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+тЛп=тИСn=1NarnтИТ1a + ar + ar^2 + \cdots = \sum_{n=1}^{N} ar^{n-1}

Finite sum of the first nn terms, valid for any rтЙа1r \neq 1:

Sn=aтАЙ1тИТrn1тИТr=aтАЙrnтИТ1rтИТ1S_n = a\,\frac{1 - r^n}{1 - r} = a\,\frac{r^n - 1}{r - 1}

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

SтИЮ=a1тИТrS_\infty = \frac{a}{1 - r}

The restriction is not decoration. If тИгrтИгтЙе1|r| \ge 1 the terms never shrink to zero, rnr^n does not vanish, and the series diverges тАФ 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 тАФ 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, SтИЮS_\infty only after checking тИгrтИг<1|r| < 1.
  5. Substitute and simplify, keeping exact fractions where possible.

Reading sigma notation

тИСk=0nтИТ1arkandтИСk=1narkтИТ1\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 тАФ that removes the off-by-one error entirely.

Working backwards

Given two terms apa_p and aqa_q, the ratio follows from rqтИТp=aq/apr^{q-p} = a_q / a_p, and then a=ap/rтАЙpтИТ1a = a_p / r^{\,p-1}.

Common Mistakes to Avoid

  • Using SтИЮS_\infty without checking тИгrтИг<1|r| < 1: with r=2r = 2 the formula returns тИТa-a, a confidently wrong finite number for a divergent series.
  • Off-by-one in nn: тИСk=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 arnтИТ1ar^{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=тИТ25r = -\frac{2}{5}, 1тИТr=1+251 - r = 1 + \frac{2}{5}. Alternating series still converge whenever тИгrтИг<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 rтЙа1r \neq 1, use Sn=arnтИТ1rтИТ1S_n = a\dfrac{r^n - 1}{r - 1} with n=8n = 8
Step 3: r8=28=256r^8 = 2^8 = 256
Step 4: S8=3тЛЕ256тИТ12тИТ1=3тЛЕ255S_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: тИгrтИг=13<1|r| = \frac{1}{3} < 1, so the infinite sum exists
Step 3: Apply SтИЮ=a1тИТr=51тИТ13S_\infty = \dfrac{a}{1-r} = \dfrac{5}{1 - \frac{1}{3}}
Step 4: 1тИТ13=231 - \frac{1}{3} = \frac{2}{3}, so SтИЮ=5тЛЕ32S_\infty = 5 \cdot \dfrac{3}{2}
Answer: SтИЮ=152=7.5S_\infty = \dfrac{15}{2} = 7.5

Step 1: Substitute n=1n = 1 into the general term: a=4(тИТ2/5)0=4a = 4(-2/5)^0 = 4
Step 2: The ratio is r=тИТ25r = -\dfrac{2}{5}, so тИгrтИг=0.4<1|r| = 0.4 < 1 and the series converges
Step 3: 1тИТr=1тИТ(тИТ25)=751 - r = 1 - \left(-\dfrac{2}{5}\right) = \dfrac{7}{5}
Step 4: SтИЮ=47/5=4тЛЕ57S_\infty = \dfrac{4}{7/5} = 4 \cdot \dfrac{5}{7}
Answer: SтИЮ=207тЙИ2.857S_\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 тАФ 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 a┬╖r^(n-1) gives the n-th term of the sequence.

No тАФ 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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