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10.1 Sequences and Series

A sequence is a list of numbers in a particular order.

For example,

\[ \{1, 2, 3, 4, 5\} \]

first five natural numbers

finite sequence because the sequence ends

\[ \{1, 2, 3, 4, 5, \dots\} \]

all natural numbers

infinite sequence

\[ \{2, 4, 6, 8, 10, \dots\} \]

even numbers

We can also list using explicit formula

\[ \{1, 2, 3, 4, 5, \dots\} = \{a_n\}_{n=1}^{\infty} \]
  • • "a" is the name of the sequence
  • • \(\infty\) where we end
  • • n=1 where do we start counting?
  • • a_n is the \(n^{th}\) term

Mapping terms:

\(a_1 = 1, a_2 = 2, a_3 = 3\)

\[ = \{n\}_{n=1}^{\infty} \]
\[ \{2, 4, 6, 8, 10, \dots\} = \{2n\}_{n=1}^{\infty} \]
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Sometimes we use the recurrence relation

\[ \{1, 2, 3, 4, 5, \dots\} \]each is one more than the one before

\(a_1 = 1, a_{n+1} = a_n + 1\)

for example,

\(a_2 = a_1 + 1 = 1 + 1 = 2\)

\(a_3 = a_2 + 1 = 2 + 1 = 3\)

\(\vdots\)

\[ \{2, 4, 6, 8, 10, \dots\} \]

\(a_1 = 2, a_{n+1} = a_n + 2\)

\[ \{1, 1, 2, 3, 5, 8, 13, 21, \dots\} \]

Fibonacci Sequence

each term is sum of previous two

\(a_1 = 1, a_2 = 1, a_{n+2} = a_{n+1} + a_n\)

e.g. \(a_3 = a_2 + a_1\)

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Sequence Convergence and Divergence

A sequence is said to converge if \(\lim_{n \to \infty} a_n\) exists.

If \(\lim_{n \to \infty} a_n\) DNE, then the sequence diverges.

Example

\[a_n = \frac{(-1)^n n}{2n^2 + 1} \quad n = 1, 2, 3, \dots\]

First few terms:

  • \(a_1 = \frac{(-1)^1 \cdot 1}{2(1)^2 + 1} = -\frac{1}{3}\)
  • \(a_2 = \frac{2}{9}\)
  • \(a_3 = -\frac{3}{19}\)
  • \(a_4 = \frac{4}{33}\)
  • \(a_5 = -\frac{5}{51}\)

The magnitude appears to decrease as \(n\) increases.

A scatter plot of sequence terms a_n versus n, showing values alternating signs and decaying toward zero.
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\[a_n = \frac{(-1)^n n}{2n^2 + 1}\]
\[\lim_{n \to \infty} \frac{(-1)^n n}{2n^2 + 1} = \underbrace{\lim_{n \to \infty} (-1)^n}_{\text{no effect on magnitude}} \cdot \underbrace{\lim_{n \to \infty} \frac{n}{2n^2 + 1}}_{\text{magnitude here, goes to 0}} = 0\]

Since \(\lim_{n \to \infty} a_n\) exists, this sequence converges (or is convergent).

What about

\[a_n = \frac{2n}{n+1} \quad n = 1, 2, 3, \dots\]

Converges?

\(\lim_{n \to \infty} \frac{2n}{n+1} = 2\) since limit exists, sequence converges.

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Limits (review)

\[ \lim_{n \to \infty} \frac{2n}{n+1} \]

When \( n \) is large, \( n+1 \approx n \) (e.g., \( n = 100,000 \)).

\[ \frac{2n}{n+1} \approx \frac{2n}{n} = 2 \]

Or, we can use l'Hospital's Rule

When limit \( \to \frac{\infty}{\infty} \) or \( \frac{0}{0} \), then:

\[ \lim_{n \to \infty} \frac{f(n)}{g(n)} = \lim_{n \to \infty} \frac{f'(n)}{g'(n)} \]
\[ \lim_{n \to \infty} \frac{n}{2n^2+1} \to \frac{\infty}{\infty} \]

By l'Hospital's Rule:

\[ = \lim_{n \to \infty} \frac{1}{4n} = 0 \]
  • Numerator: deriv. of \( n \) is 1
  • Denominator: deriv. of \( 2n^2+1 \) is \( 4n \)
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A series is the sum of the terms in a sequence.

\( \{1, 2, 3, 4, 5, \dots\} \) sequence

\( 1 + 2 + 3 + 4 + 5 + \dots \) series

Like with sequences, we can express series compactly

\[ 1 + 2 + 3 + 4 + 5 + \dots = \sum_{n=1}^{\infty} n \]

\( \sum \) (sigma) means sum

\( \infty \) means end at infinity

\( n=1 \) means start adding at \( n=1 \)

\( n \) because each term is equal to \( n \)

\[ 2 + 4 + 6 + 8 + \dots = \sum_{n=1}^{\infty} 2n \]

(so \( a_n = 2n \))

infinite series → no end to adding

\[ 1 + 2 + 3 = \sum_{n=1}^{3} n \]

there is an end

finite series

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Example: Infinite Series

\[ \sum_{n=1}^{\infty} \frac{1}{2^n} \]
\[ = \frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^5} + \dots = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \dots \]

The sum of the first \( n \) terms is called the \( n^{\text{th}} \) partial sum, \( S_n \).

first partial sum\( S_1 = \frac{1}{2} \)

\( S_2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \)

\( S_3 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8} \)

\( S_4 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{15}{16} \)

\( \vdots \)

\( S_{10} = \dots = \frac{1023}{1024} \)

Note the partial sums appear to approach 1 \( \rightarrow \lim_{n \to \infty} S_n = 1 \)

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If the partial sums appear to approach some finite number, we say the series converges (or is convergent).

Otherwise, the series diverges (or is divergent).

\( \Rightarrow \) does \( \lim_{n \to \infty} S_n \) exist?

yes: converges

no: diverges

Examples of Convergence and Divergence

So, \( \sum_{n=1}^{\infty} \frac{1}{2^n} \) converges.

\( \sum_{n=1}^{\infty} n \) diverges because \( 1 + 2 + 3 + 4 + \dots \) does not have a limit.

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Example: Infinite Series Convergence

\[ \sum_{n=1}^{\infty} \cos(n\pi) \]

Expanding the series by evaluating terms for \( n = 1, 2, 3, 4, \dots \):

\[ = \underbrace{\cos(\pi)}_{n=1} + \underbrace{\cos(2\pi)}_{n=2} + \underbrace{\cos(3\pi)}_{n=3} + \underbrace{\cos(4\pi)}_{n=4} + \dots \]
\[ = -1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + \dots \]

Partial Sums

  • \( S_1 = -1 \)
  • \( S_2 = 0 \)
  • \( S_3 = -1 \)
  • \( S_4 = 0 \)
  • \( S_5 = -1 \)
  • \( \vdots \)

Do these appear to settle down around some finite number?

\( \lim_{n \to \infty} S_n \) exists? NO!

This series DIVERGES

Sequence of partial sums:

\( \{ -1, 0, -1, 0, -1, \dots \} \)