{1 - r^n}{1 - r} = a\,\frac{r^n - 1}{r - 1}

Infinite sum, valid only when โˆฃrโˆฃ<1|r| < 1:

Sโˆž=a1โˆ’rS_\infty = \frac{a}{1 - r}

The restriction is not decoration. If โˆฃrโˆฃโ‰ฅ1|r| \ge 1 the terms never shrink to zero, rnr^n does not vanish, and the series diverges โ€” there is no sum to find. When r=1r = 1 the series is just aa added nn times, so Sn=naS_n = na.

Here aa is the first term of the series you are actually summing, not necessarily the first term of the underlying sequence.

How to Sum a Geometric Series

Step by step

  1. Find aa: the first term you are told to include.
  2. Find rr: divide any term by the one before it, r=an+1anr = \dfrac{a_{n+1}}{a_n}. Test two different pairs โ€” if they disagree, the series is not geometric and none of these formulas apply.
  3. Count nn: the number of terms, not the last index. Summing from k=0k = 0 to k=5k = 5 is six terms.
  4. Choose the formula: finite SnS_n for a fixed count, SโˆžS_\infty only after checking โˆฃrโˆฃ<1|r| < 1.
  5. Substitute and simplify, keeping exact fractions where possible.

Reading sigma notation

โˆ‘k=0nโˆ’1arkandโˆ‘k=1narkโˆ’1\sum_{k=0}^{n-1} ar^{k} \quad\text{and}\quad \sum_{k=1}^{n} ar^{k-1}

are the same series written two ways. Substitute the lower limit into the general term to read off aa directly โ€” that removes the off-by-one error entirely.

Working backwards

Given two terms apa_p and aqa_q, the ratio follows from rqโˆ’p=aq/apr^{q-p} = a_q / a_p, and then a=ap/rโ€‰pโˆ’1a = a_p / r^{\,p-1}.

Common Mistakes to Avoid

  • Using SโˆžS_\infty without checking โˆฃrโˆฃ<1|r| < 1: with r=2r = 2 the formula returns โˆ’a-a, a confidently wrong finite number for a divergent series.
  • Off-by-one in nn: โˆ‘k=05\sum_{k=0}^{5} has six terms. Count them, or convert the index so the sum starts at 11.
  • Confusing sequence and series: the nn-th term is arnโˆ’1ar^{n-1}; the sum of the first nn terms is SnS_n. "Find the sum of the sequence" means the series.
  • Dropping the sign of a negative ratio: for r=โˆ’25r = -\frac{2}{5}, 1โˆ’r=1+251 - r = 1 + \frac{2}{5}. Alternating series still converge whenever โˆฃrโˆฃ<1|r| < 1.
  • Mixing in an arithmetic pattern: constant differences mean arithmetic, constant ratios mean geometric. Check by division, not subtraction.

Examples

Step 1: First term a=3a = 3; ratio r=63=2r = \frac{6}{3} = 2 (confirmed by 126=2\frac{12}{6} = 2)
Step 2: Since rโ‰ 1r \neq 1, use Sn=arnโˆ’1rโˆ’1S_n = a\dfrac{r^n - 1}{r - 1} with n=8n = 8
Step 3: r8=28=256r^8 = 2^8 = 256
Step 4: S8=3โ‹…256โˆ’12โˆ’1=3โ‹…255S_8 = 3 \cdot \dfrac{256 - 1}{2 - 1} = 3 \cdot 255
Step 5: S8=765S_8 = 765
Answer: S8=765S_8 = 765

Step 1: a=5a = 5 and r=5/35=13r = \dfrac{5/3}{5} = \dfrac{1}{3}
Step 2: Convergence check: โˆฃrโˆฃ=13<1|r| = \frac{1}{3} < 1, so the infinite sum exists
Step 3: Apply Sโˆž=a1โˆ’r=51โˆ’13S_\infty = \dfrac{a}{1-r} = \dfrac{5}{1 - \frac{1}{3}}
Step 4: 1โˆ’13=231 - \frac{1}{3} = \frac{2}{3}, so Sโˆž=5โ‹…32S_\infty = 5 \cdot \dfrac{3}{2}
Answer: Sโˆž=152=7.5S_\infty = \dfrac{15}{2} = 7.5

Step 1: Substitute n=1n = 1 into the general term: a=4(โˆ’2/5)0=4a = 4(-2/5)^0 = 4
Step 2: The ratio is r=โˆ’25r = -\dfrac{2}{5}, so โˆฃrโˆฃ=0.4<1|r| = 0.4 < 1 and the series converges
Step 3: 1โˆ’r=1โˆ’(โˆ’25)=751 - r = 1 - \left(-\dfrac{2}{5}\right) = \dfrac{7}{5}
Step 4: Sโˆž=47/5=4โ‹…57S_\infty = \dfrac{4}{7/5} = 4 \cdot \dfrac{5}{7}
Answer: Sโˆž=207โ‰ˆ2.857S_\infty = \dfrac{20}{7} \approx 2.857

Frequently Asked Questions

Only when the common ratio satisfies |r| < 1. Then the powers r^n shrink to zero and the partial sums settle at a/(1 - r). If |r| is 1 or larger the terms do not tend to zero, so the series diverges and no finite sum exists.

The first term is a. For r, divide any term by the one before it, then confirm with a second pair โ€” a genuine geometric series gives the same ratio every time. If the ratios differ, the series is not geometric.

The sequence is the list of terms a, ar, ar^2, and so on; the series is what you get by adding them. Sum formulas like S_n = a(1 - r^n)/(1 - r) apply to the series, while aยทr^(n-1) gives the n-th term of the sequence.

No โ€” despite the shared word, this page is about geometric series, meaning sums with a constant ratio. For areas, volumes and shapes, use the geometry solvers instead.

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