Skip to main content

Sequences and Series

A sequence is simply an ordered list of numbers that follows a rule, and a series is what you get when you add those numbers up. That sounds modest, but this one idea underlies compound interest, computer algorithms, the way a bouncing ball eventually comes to rest, and even the number π\pi. Once you can spot the pattern in a list and turn it into a formula, you stop adding numbers one painful term at a time and start reasoning about infinitely many of them at once.

This page will teach you to recognize the two most important families — arithmetic and geometric — to write sums compactly with sigma notation, to compute partial sums with formulas instead of a calculator, and to understand the beautiful, counterintuitive fact that an infinite list of positive numbers can add up to a finite total.

Learning Objectives

  • Distinguish a sequence from a series, and a term from a sum.
  • Identify arithmetic sequences (constant difference) and geometric sequences (constant ratio) and write their nnth-term formulas.
  • Read and write sums using sigma (Σ\Sigma) notation.
  • Compute partial sums of arithmetic and geometric series with closed-form formulas.
  • Determine when an infinite geometric series converges, and find its sum.
  • Apply these tools to real problems in finance, physics, and computing.

Quick Answer

A sequence lists terms; a series adds them. An arithmetic sequence adds a fixed number dd each step, so its nnth term is an=a1+(n1)da_n = a_1 + (n-1)d. A geometric sequence multiplies by a fixed ratio rr each step, so an=a1rn1a_n = a_1 r^{\,n-1}. The sum of the first nn terms is Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n) for arithmetic series and Sn=a11rn1rS_n = a_1\frac{1 - r^n}{1 - r} for geometric series. An infinite geometric series converges only when r<1|r| < 1, and then its sum is S=a11rS = \frac{a_1}{1 - r}. Sigma notation, written k=1nak\sum_{k=1}^{n} a_k, is just shorthand for "add these terms up."

Where It Came From

People needed to add long, patterned lists long before they had formulas, and the history of this topic is really the history of clever shortcuts.

The most famous story belongs to Carl Friedrich Gauss. As a schoolboy around 1785, he was reportedly told (as busywork) to add every whole number from 1 to 100. Within moments he wrote the answer: 5050. He had noticed that if you pair the first and last terms (1+100=101 1 + 100 = 101), the second and second-to-last (2+99=101 2 + 99 = 101), and so on, you get 50 pairs each summing to 101, giving 50×101=5050 50 \times 101 = 5050. That single insight — pairing terms from the ends inward — is the arithmetic series formula, and it works for any arithmetic list, not just 1 to 100. The need it answered was ancient and practical: surveyors, accountants, and astronomers constantly faced sums of evenly spaced quantities.

Infinite series carry an even older and more philosophical origin: the paradoxes of Zeno of Elea (around 450 BCE). Zeno argued that motion is impossible. To walk across a room you must first cover half the distance, then half of what remains, then half of that, forever — infinitely many steps, so (he claimed) you can never arrive. The resolution, made rigorous only two thousand years later, is that the infinite sum 12+14+18+\tfrac12 + \tfrac14 + \tfrac18 + \cdots adds up to exactly 1. Infinitely many pieces can have a finite total. Mathematicians from Archimedes (who summed geometric series to find areas) through the calculus of Newton, Leibniz, and finally the rigorous limit definitions of Cauchy in the 1820s turned Zeno's puzzle into the modern theory of convergence.

Arithmetic Sequences and Series

An arithmetic sequence changes by the same amount, the common difference dd, at every step: 3,7,11,15,19, 3, 7, 11, 15, 19, \dots has d=4d = 4. Because each term is the first term plus some number of steps of size dd:

an=a1+(n1)da_n = a_1 + (n-1)d

To sum the first nn terms, use Gauss's pairing trick. Writing the sum forwards and backwards and adding gives 2Sn=n(a1+an) 2S_n = n(a_1 + a_n), so:

Sn=n2(a1+an)=n2(2a1+(n1)d)S_n = \frac{n}{2}\,(a_1 + a_n) = \frac{n}{2}\,\big(2a_1 + (n-1)d\big)

Worked example. Find the 20th term and the sum of the first 20 terms of 3,7,11,15, 3, 7, 11, 15, \dots

Here a1=3a_1 = 3 and d=4d = 4. The 20th term is

a20=3+(201)×4=3+76=79.a_{20} = 3 + (20-1)\times 4 = 3 + 76 = 79.

The sum is

S20=202(3+79)=10×82=820.S_{20} = \frac{20}{2}\,(3 + 79) = 10 \times 82 = 820.

Check Gauss's own case with this formula: a1=1 a_1 = 1, a100=100a_{100} = 100, so S100=1002(1+100)=50×101=5050 S_{100} = \frac{100}{2}(1 + 100) = 50 \times 101 = 5050. It matches.

Geometric Sequences and Series

A geometric sequence multiplies by the same common ratio rr at each step: 2,6,18,54, 2, 6, 18, 54, \dots has r=3r = 3. The nnth term is

an=a1rn1.a_n = a_1\, r^{\,n-1}.

For the sum, there is an elegant trick. Write Sn=a1+a1r++a1rn1 S_n = a_1 + a_1 r + \cdots + a_1 r^{\,n-1}, multiply the whole thing by rr, and subtract. Almost every term cancels, leaving:

Sn=a11rn1r(r1).S_n = a_1\,\frac{1 - r^{\,n}}{1 - r}\qquad (r \neq 1).

Worked example. Sum the first 6 terms of 2,6,18,54, 2, 6, 18, 54, \dots

Here a1=2a_1 = 2, r=3r = 3, n=6n = 6:

S6=2×13613=2×17292=2×7282=2×364=728.S_6 = 2 \times \frac{1 - 3^{6}}{1 - 3} = 2 \times \frac{1 - 729}{-2} = 2 \times \frac{-728}{-2} = 2 \times 364 = 728.

Verify by hand: 2+6+18+54+162+486=728 2 + 6 + 18 + 54 + 162 + 486 = 728. Exactly right.

Notice how geometric growth explodes: six terms already exceed 700, while an arithmetic sequence with the same start would still be small. This gap is why geometric reasoning governs compound interest, population growth, and viral spread.

Sigma Notation

Writing out long sums is tedious and error-prone, so mathematicians use the capital Greek letter sigma, Σ\Sigma, as a compact instruction to add:

k=1nak=a1+a2++an.\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n.

The letter kk is the index, it starts at the number below the sigma and increases by 1 up to the number on top. For example:

k=15(2k+1)=3+5+7+9+11=35.\sum_{k=1}^{5} (2k + 1) = 3 + 5 + 7 + 9 + 11 = 35.

Here each term is generated by plugging k=1,2,3,4,5k = 1, 2, 3, 4, 5 into the rule 2k+1 2k + 1. This is an arithmetic series with a1=3a_1 = 3, a5=11a_5 = 11, so the formula confirms 52(3+11)=52×14=35 \frac{5}{2}(3 + 11) = \frac{5}{2}\times 14 = 35.

Two facts make sigma notation powerful: you can pull constants out, cak=cak\sum c\,a_k = c\sum a_k, and split sums, (ak+bk)=ak+bk\sum (a_k + b_k) = \sum a_k + \sum b_k. These let you rewrite messy sums into standard forms you already know how to compute.

Infinite Geometric Series and Convergence

What happens if a geometric series never stops? Look again at rnr^{\,n} in the sum formula. If r<1|r| < 1, then rnr^{\,n} shrinks toward 0 as nn grows without bound. Dropping that vanishing term leaves a finite total:

S=a11rvalid only when r<1.S = \frac{a_1}{1 - r}\qquad \text{valid only when } |r| < 1.

If r1|r| \geq 1 the terms do not shrink, the partial sums grow forever, and the series diverges (no finite sum).

Worked example — Zeno resolved. The distances 12,14,18,\tfrac12, \tfrac14, \tfrac18, \dots form a geometric series with a1=12a_1 = \tfrac12 and r=12r = \tfrac12. Since r<1|r| < 1:

S=12112=1212=1.S = \frac{\tfrac12}{1 - \tfrac12} = \frac{\tfrac12}{\tfrac12} = 1.

The infinitely many steps add to exactly 1 — you do cross the room. Zeno's paradox dissolves the moment you accept that an unending sum can converge.

Worked example — repeating decimals. The decimal 0.4444 0.4444\dots equals 410+4100+41000+\tfrac{4}{10} + \tfrac{4}{100} + \tfrac{4}{1000} + \cdots, a geometric series with a1=410a_1 = \tfrac{4}{10} and r=110r = \tfrac{1}{10}:

S=4/1011/10=4/109/10=49.S = \frac{4/10}{1 - 1/10} = \frac{4/10}{9/10} = \frac{4}{9}.

So 0.4=49 0.\overline{4} = \tfrac{4}{9} — a satisfying proof that repeating decimals are always fractions.

Real-World Applications

  • Compound interest and savings. Depositing a fixed amount each year that then earns interest produces a geometric series; the "future value of an annuity" formula is exactly Sn=a11rn1rS_n = a_1\frac{1 - r^n}{1 - r} in disguise.
  • Loan and mortgage payments. Amortization schedules sum geometric series to figure out the level monthly payment that pays off a balance.
  • Physics of damped motion. A bouncing ball that rebounds to a fixed fraction of its previous height travels a total distance found by an infinite geometric series.
  • Computer science. Analyzing algorithms (like the cost of doubling an array, or a binary search's halving) relies on geometric-series sums; so does estimating the total work in recursive processes.
  • Medicine. Drug dosing at regular intervals reaches a steady-state concentration modeled by a convergent geometric series.

Common Mistakes

  1. Confusing a sequence with a series. A sequence is the list (2,4,6, 2, 4, 6, \dots); a series is the sum (2+4+6+ 2 + 4 + 6 + \cdots). Why it matters: the questions "what is the 10th term?" and "what is the sum of 10 terms?" have completely different answers. Correction: decide first whether you are locating a term (ana_n) or totaling terms (SnS_n).

  2. Applying the infinite-sum formula when r1|r| \geq 1. Students plug into a11r\frac{a_1}{1-r} even for, say, r=2r = 2, getting a negative "sum" for a series of growing positive terms. Why it's wrong: the series diverges, so no finite sum exists; the formula is meaningless there. Correction: always check r<1|r| < 1 before using S=a11rS = \frac{a_1}{1-r}.

  3. Off-by-one errors in an=a1+(n1)da_n = a_1 + (n-1)d. People multiply dd by nn instead of n1n-1. Why it's wrong: the first term needs zero steps of dd, the nnth term needs n1n-1 steps. Correction: test on term 1 — the formula must return a1a_1, which it does only with (n1)(n-1).

Comparison and Connections

Arithmetic and geometric progressions are the two building blocks; the key contrast is addition versus multiplication.

FeatureArithmeticGeometric
Step ruleadd ddmultiply by rr
nnth terma1+(n1)da_1 + (n-1)da1rn1a_1 r^{\,n-1}
Partial sum SnS_nn2(a1+an)\frac{n}{2}(a_1 + a_n)a11rn1ra_1\frac{1-r^n}{1-r}
Infinite sumnever converges (unless all terms 0)converges to a11r\frac{a_1}{1-r} if r<1\vert r\vert < 1
Growth shapestraight lineexponential curve

Arithmetic sequences are the discrete cousins of linear functions; geometric sequences are the discrete cousins of exponential functions. Infinite series are the gateway to calculus, where limits, Taylor series, and integrals all rest on summing infinitely many pieces.

Practice Questions

Recall

State the nnth-term formula for an arithmetic sequence and for a geometric sequence. Answer: Arithmetic: an=a1+(n1)da_n = a_1 + (n-1)d. Geometric: an=a1rn1a_n = a_1 r^{\,n-1}.

Understanding

Explain why the infinite series 1+2+4+8+ 1 + 2 + 4 + 8 + \cdots has no finite sum, while 1+12+14+ 1 + \tfrac12 + \tfrac14 + \cdots does. Guidance: The first has r=2r = 2 so r1|r| \geq 1; terms grow and partial sums increase without bound (diverges). The second has r=12r = \tfrac12 so r<1|r| < 1; terms shrink to 0 and the sum converges to 111/2=2\frac{1}{1 - 1/2} = 2.

Application

A theater has 18 seats in the front row and 3 more in each successive row, for 25 rows. How many seats total? Answer: Arithmetic with a1=18a_1 = 18, d=3d = 3, n=25n = 25. Then a25=18+24×3=90a_{25} = 18 + 24\times 3 = 90, and S25=252(18+90)=252×108=1350S_{25} = \frac{25}{2}(18 + 90) = \frac{25}{2}\times 108 = 1350 seats.

Analysis

A ball is dropped from 10 m and each bounce reaches 60% of the previous height. Find the total vertical distance it travels before coming to rest. Guidance: It falls 10 m, then each bounce goes up and back down. Up-distances form a geometric series 6+3.6+ 6 + 3.6 + \cdots with a1=6a_1 = 6, r=0.6r = 0.6, summing to 610.6=15\frac{6}{1 - 0.6} = 15; the down-distances after the first drop equal the same 15. Total =10+15+15=40 = 10 + 15 + 15 = 40 m.

FAQ

Is zero a valid common ratio? No — if r=0r = 0 every term after the first is 0, which is trivial, and the formulas assume r0r \neq 0 (and r1r \neq 1 for the partial-sum formula, since that would divide by zero).

Can a common difference or ratio be negative? Yes. A negative dd gives a decreasing arithmetic sequence (10,7,4, 10, 7, 4, \dots). A negative rr gives an alternating geometric sequence (3,6,12, 3, -6, 12, \dots); it still converges if r<1|r| < 1, e.g. r=12r = -\tfrac12.

What's the difference between SnS_n and SS? SnS_n is a partial sum — the total of the first nn terms (always finite). SS (no subscript) is the infinite sum, the limit of SnS_n as nn \to \infty, which only exists for convergent series.

Why does the index letter not matter? In k=1nak\sum_{k=1}^{n} a_k the kk is a "dummy variable" — using ii or jj instead gives the identical sum. It's just a counter that disappears once the addition is done.

How is this connected to calculus? Calculus is built on infinite sums. Integrals are limits of sums of infinitely many thin slices, and functions like exe^x and sinx\sin x are defined by infinite series. Mastering convergence here is direct preparation.

Quick Revision

  • Sequence = ordered list; series = sum of the list.
  • Arithmetic: constant difference dd; an=a1+(n1)da_n = a_1 + (n-1)d; Sn=n2(a1+an)S_n = \frac{n}{2}(a_1 + a_n).
  • Geometric: constant ratio rr; an=a1rn1a_n = a_1 r^{\,n-1}; Sn=a11rn1rS_n = a_1\frac{1 - r^n}{1 - r}.
  • Infinite geometric sum: S=a11rS = \frac{a_1}{1 - r}, only if r<1|r| < 1; otherwise it diverges.
  • Sigma notation: k=1nak\sum_{k=1}^{n} a_k means add terms from k=1k=1 to k=nk=n.
  • Gauss's trick: pair ends inward. Zeno's series 12+14+=1\tfrac12 + \tfrac14 + \cdots = 1.

Prerequisites

  • Exponential and logarithmic functions (geometric growth in continuous form)
  • Linear functions (the continuous analog of arithmetic sequences)

Next Topics