Arithmetic sequences, geometric series, infinite series convergence, sigma notation, telescoping series, recursive sequences, and real-world applications — complete Stewart Ch. 12 coverage.
An ordered list of numbers following a pattern. Each number is a term.
2, 5, 8, 11, 14, ...
Terms: a_1=2, a_2=5, a_3=8 ...
The sum of the terms of a sequence. Partial sums stop; infinite series go forever.
2 + 5 + 8 + 11 + 14 + ...
S_5 = 40 (partial), S_infinity diverges
| Type | Pattern | Example | Graph |
|---|---|---|---|
| Arithmetic | Add constant d each step | 3, 7, 11, 15 (d=4) | Linear |
| Geometric | Multiply by constant r each step | 2, 6, 18, 54 (r=3) | Exponential |
| Fibonacci (recursive) | Each term = sum of two before it | 1, 1, 2, 3, 5, 8, 13 ... | Exponential-ish |
| Alternating | Signs switch each term | 1, -1, 1, -1 ... or 1, -1/2, 1/4 ... | Oscillating |
| Telescoping | Consecutive terms cancel in sum | 1/(1x2) + 1/(2x3) + ... | Converging staircase |
Add (or subtract) the same number to every term. The constant gap is called the common difference d. Arithmetic sequences are the discrete version of linear functions — they graph as equally spaced points along a line.
| Formula | Expression | When to use it |
|---|---|---|
| nth term | a_n = a_1 + (n minus 1) times d | d = common difference, constant gap between terms |
| Common difference | d = a_n minus a_(n minus 1) | subtract any term from the next term |
| Partial sum (two-term form) | S_n = n divided by 2, times (a_1 + a_n) | need first and last terms |
| Partial sum (one-term form) | S_n = n divided by 2, times (2a_1 + (n minus 1)d) | need first term and d only |
| Middle term property | a_m = (a_j + a_k) divided by 2 if j + k = 2m | arithmetic mean of symmetric terms |
Young Carl Gauss discovered this as a schoolboy: write the sum forwards and backwards, then add the columns. Every column sums to (a_1 + a_n), and there are n such columns, giving 2S_n = n times (a_1 + a_n).
S_n = a_1 + (a_1+d) + (a_1+2d) + ... + a_n
S_n = a_n + (a_n-d) + (a_n-2d) + ... + a_1
2S_n = (a_1+a_n) + (a_1+a_n) + ... + (a_1+a_n) [n times]
S_n = n divided by 2, times (a_1 + a_n)
Sequence: 5, 9, 13, 17, ... Find a_20.
Identify: a_1 = 5, d = 9 minus 5 = 4
Apply formula: a_20 = 5 + (20 minus 1)(4)
= 5 + (19)(4) = 5 + 76
a_20 = 81
Continuing from Example 1: a_1 = 5, a_20 = 81, n = 20
S_20 = 20 divided by 2, times (5 + 81)
= 10 times 86
S_20 = 860
Check with alternate form: S_20 = 20/2 times (2(5) + 19(4)) = 10 times (10 + 76) = 10 times 86 = 860 confirmed
The sum of the first n terms of 2, 5, 8, 11, ... is 155. Find n.
a_1 = 2, d = 3. Use S_n = n/2 times (2a_1 + (n-1)d)
155 = n/2 times (4 + 3(n-1)) = n/2 times (3n + 1)
310 = n(3n + 1) = 3n^2 + n
3n^2 + n minus 310 = 0 → use quadratic formula
n = (minus 1 + sqrt(1 + 4 times 3 times 310)) / 6 = (minus 1 + sqrt(3721)) / 6 = (minus 1 + 61) / 6
n = 10 terms
In an arithmetic sequence, a_3 = 11 and a_8 = 31. Find a_n.
The gap from term 3 to term 8 spans 5 differences: 31 minus 11 = 20
So d = 20 / 5 = 4
Back-compute a_1: a_1 = a_3 minus 2d = 11 minus 8 = 3
a_n = 3 + (n minus 1)(4) = 4n minus 1
Multiply every term by the same number. The constant multiplier is called the common ratio r. Geometric sequences are the discrete version of exponential functions. When r is positive the sequence stays positive; when r is negative the signs alternate.
| Formula | Expression | Notes |
|---|---|---|
| nth term | a_n = a_1 times r^(n minus 1) | r = common ratio, constant multiplier between terms |
| Common ratio | r = a_(n+1) divided by a_n | divide any term by the one before it |
| Partial sum (finite) | S_n = a_1 times (1 minus r^n) divided by (1 minus r) | valid when r is not equal to 1 |
| Partial sum (r = 1) | S_n = n times a_1 | all terms are equal, just multiply |
| Infinite sum | S = a_1 divided by (1 minus r) | only valid when the absolute value of r is less than 1 |
Sequence: 3, 6, 12, 24, ... Find a_10.
r = 6 / 3 = 2, a_1 = 3
a_10 = 3 times 2^(10 minus 1) = 3 times 2^9 = 3 times 512
a_10 = 1536
Series: 5 + 10 + 20 + 40 + ... Find S_8.
a_1 = 5, r = 2
S_8 = 5 times (1 minus 2^8) / (1 minus 2)
= 5 times (1 minus 256) / (minus 1)
= 5 times (minus 255) / (minus 1) = 5 times 255
S_8 = 1275
Geometric sequence: a_2 = 6 and a_5 = 162. Find r and a_1.
a_5 = a_2 times r^3 → 162 = 6 times r^3 → r^3 = 27
r = cube root of 27 = 3
a_1 = a_2 / r = 6 / 3 = 2
Sequence: 2, 6, 18, 54, 162, ...
Sequence: 4, minus 8, 16, minus 32, ... Find a_n and S_6.
r = (minus 8) / 4 = minus 2
a_n = 4 times (minus 2)^(n minus 1)
S_6 = 4 times (1 minus (minus 2)^6) / (1 minus (minus 2))
= 4 times (1 minus 64) / 3 = 4 times (minus 63) / 3 = minus 252 / 3
S_6 = minus 84
Verify: 4 + (minus 8) + 16 + (minus 32) + 64 + (minus 128) = minus 84 confirmed
| Property | Arithmetic | Geometric |
|---|---|---|
| How terms change | Add constant d | Multiply by constant r |
| Identify from list | Constant differences | Constant ratios |
| nth term formula | a_1 + (n-1)d | a_1 times r^(n-1) |
| Partial sum | n/2 times (a_1 + a_n) | a_1(1 - r^n) / (1 - r) |
| Infinite sum | Diverges (unless all zeros) | Converges only when |r| is less than 1 |
| Graph shape | Linear (straight line) | Exponential (curve) |
| Real-world model | Salary with fixed annual raise | Account with percentage interest |
A finite geometric series always has a sum. An infinite geometric series sometimes converges (reaches a finite total) and sometimes diverges (the sum blows up). The deciding factor is the absolute value of r.
Absolute value of r is less than 1
CONVERGES
S = a_1 divided by (1 minus r)
r equals 1 (every term is the same)
DIVERGES
Sum grows without bound
r equals negative 1 (alternating signs, same magnitude)
DIVERGES
Partial sums oscillate between a_1 and 0
Absolute value of r is greater than 1 (growing terms)
DIVERGES
Terms grow, sum is infinite
Start with S_n = a_1(1 minus r^n) / (1 minus r). As n grows without bound, if |r| is less than 1 then r^n approaches 0. So the formula simplifies to a_1 times (1 minus 0) / (1 minus r) = a_1 / (1 minus r). This works only when r^n shrinks — which requires |r| to be less than 1.
S_n = a_1(1 minus r^n) / (1 minus r)
As n goes to infinity and |r| is less than 1: r^n goes to 0
S = a_1(1 minus 0) / (1 minus r) = a_1 / (1 minus r)
Find the sum: 16 + 8 + 4 + 2 + 1 + ...
a_1 = 16, r = 8/16 = 1/2. |r| = 0.5, which is less than 1 → converges
S = 16 / (1 minus 1/2) = 16 / (1/2)
S = 32
Express 0.272727... as a fraction using infinite series.
= 27/100 + 27/10000 + 27/1000000 + ...
a_1 = 27/100, r = 1/100. |r| is less than 1 → converges
S = (27/100) / (1 minus 1/100) = (27/100) / (99/100)
= 27/100 times 100/99 = 27/99
= 3/11
Check: 3 divided by 11 = 0.272727... confirmed
An infinite geometric series has sum 15 and r = 2/3. Find a_1.
S = a_1 / (1 minus r) → 15 = a_1 / (1 minus 2/3) = a_1 / (1/3)
a_1 = 15 times (1/3)
a_1 = 5
Sigma (the Greek letter for S, standing for Sum) is shorthand for writing long sums compactly. Mastering sigma notation is essential for college-level math and for reading Stewart clearly.
The notation has three parts:
Sigma from i=1 to 4 of (2i + 1):
= (2(1)+1) + (2(2)+1) + (2(3)+1) + (2(4)+1)
= 3 + 5 + 7 + 9
= 24
| Rule name | Formula |
|---|---|
| Constant factor rule | Sigma of (c times a_n) = c times Sigma of a_n |
| Sum rule | Sigma of (a_n + b_n) = Sigma of a_n + Sigma of b_n |
| Constant sum | Sigma from i=1 to n of c = n times c |
| Sum of first n integers | Sigma from i=1 to n of i = n(n+1) divided by 2 |
| Sum of squares | Sigma from i=1 to n of i^2 = n(n+1)(2n+1) divided by 6 |
| Sum of cubes | Sigma from i=1 to n of i^3 = [n(n+1) divided by 2]^2 |
You can re-index a sigma expression to change where the index starts, which is useful when combining two sums or matching a pattern. Replace every occurrence of i with (i minus 1) and adjust the bounds accordingly.
Original: Sigma from i=1 to n of (2i minus 1)
Let j = i minus 1, so i = j + 1 and when i=1 we have j=0, when i=n we have j=n-1
Shifted: Sigma from j=0 to (n-1) of (2(j+1) minus 1) = Sigma from j=0 to (n-1) of (2j + 1)
Both expressions produce the same sum — just with different index labeling
Evaluate: Sigma from i=1 to 5 of (3i^2 minus 2i + 4)
Split: 3 times Sigma(i^2) minus 2 times Sigma(i) + Sigma(4)
Sigma(i^2) from 1 to 5 = 5(6)(11)/6 = 55
Sigma(i) from 1 to 5 = 5(6)/2 = 15
Sigma(4) from 1 to 5 = 5 times 4 = 20
Total = 3(55) minus 2(15) + 20 = 165 minus 30 + 20
= 155
Write in sigma notation: 1/2 + 2/3 + 3/4 + 4/5 + ... + 20/21
Notice: term number i has numerator i and denominator i+1
General term: i / (i + 1)
Sigma from i=1 to 20 of [i / (i + 1)]
A telescoping series is one where consecutive terms cancel, leaving only the first and last pieces. The name comes from the old telescopes that collapse — most terms disappear when you write the partial sum.
Use partial fractions to split a complicated fraction into two simpler ones. Then most terms cancel when you expand the sum.
Partial fractions: 1/(i(i+1)) = 1/i minus 1/(i+1)
Sigma from i=1 to n of 1/(i(i+1)) = Sigma of [1/i minus 1/(i+1)]
Expand: (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))
Middle terms cancel in pairs (telescope!):
= 1 minus 1/(n+1) = n/(n+1)
As n goes to infinity: sum converges to 1
Find S_8 of the series: Sigma from i=1 to 8 of 1/(i(i+1))
From the formula above: S_n = n / (n + 1)
S_8 = 8 / (8 + 1) = 8/9
S_8 = 8/9
Check (expand 2 terms): 1/(1x2) + 1/(2x3) = 1/2 + 1/6 = 2/3. S_2 = 2/3. Formula gives 2/3. Confirmed.
A recursive sequence defines each term in relation to previous terms, plus initial conditions. The Fibonacci sequence is the most famous example.
Requires an initial condition (starting value) plus a recurrence relation telling how to get each term from the previous.
Arithmetic: a_1 = 3, a_n = a_(n-1) + 4
Geometric: a_1 = 2, a_n = 3 times a_(n-1)
Fibonacci: F_1 = 1, F_2 = 1, F_n = F_(n-1) + F_(n-2)
Explicit formulas let you compute a_100 directly. Recursive formulas require computing all 99 terms before it.
Recursive: need a_99 to find a_100
Explicit: a_n = 4n minus 1 → a_100 = 399 immediately
Convert recursive to explicit when possible
The ratio of consecutive Fibonacci terms approaches the golden ratio phi = (1 + sqrt(5)) / 2 approximately equal to 1.618. This appears in spirals of sunflower seeds, nautilus shells, and the proportions of many natural structures.
Recurrence
F_n = F_(n-1) + F_(n-2)
Initial conditions
F_1 = 1, F_2 = 1
Ratio convergence
F_(n+1) / F_n approaches phi = 1.618...
Sequence: a_1 = 2, a_n = 3a_(n-1) minus 1. Find the first 5 terms.
a_1 = 2
a_2 = 3(2) minus 1 = 5
a_3 = 3(5) minus 1 = 14
a_4 = 3(14) minus 1 = 41
a_5 = 3(41) minus 1 = 122
First 5 terms: 2, 5, 14, 41, 122
An alternating series switches sign with each term. The cleanest way to create one is to include a factor of (minus 1)^n or (minus 1)^(n+1) in the general term.
Sign starts negative (n=1)
General term: (minus 1)^n times a_n
n=1: (minus 1)^1 = minus 1
n=2: (minus 1)^2 = plus 1
Example: minus 1 + 1/2 minus 1/4 + 1/8 ...
Sign starts positive (n=1)
General term: (minus 1)^(n+1) times a_n
n=1: (minus 1)^2 = plus 1
n=2: (minus 1)^3 = minus 1
Example: 1 minus 1/2 + 1/4 minus 1/8 ...
Find the sum: 1 minus 1/3 + 1/9 minus 1/27 + ...
This is geometric with a_1 = 1 and r = minus 1/3
|r| = 1/3, which is less than 1 → converges
S = 1 / (1 minus (minus 1/3)) = 1 / (1 + 1/3) = 1 / (4/3)
S = 3/4
Critical insight for exams:
A necessary condition for an infinite series to converge is that the individual terms must approach 0. If a_n does not approach 0, the series definitely diverges. But terms approaching 0 is not sufficient for convergence (the harmonic series 1 + 1/2 + 1/3 + ... diverges even though terms approach 0). For geometric series, |r| less than 1 is both necessary and sufficient.
Sequences and series model many real-world phenomena. Stewart emphasizes these connections throughout Chapter 12.
If you invest P dollars at annual interest rate r (as a decimal), compounding annually, then after n years you have a geometric sequence.
Year 0: P
Year 1: P(1 + r) = P times (1+r)
Year 2: P(1+r)^2
Year n: P(1+r)^n → this is a_n of a geometric sequence with a_1 = P(1+r) and ratio = (1+r)
Monthly compounding: A = P times (1 + r/12)^(12n)
Example: $5000 at 6% annual, compounded monthly for 10 years:
A = 5000 times (1 + 0.06/12)^(12 times 10) = 5000 times (1.005)^120
A approximately equals $9093.49
An annuity is a sequence of equal payments. Each payment earns interest for a different number of periods. The future value is a geometric series sum.
Payment of R per period, interest rate i per period, n periods:
FV = R times [(1+i)^n minus 1] / i
This is S_n of a geometric series with first term R(1+i) and ratio (1+i)
Example: Deposit $200/month for 5 years at 6% annual (0.5% monthly):
FV = 200 times [(1.005)^60 minus 1] / 0.005
= 200 times [1.3489 minus 1] / 0.005 = 200 times 69.77
FV approximately equals $13,954 saved after 5 years
When a population grows by a fixed percentage each year it follows a geometric sequence. When it grows by a fixed amount, it follows an arithmetic sequence.
Arithmetic growth (fixed amount)
Population adds 500 per year
Start: P_0 = 10000, d = 500
P_n = 10000 + 500n
After 20 years: 20,000
Geometric growth (fixed percent)
Population grows 3% per year
Start: P_0 = 10000, r = 1.03
P_n = 10000 times (1.03)^n
After 20 years: about 18,061
The ancient Greek philosopher Zeno argued that motion is impossible: to cross a room you must first reach the halfway point, then half the remaining distance, then half of that, forever. How can infinitely many steps take a finite time?
Total distance = 1/2 + 1/4 + 1/8 + 1/16 + ...
a_1 = 1/2, r = 1/2, |r| = 0.5 which is less than 1 → converges
S = (1/2) / (1 minus 1/2) = (1/2) / (1/2) = 1
The infinite geometric series sums to exactly 1 — the full distance. Infinitely many steps can add to a finite total if the steps shrink fast enough. This resolved Zeno's paradox mathematically and was a key motivation for the development of calculus.
The chapter typically follows this order. Know where each topic lives so you can navigate efficiently.
Sequences
Definitions, explicit and recursive formulas, limits of sequences, monotone sequences
Series
Partial sums, convergence and divergence, geometric series test, sum of a series
Arithmetic Sequences
Common difference, nth term, arithmetic series sum formulas, applications
Geometric Sequences
Common ratio, nth term, partial sums, infinite geometric series, applications
Mathematical Induction
Proving formulas for all natural numbers by induction
The Binomial Theorem
Binomial coefficients, Pascal triangle, expanding (a+b)^n
Mistake: Using the geometric sum formula when r = 1
Fix: If r = 1, all terms are equal. S_n = n times a_1. The standard formula gives division by zero.
Mistake: Forgetting absolute value in convergence test
Fix: The condition is |r| less than 1, not r less than 1. A ratio of r = minus 0.9 still converges!
Mistake: Off-by-one in nth term formula
Fix: a_n = a_1 times r^(n minus 1), not r^n. Term 1 uses r^0 = 1. Double-check by plugging in n=1.
Mistake: Confusing sequence and series
Fix: Sequence: the list. Series: the sum. 'Find the 10th term' is a sequence question. 'Find S_10' is a series question.
Mistake: Misidentifying the type (arithmetic vs geometric)
Fix: Check differences first (subtract consecutive terms). If not constant, check ratios (divide consecutive terms). If neither, it may be recursive or neither.
Mistake: Wrong sigma index manipulation
Fix: When the index starts at 0 instead of 1, the formula Sigma i = n(n+1)/2 does not apply directly. Re-index or adjust accordingly.
An arithmetic sequence adds a fixed number (the common difference d) to each term: 3, 7, 11, 15 ... (d = 4). A geometric sequence multiplies each term by a fixed number (the common ratio r): 3, 6, 12, 24 ... (r = 2). To tell them apart: check differences first (constant? arithmetic), then check ratios (constant? geometric). Arithmetic sequences graph as lines; geometric sequences graph as exponential curves.
The nth term of an arithmetic sequence is a_n equals a_1 plus (n minus 1) times d, where a_1 is the first term, d is the common difference, and n is the term number. Example: For the sequence 5, 9, 13, 17 ..., a_1 = 5 and d = 4, so a_20 = 5 + (20 minus 1)(4) = 5 + 76 = 81.
The sum of the first n terms of an arithmetic series is S_n = n divided by 2, times (a_1 plus a_n), where a_1 is the first term and a_n is the last term. Equivalently, S_n = n divided by 2, times (2 times a_1 plus (n minus 1) times d). The formula works because you can pair the first and last terms, the second and second-to-last, and so on, each pair summing to the same total.
An infinite geometric series converges (has a finite sum) only when the absolute value of r is less than 1. When it converges, the sum S equals a_1 divided by (1 minus r). If the absolute value of r is greater than or equal to 1, the series diverges. For example, 1 + 1/2 + 1/4 + 1/8 + ... has r = 1/2, so it converges to S = 1 divided by (1 minus 1/2) = 2.
Sigma notation uses the Greek letter sigma to represent a sum. The expression written below sigma is the starting index value, the number above sigma is the ending index value, and the expression to the right of sigma is what you add for each index value. For example, sigma from i equals 1 to 5 of (2i) means you add 2(1) + 2(2) + 2(3) + 2(4) + 2(5) = 2 + 4 + 6 + 8 + 10 = 30.
A recursive sequence defines each term using one or more previous terms. The Fibonacci sequence is defined by F_1 = 1, F_2 = 1, and F_n = F_(n-1) plus F_(n-2) for n greater than 2. This gives: 1, 1, 2, 3, 5, 8, 13, 21, 34, ... Each term is the sum of the two before it. Recursive definitions are exact but require computing every term before the one you want; explicit formulas let you jump directly to any term.
When interest compounds annually at rate r, the account balance after each year forms a geometric sequence with common ratio (1 + r). If you start with principal P, after n years you have P times (1 + r) to the power n. For monthly compounding at annual rate r, the monthly ratio is (1 + r/12) and after n months you have P times (1 + r/12) to the power n. This is exactly the nth term formula for a geometric sequence.
Zeno argued that to walk across a room you must first walk half the distance, then half the remaining distance, then half again, infinitely many times -- so how can you ever arrive? The resolution is that the infinite geometric series 1/2 + 1/4 + 1/8 + ... converges to exactly 1. The common ratio is 1/2 (absolute value less than 1), so S equals (1/2) divided by (1 minus 1/2) = 1. Infinitely many steps can add up to a finite distance if the steps shrink fast enough.
Comprehensive arithmetic, geometric, and sigma notation reference
nth term, sums, convergence, and real-world applications
Proving series formulas for all natural numbers
Index notation, summation rules, and manipulations
Geometric sequences are discrete exponential functions
Infinite series convergence is fundamentally about limits
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