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PAGE 1

10.2 Sequences

this section is not on exam 2

Definition

sequence: list of numbers in some order

\[ \{a_n\}_{n=1}^{\infty} = \{a_1, a_2, a_3, \dots\} \]

Convergence

sequence converges if \( \lim_{n \to \infty} a_n \) exists

Example

for example, \( \{ \frac{n}{n+1} \}_{n=1}^{\infty} = \{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \dots, \frac{99,999}{100,000}, \dots \} \)

the terms appear to approach 1

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

so \( \{ \frac{n}{n+1} \}_{n=1}^{\infty} \) converges to 1

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Example

\[ \{n^{13/n}\}_{n=1}^{\infty} \text{ converges?} \]
\[ = \{1^{13}, 2^{13/2}, 3^{13/3}, \dots, 13^{13/13}, \dots\} \]

the terms appear to initially get bigger

does the sequence converge?

listing numbers does not seem to answer that question

\[ \text{try } \lim_{n \to \infty} a_n = \lim_{n \to \infty} n^{13/n} \to \infty^0 \]

indeterminate

limit could be anything

Applying l'Hospital's Rule

l'Hospital's Rule can give us the limit

let \( y = n^{13/n} \to \) we want \( \lim_{n \to \infty} y \)

\[ \ln y = \ln n^{13/n} = \frac{13}{n} \ln(n) = \frac{13 \ln(n)}{n} \to \frac{\infty}{\infty} \text{ as } n \to \infty \]

we will use l'Hospital's Rule on this

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Limit Calculation for Sequences

\[ \lim_{n \to \infty} \frac{13 \ln(n)}{n} \]
\[ = \lim_{n \to \infty} \frac{13 \cdot \frac{1}{n}}{1} = \lim_{n \to \infty} \frac{13}{n} = 0 \]

this means \( \lim_{n \to \infty} \ln y = 0 \) but we want \( \lim_{n \to \infty} y \) where \( y = n^{13/n} \)

remember \( e^{\ln y} = y \)

so, \( \lim_{n \to \infty} \ln y = 0 \) becomes

\[ \lim_{n \to \infty} e^{\ln y} = e^0 = 1 \]

this is the limit of \( \{n^{13/n}\}_{n=1}^{\infty} \)

\[ \{1, 2^{13/2}, 3^{13/3}, \dots, 1,000,000^{13/1000000}, \dots\} \]

wait long enough, very close to 1

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Visualizing the Sequence Convergence

Sequence Plot of \( a_n = n^{13/n} \)

The following graph illustrates the values of the sequence \( a_n \) as \( n \) increases. Note the behavior at \( n=1 \) and the peak at \( n=2 \) before the sequence gradually converges toward 1.

Scatter plot of sequence a_n = n to the power of 13/n, showing points converging to 1 as n goes to 100.

Annotations on graph:

  • \( n=1 \) at the first point near the origin.
  • \( n=2 \) at the peak value.
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Sequence Visualization

\[ \left\{ \frac{n}{n+1} \right\} \]

The following graph illustrates the behavior of the sequence \( \left\{ \frac{n}{n+1} \right\} \) as \( n \) increases. The values approach a horizontal asymptote at 1.

A scatter plot showing points of the sequence n/(n+1) increasing from 0.5 and asymptotically approaching 1.0.
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Monotonic Sequences

\( \{n^{13/n}\} \) increases initially, then decreases and settles down around 1 whereas \( \left\{ \frac{n}{n+1} \right\} \) always increases.

If a sequence either always increases or always decreases (for all \( n \)) then the sequence is said to be monotonic.

Examples

  • \( \{1, 2, 3, 4, 5, \dots\} \) is monotonic
  • \( \{-1, -2, -3, -5, -7, -9, -13, \dots\} \) is monotonic
  • \( \{-1, \frac{1}{2}, -\frac{1}{3}, \frac{1}{4}, -\frac{1}{5}, \frac{1}{6}, \dots\} \) is not monotonic
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\[ \left\{ \frac{n}{n+1} \right\}_{n=1}^{\infty} = \left\{ \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots \right\} \]

The term \( \frac{1}{2} \) is the smallest term in this sequence. The subsequent terms \( \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots \) increase to approach 1.

This sequence is bounded because all terms are no smaller than something \( (\frac{1}{2}) \) and no bigger than something else \( (1) \).

If a sequence is bounded and monotonic, then it converges.

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Geometric Sequences

An important sequence is the geometric sequence \( \rightarrow \{r^n\} \), where \( r \) is constant.

\[ \left\{ \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \dots \right\} \]\[ = \left\{ (\frac{1}{2})^1, (\frac{1}{2})^2, (\frac{1}{2})^3, (\frac{1}{2})^4, \dots \right\} = \left\{ (\frac{1}{2})^n \right\}_{n=1}^{\infty} \]

This sequence converges to 0. Here, \( r = \frac{1}{2} \) is the common ratio between succeeding terms.

\[ \left\{ -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27}, \frac{1}{81}, \dots \right\} = \left\{ (-\frac{1}{3})^n \right\}_{n=1}^{\infty} \quad \text{converges to 0} \]\[ \left\{ 2, 4, 8, 16, 32, 64, \dots \right\} = \left\{ 2^n \right\}_{n=1}^{\infty} \quad \text{diverges} \]

The ratio \( r \) appears to determine convergence. The geometric sequence \( \{r^n\} \) converges if:

\[ -1 < r \leq 1 \]
PAGE 9

Convergence of Geometric Sequences

Why is 1 included but -1 is not?

\[ \{r^n\}_{n=1}^{\infty} \]

If \( r = 1 \) :

\[ \{1, 1, 1, 1, 1, \dots\} \text{ converges to } 1 \]

If \( r = -1 \) :

\[ \{-1, 1, -1, 1, -1, 1, \dots\} \text{ no limit, diverges} \]
PAGE 10

Factorials in Sequences and Series

Factorial (!) shows up a lot in sequences and series.

\[ n! = (n)(n-1)(n-2) \dots (1) \]\[ 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \]\[ 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \]

Example

\[ \left\{ \frac{n}{n!} \right\}_{n=1}^{\infty} \]\[ = \left\{ 1, \frac{2}{2!}, \frac{3}{3!}, \frac{4}{4!}, \dots \right\} \]\[ = \left\{ 1, 1, \frac{1}{2}, \frac{1}{6}, \dots \right\} \]

Converges? Check \( \lim_{n \to \infty} \frac{n}{n!} \)

As \( n \to \infty \), numerator is large, denominator is very LARGE. Limit appears to be zero.

Let's check properly:

\[ \lim_{n \to \infty} \frac{n}{n!} = \lim_{n \to \infty} \frac{n}{(n)(n-1)(n-2) \dots (1)} \]
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\[ = \lim_{n \to \infty} \frac{1}{(n-1)(n-2) \cdots (1)} = 0 \quad \text{so, } \left\{ \frac{n}{n!} \right\} \text{ converges to } 0 \]

Example

\[ \left\{ \frac{e^n}{n!} \right\}_{n=1}^{\infty} \]\[ = \left\{ \frac{e^1}{1!}, \frac{e^2}{2!}, \frac{e^3}{3!}, \dots \right\} \]

Initially, \( e^n \) dominates \( n! \)

As \( n \) becomes large, for example:

\( n = 10 \):

\( e^{10} \approx 22,026 \)

\( 10! \approx 3,628,800 \)

\( n = 100 \):

\( e^{100} \approx 3 \times 10^{43} \)

\( 100! \approx 9 \times 10^{157} \)

\( n \to \infty \), denominator wins, limit appears to be zero

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Can we be more sure?

\[ \frac{e^n}{n!} = \frac{e^n}{(n)(n-1)(n-2) \cdots (1)} \]

As \( n \) becomes large, the denominator is polynomial w/ leading term \( n^n \) which dominates.

\[ \text{So, as } n \to \infty, \quad \frac{e^n}{n!} \approx \frac{e^n}{n^n} = \left( \frac{e}{n} \right)^n \]

looks like \( r^n \)

with \( r \to 0 \) as \( n \to \infty \)

and since \( -1 < r \le 1 \)

this behaves like a geo. seq. as \( n \to \infty \)

therefore converges.