Sequences

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 13 November 2011

Sequences

Let $Y$ be a set. A sequence $(y_1,y_2,y_3,\dots )$ in $Y$ is a function $$\begin{array}{rcl}{\mathbb{Z}}_{>0}& ⟶& Y\\ n& ⟼& y_n.\end{array}$$

Let $Y$ be a set with a partial order $\le$ and let $(y_1,y_2,y_3,\dots )$ be a sequence in $Y$.

• The sequence $(y_1,y_2,y_3,\dots )$ is increasing if $(y_1,y_2,y_3,\dots )$ satisfies $$\text{if}i\in {\mathbb{Z}}_{>0}\text{then}{y}_i\le y_{i+1}.$$
• The sequence $(y_1,y_2,y_3,\dots )$ is decreasing if $(y_1,y_2,y_3,\dots )$ satisfies $$\text{if}i\in {\mathbb{Z}}_{>0}\text{then}{y}_i\ge y_{i+1}.$$
• The sequence $(y_1,y_2,y_3,\dots )$ is monotone if it is increasing or decreasing.

Let $Y$ be a metric space and let $(y_1,y_2,y_3,\dots )$ be a sequence in $Y$.

• The sequence $(y_1,y_2,y_3,\dots )$ is bounded if the set $\{y_1,y_2,y_3,\dots \}$ is bounded.
• The sequence $(y_1,y_2,y_3,\dots )$ is contractive if $(y_1,y_2,y_3,\dots )$ satisfies: There exists $\alpha \in (0,1)$ such that $$\text{if}i\in {\mathbb{Z}}_{>0}\text{then}d(y_i,y_{i+1})\le \alpha d(y_{i-1},y_i).$$
• The sequence $(y_1,y_2,y_3,\dots )$ is Cauchy if $(y_1,y_2,y_3,\dots )$ satisfies:
  1. if $ϵ\in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N$ then $d(y_m,y_n)<ϵ$.
• let $l\in Y$. The sequence $(y_1,y_2,y_3,\dots )$ converges to $l$ if $$\underset{n\to \infty }{\lim }{y}_n=l.$$ i.e., if $(y_1,y_2,y_3,\dots )$ satisfies
  1. if $ϵ\in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $d(y_n,l)<ϵ$.

Let $(y_1,y_2,y_3,\dots )$ be a sequence in $ℝ$ (or, more generally, any totally ordered set with the order topology).

• The supremum of $(y_1,y_2,y_3,\dots )$ is $\sup \{y_1,y_2,y_3,\dots \}$
• The infimum of $(y_1,y_2,y_3,\dots )$ is $\inf \{y_1,y_2,y_3,\dots \}$
• The upper limit of $(y_1,y_2,y_3,\dots )$ is $$\mathrm{lim\; sup}{y}_n=\underset{n\to \infty }{\lim }\sup \{y_n,y_{n+1},\dots ,\}.$$
• The lower limit of $(y_1,y_2,y_3,\dots )$ is $$\mathrm{lim\; inf}{y}_n=\underset{n\to \infty }{\lim }\inf \{y_n,y_{n+1},\dots ,\}.$$

Example. If $y_n={(-1)}^n(1-\frac{1}{n})$ then $$\mathrm{lim\; sup}{y}_n=1\text{and}\mathrm{lim\; inf}{y}_n=-1.$$

Let $(y_1,y_2,y_3,\dots )$ be a sequence in $ℝ$. Then $$$ \mathrm{lim\; sup}{y}_n=\sup \left\{\text{cluster points of}\{y_1,y_2,y_3,\dots \}\right\}\text{and}\mathrm{lim\; inf}{y}_n=\inf \left\{\text{cluster points of}\{y_1,y_2,y_3,\dots \}\right\}.$$$

The interest sequence

Example. If you borrow $500 on your credit card at 14% interest, find the amounts due at the end of two years if the interest is compounded

1. annually,
2. quarterly,
3. monthly,
4. daily,
5. hourly,
6. every second,
7. every nanosecond,
8. continuously.

		Answers.
	You owe $$500+500\left(.14\right)=500(1+.14)\text{after one year and}500(1+.14)(1+.14)$$ after two years. You owe $$500+500\left(\frac{.14}{12}\right)=500(1+\frac{.14}{12})\text{after one month.}$$ You owe $$500(1+\frac{.14}{12})(1+\frac{.14}{12})\text{after two months.}$$ You owe $$500{(1+\frac{.14}{12})}^{24}\text{after two years.}$$ You owe $$500+500\left(\frac{.14}{365\cdot 24\cdot 3600}\right)\text{after 1 second,}$$ and $$500{(1+\frac{.14}{365\cdot 24\cdot 3600})}^{2\cdot 365\cdot 24\cdot 3600}\text{after two years.}$$ You owe $$\underset{n\to \infty }{\lim }500{(1+\frac{.14}{n})}^{2n}\text{after two years.}$$ In fact, $$\begin{array}{rcl}\underset{n\to \infty }{\lim }500{(1+\frac{.14}{n})}^{2n}& =& 500\underset{n\to \infty }{\lim }{\left(e^{\log (1+\frac{.14}{n})}\right)}^{2n}\\ & =& 500\underset{n\to \infty }{\lim }{e}^{2n\phantom{,}\log (1+\frac{.14}{n})}\\ & =& 500\underset{n\to \infty }{\lim }{e}^{2\cdot .14\phantom{,}\frac{\log (1+\frac{.14}{n})}{\frac{.14}{n}}}\\ & =& 500\underset{n\to \infty }{\lim }{e}^{.28\phantom{,}\frac{\log (1+\frac{.14}{n})}{\frac{.14}{n}}}\end{array}$$ Recall $$\underset{x\to 0}{\lim }\frac{\log (1+x)}{x}=1.$$ So you owe $500e^{.28}$ after two years if the interest is compounded continuously. Note:   $500{(1+.14)}^2=649.80,500{(1+.14)}^{24}\approx 660.49,\text{and}500e^{.28}\approx 661.56$. $\square$

References

[Ru1] W. Rudin, Principles of mathematical analysis, McGraw-Hill, 1976., MR????? (???)

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