STEP-BY-STEP LESSON · §5.1
Sequences
Read an index and calculate sequence terms without confusing position and value.
Before you begin
Variables
In x+y=0, each variable keeps its chosen value throughout the statement. If x=3, only y=−3 satisfies this condition.
x+y=0; x=3 ⇒ y=−3
The condition x>0 depends on the permitted values of x. Specify the domain: integers.
x∈ℤ
Symbols
- a_n
- term with index n
- Σ
- sum of specified terms
- n≥0
- n is a nonnegative integer index
- Π
- product of all indicated factors
Definitions and notation for this topic
Summation
∑_{k=m}^{n} a_kFor integers m≤n, add the terms a_m,a_(m+1),…,a_n. The index k is bound locally by the summation; m and n are the limits.
Renaming the bound index does not change the sum. ∑ is the summation operator; capital Σ can instead name an alphabet in formal-language notation.
Separate glossary example
∑_{k=1}^{4} k²=1+4+9+16=30.
∑ᵢ₌₁³ i = 1+2+3 = 6
Finite product
∏_{k=m}^{n} a_kFor integer limits m≤n, multiply a_m,a_(m+1),…,a_n. The index k is local to the product.
The standard empty-product convention is 1; the corresponding empty sum is 0. ∏ denotes multiplication of terms, whereas ∑ denotes addition.
Separate glossary example
∏_{k=1}^{4}(k+1)=2·3·4·5=120.
Sequence
aₙ
Objects arranged by their indices.
The index identifies a position in the sequence.
State the index domain and its first value, for example n=0,1,2,… . The symbol a_n denotes one term, whereas (a_n) denotes the sequence; terms may repeat.
Separate glossary example
aₙ=2n gives 2,4,6,… for n≥1.
Step by step
Step 1 / 6
Specify the index set
A sequence assigns one value to each permitted index. Whether it starts at 0 or 1 is part of its definition.
a_n = 2n + 1, n ≥ 0
Without the starting index, “the first three” is ambiguous.
Worked example
For a_n=2n+1, n≥0, find the first three terms and their sum.
- Use indices 0, 1, 2 because the sequence begins at 0.
- Substitution gives a_0=1, a_1=3, a_2=5.
- Add the resulting values: 1+3+5=9.
Rename an index without changing the terms
A bound belongs to its index variable. When introducing a new index, transform the lower bound, upper bound and term formula together. Reindexing only renames the same finite sequence of terms; it does not add, delete or replace terms. The method also applies to products.
Worked example
Rewrite ∑_{i=3}^{6}(2i−1) with an index starting at 0, then evaluate it.
Choose the index shift and invert it.
j=i−3; i=j+3
Why: Each old index has exactly one new index and conversely.
Transform both bounds and the summand.
i=3→j=0; i=6→j=3; 2i−1=2(j+3)−1=2j+5
Why: Changing only the bounds would change which numbers are added.
Write the equal sum and check all four terms.
∑_{i=3}^{6}(2i−1)=∑_{j=0}^{3}(2j+5)=5+7+9+11=32Why: Both index ranges have four positions and produce the same ordered terms.
Try it yourself
Rewrite ∏_{k=2}^{5}(k+1) using j=k−1. Give the new bounds, factors and value.
Show worked answer
Invert the substitution and transform the bounds.
k=j+1; j=1,…,4; k+1=j+2
Why: The combining operation remains multiplication.
Write and evaluate the product.
∏_{j=1}^{4}(j+2)=3·4·5·6=360Why: These are exactly the four original factors.
Cancel factorials as products
For a positive integer n, n!=1·2·…·n; also 0!=1. A factorial quotient can often be simplified by displaying the common product. Cancel only a nonzero common factor.
Worked example
Simplify (n+2)!/n! for integers n≥0.
Separate the two extra factors.
(n+2)!=(n+2)(n+1)n!
Why: The product through n is n!; only n+1 and n+2 are new. The identity also works at n=0 because 0!=1.
Cancel the common nonzero factorial.
(n+2)!/n!=(n+2)(n+1)
Why: For every allowed n, n! is positive. It is a multiplicative factor, so the cancellation is valid.
Try it yourself
Simplify n!/(n−2)! for integers n≥2 and evaluate it at n=7.
Show worked answer
Display the shared product.
n!=n(n−1)(n−2)!
Why: The denominator is defined and positive because n−2≥0.
Cancel and substitute.
n!/(n−2)!=n(n−1); 7·6=42
Why: The simplification avoids calculating both large factorials separately.
Optional self-check
If b_n=n² for n≥1, what is b_3?
Show answer
9: substitute index 3 and square it.