8.1 Sequences and Series

A sequence is a list of numbers. A series is a sum of a list of numbers.

1, 3, 5, 7, 9 is a sequence.

1 + 3 + 5 + 7 + 9 is a series.

Sequences and series may be finite, as the two examples above, or they can be infinite, as in

1, 3, 5, 7, 9, . . .

1 + 3 + 5 + 7 + 9 + . . .

It is customary to use subscripted variables to denote the terms of a sequence or series as in

c₁, c₂, c₃, . . . or

c₁ + c₂ + c₃ + . . .

The subscript is referred to as the index of the term.

Ideally, formulas can be found for the terms of a sequence or series in terms of the index numbers. For example, the sequence 1, 3, 5, 7, 9, . . . can be represented by the formula

c_n = 2 n – 1 for n = 1, 2, 3, . . ., or by the notation { 2 n – 1 }, where the expression between the braces is the formula for the nth term of the sequence.

Exercise 8.1.1

Write the first five terms of the sequence { n² – n }.

Solution

Exercise 8.1.2

Find a general formula for the nth term of the sequence 2, 5, 10, 17, 26, . . .

Solution

Another way to specify the terms of a sequence is using a recursive definition. In a recursive definition, the first term or so is given. Then a formula is given describing how to generate other terms of the sequence from terms already known.

For example,

c₁ = 2

c_n+1 = 3c_n – 1

recursively defines the sequence 2, 5, 14, 41, 122, . . .

Exercise 8.1.3

Write the first five terms of the sequence defined recursively by

c₁ = 1

c₂ = 1

c_n+2 = c_n + c_n+1

This is a famous sequence called the Fibonacci Sequence.

Solution

An important issue in Calculus is whether or not a given infinite sequence converges.

Consider the two infinite sequences

1, 2, 3, 4, 5, . . . = { n }

and

0.9, 0.99, 0.999, 0.9999, 0.99999, . . . = { 1 – ( 0.1 )_n }

The terms of the first sequence get larger and larger, without any upper bound, whereas the terms of the second sequence have an upper bound. Any number larger than or equal to 1 is an upper bound on the terms of the second sequence. The least upper bound of the second sequence is 1. We say that the second sequence converges to a limit of 1, whereas the first sequence diverges.

A tail of an infinite sequence c₁, c₂, c₃, . . . is a sequence c_p, c_p+1, c_p+2, . . . for p $1.

For example, c₅, c₆, c₇, . . . is a tail of the sequence c₁, c₂, c₃, . . .

The statement that the infinite sequence c₁, c₂, c₃, . . . converges to the limit L means that any open interval containing L contains a tail of the sequence.

A arrow is sometimes used to denote convergence. Thus c₁, c₂, c₃, . . . ® L means that the sequence converges to the limit L.

To see that the sequence { c_n } = { 1 – ( 0.1 )ⁿ } converges to a limit of 1, suppose ( a, b ) is an open interval containing 1. Then a < 1. No matter how close a is to 1, there is a positive integer p such that ( 0.1 )^p is smaller than 1 – a. Then a is smaller than 1 – ( 0.1 ) ^p, which is less than 1 – ( 0.1 ) ^p+1 , etc. But these are terms of the tail c_p, c_p+1, c_p+2, . . . of { 1 – ( 0.1 )ⁿ }. So all the terms of the tail c_p, c_p+1, c_p+2, . . . lie between a and 1 and must, therefore lie in ( a, b ). Since this is true for any interval ( a, b ) containing 1, it follows that 1 is the limit of the sequence.

Exercise 8.1.4

Show that { ( 2 n + 1 ) / n } ® 2

Solution

Now that we know what it means for a sequence to converge to a limit L, we can define what it means for a series to converge to a sum L. But first, we need the concept of a sequence of partial sums.

Given the infinite series c₁ + c₂ + c₃ + . . ., define the partial sums as

s₁ = c₁

s₂ = c₁ + c₂

s₃ = c₁ + c₂ + c₃

.

.

.

s_n = c₁ + c₂ + c₃ + . . . + c_n

Then the sequence of partial sums is

s₁, s₂, s₃, . . ., s_n, . . .

To say that the series c₁ + c₂ + c₃ + . . . has the sum L means that the sequence of partial sums s₁, s₂, s₃, . . ., s_n, . . . converges to the limit L.

Let us show that 1 + 0.1 + 0.01 + 0.001 + . . . ® 10 / 9

Notice that the decimal expansion of 10 / 9 is 1.1111. . . etc.

The sequence of partial sums is an increasing sequence

= 1, 1.1, 1.11, 1.111, . .

= 1 / 1, 11 / 10, 111 / 100, 1111 / 1000, . . .

= 9 / 9, 99 / 90, 999 / 900, 9999 / 9000, . . .

= ( 10 / 9 – 1 / 9 ), ( 10 / 9 – 1 / 90 ), ( 10 / 9 – 1 / 900 ), ( 10 / 9 – 1 / 9000 ), . . .

= { 10 / 9 – ( 1 / 9 ) ( 1 / 10 )^n-1 }

Let ( a, b ) denote an interval containing the number 10 / 9. Then a < 10 / 9. Since we can find a value of p making the quantity ( 1 / 9 ) ( 1 / 10 )^p-1 as small as we please, pick p so that the quantity is smaller than 10 / 9 – a. Then it follows that

a < 10 / 9 – ( 1 / 9 ) ( 1 / 10 )^p-1, which is the first term of a tail of the sequence. Since the sequence is increasing, the interval

( a, b ) contains a tail of the sequence. Since this will be true for any interval containing 10 / 9, that must be the limit of the sequence of partial sums. Thus that is the sum of the series.

Exercise 8.1.5

Find a formula for the nth partial sum of the sequence of partial sums of the sequence

1, 1 / 2, 1 / 4, 1 / 8, . . .

Prove that 1 + 1 / 2 + 1 / 4 + 1 / 8 + . . . = 2.

Solution

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