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Integral Calculus lecture notes: Sequences

Eugene Kritchevski, Vanier College, last updated on November 16, 2017

What is a sequence ?

A sequence is an infinite, ordered list of real numbers. For example, the odd numbers form the sequence $$\set{1,3,5,7,9,\cdots}$$ and reciprocals of positive integers form the sequence $$\set{\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\cdots}.$$ A sequence $$\set{a_1,a_2,a_3,a_4,a_5,\cdots}$$ can be denoted by $\set{a_n}_{n=1}^{\infty}$ or simply $\set{a_n}$. We refer to $a_n$ as the $n^{th}$ term of the sequence. Thus $a_1$ if the $1^{st}$ term, $a_2$ is the $2^{nd}$ term, $a_3$ is the $3^{rd}$ term, and so on.

With this notation, the sequence $\set{a_n}_{n=1}^{\infty}$ where $a_n=2n-1$ gives the odd numbers: $a_1=2\cdot1-1=1,a_2=2\cdot2-1=2,a_3=2\cdot3-1=5$, and so on. The sequence $\displaystyle\set{\frac{1}{n}}$ gives the reciprocals of positive integers: $\displaystyle\set{\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots}.$

Exercise: Write out the first five terms for each of the sequences below.

(a)$\displaystyle\qquad\set{n(n+1)}_{n=1}^{\infty}$

(b) $\displaystyle\qquad\set{\frac{(-1)^{n-1}}{2^n}}_{n=1}^{\infty}$

Find a formula for the $n^{th}$ term for each of the sequences below.

(c)$\displaystyle\qquad\set{5,8,11,14,17,\cdots}$

(d)$\displaystyle\qquad\set{7,0.7,0.07,0.007,0.0007,\cdots}$
Solution:

(a)$\displaystyle\qquad\set{1,2,6,12,20,\cdots}$

(b) $\displaystyle\qquad\set{\frac{1}{2},\frac{-1}{4},\frac{1}{8},\frac{-1}{16},\frac{-1}{32},\cdots}$

(c)$\displaystyle\qquad\set{5+3(n-1)}_{n=1}^{\infty}$

(d)$\displaystyle\qquad\set{7\times 10^{n-1}}_{n=1}^{\infty}$

Remarks about the notation
  1. The sequence $\set{a_n}_{n=1}^{\infty}$ can as well be denoted by $\set{a_i}_{i=1}^{\infty}$, $\set{a_k}_{k=1}^{\infty}$, $\set{a_m}_{m=1}^{\infty}$ or with any other letter playing the role of an index as long as confusing notations such as $\set{a_a}_{a=1}^{\infty}$ are avoided. For instance, $\set{2i-1}_{i=1}^{\infty}$ and $\set{2n-1}_{n=1}^{\infty}$ yield exactly the same sequence of odd integers.
  2. It is sometimes convenient to start sequences with index $0$ instead of $1$. The sequence $$\set{a_0,a_1,a_2,a_3,a_4,\cdots}$$ is then denoted by $\set{a_n}_{n=0}^{\infty}$. Changing the starting index will in general change the formula for the $n^{ th}$ term. Example: $$\set{1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16},\cdots}=\set{\frac{1}{2^n}}_{n=0}^{\infty}=\set{\frac{1}{2^{n-1}}}_{n=1}^{\infty}$$

The graph of sequence

The graph of a sequence $\set{a_n}$ consists of all the points $(n,a_n)$ in the plane. If the sequence $\set{a_n}$ is defined by the formula $a_n=f(n)$, where $f(x)$ is a function, then the graph of the sequence consists of points on the graph of $y=f(x)$ with integer $x$-coordinates.

Informal definition of the limit of a sequence
We say that the sequence $\set{a_n}$ converges to a real number $L$ and write $$\lim_{n\to\infty}a_n = L,$$ if $a_n$ gets arbitrarily close to $L$ provided $n$ is sufficiently large.
Example: From the graph of the sequence (see above applet) it appears that

(a) $\displaystyle\lim_{n\to\infty}\frac{1}{n}=0$

(b) $\displaystyle\lim_{n\to\infty}2+\frac{\sin(n)}{n}=2$

(c) $\displaystyle\lim_{n\to\infty}\frac{n+1}{n+2}=1$

(d) $\displaystyle\lim_{n\to\infty}(8/9)^n=0$

(e) $\displaystyle\lim_{n\to\infty}\frac{2n}{n+3}=2$

(f) $\displaystyle\lim_{n\to\infty}\sin(n) \qquad DNE$

(h) $\displaystyle\lim_{n\to\infty}(2n-1)=+\infty$
Instead of $\displaystyle\lim_{n\to\infty}a_n = L$, we sometimes write $$a_n\to L \qquad\textrm{ as } \qquad n\to\infty,$$ or, more compactly, $$a_n\to L.$$

Sequences defined recursively

Instead of giving the formula for the $n^{th}$ term $a_n$, one can define a sequence by specifying the initial term and a procedure for computing $a_{n}$ from the previous term $a_{n-1}$.

Example: $a_1=1$ and $a_n=a_{n-1}+2$

Let us compute the first few terms: $$a_1=1$$ $$a_2=a_1+2=1+2=3$$ $$a_3=a_2+2=3+2=5$$ $$a_4=a_3+2=5+2=7$$ $$\cdots$$ We see that $a_n$ is the sequence of odd integers. WARNING! $a_{n-1}$ DOES NOT MEAN $a_n-1$

Example: (Arithmetic sequences) The numbers $a$ and $d$ are fixed and the sequence $\set{a_n}_{n=0}^{\infty}$ is defined by $a_0=a$ and $a_n=a_{n-1}+d$. Then $$a_0=a$$ $$a_1=a_0+d=a+d$$ $$a_2=a_1+d=a+2d$$ $$a_3=a_2+d=a+3d$$ $$\cdots$$ We see that $a_n=a+nd$. Let us summarize the formula for an arithmetic sequence, with strating index $0$ and also with strating index $1$.
Arithmetic sequence formula
$$\set{a,a+d,a+2d,a+3d,\cdots}=\set{a+nd}_{n=0}^{\infty}=\set{a+(n-1)d}_{n=1}^{\infty}$$




Example: (Geometric sequences) The numbers $a$ and $r$ are fixed and the sequence $\set{a_n}_{n=0}^{\infty}$ is defined by $a_0=a$ and $a_n=r a_{n-1}$. Then $$a_0=a$$ $$a_1=r a_0=ar$$ $$a_2=r a_1=ar^2$$ $$a_3=r a_2=ar^3$$ $$\cdots$$ We see that $a_n=r a^n$. Let us summarize the formula for a geometric sequence, with strating index $0$ and also with strating index $1$.
Geometric sequence formula
$$\set{a,ar,ar^2,ar^3,\cdots}=\set{ar^n}_{n=0}^{\infty}=\set{ar^{n-1}}_{n=1}^{\infty}$$


An application to fractals - The Sierpinski triangle We start from an equilateral triangle $T_0$ of area $a$, subdivide $T_0$ into 4 equilaterial sub-triangles and remove the triangle one in the center. We are left with the figure $T_1$ consisting of $3$ triangles. Each of the 3 triangles is again subdivided into 4 sub-triagles and the middle ones are removed. This gives the figure $T_2$ consisting of $9$ triangles. The procedure is repeated with each of the 9 triangles, and so on. After repeating the procedure $n$ times, we get the figure $T_n$ made out of $3^n$ (small) equilateral triangles.


The original applet by Michael Ortiz can be viewed here

Let $a_n$ denote the area of $T_n$. Then $a_0=a$, we have the recursion $a_n=\frac{3}{4} a_{n-1}$, and so $$a_n=a\of{\frac{3}{4}}^n$$ We see that as $n\to\infty$, $a_n\To 0$ and so $area(T_n)=a_n\To 0$. The Sierpinski triangle $T$ is the limiting figure. Since $T\subset T_n$ for each $n$, $area(T)\leq area(T_n)$ and therefore $area(T)=0$.