Chapter 2 Sequences and Series
2.1 Sequences
- A Sequence is a list of numbers written in a definite order: \[a_{1}, a_{2}, a_{3}, a_{4}, ..., a_{n}\]
- An element of a sequence is called a term . The number \(a_{1}\) is called the first term, \(a_{2}\) is the second term, and in general \(a_{n}\) is called the \(n^{th}\) term.
- As can be seen in the above sequence, for every positive integer \(n\) there is a corresponding number \(a_{n}\).Therefore a sequence can be defined as a function whose domain is the set of positive integers.
- But we usually write \(a_{n}\) instead of the function notation \(f(n)\) for the value of the function at the number \(n\).
- When dealing with infinite sequences, each term \(a_{n}\) will have a successor \(a_{n+1}\)
- Notation: The sequence \(\{a_{1}, a_{2}, a_{3},...\}\) is also denoted by \[\{a_{n}\}\] or \[\{a_{n}\}_{n=1}^{\infty}\]
- Some sequences can be defined by giving a formula for the \(n\)th term.
- In the following examples we give three descriptions of the sequences:
- by using the preceding notation
- by using the defining formula
- by writing out the terms of the sequence
- Notice that \(n\) doesn’t have to start at 1.
\(\{\frac{n}{n+1}\}_{n=1}^{\infty} \hspace{5em} a_{n} = \frac{n}{n+1}\hspace{5em}\{\frac{1}{2},\frac{2}{3}, \frac{3}{4}, \frac{4}{5},..., \frac{n}{n+1},... \}\)
\(\{\frac{(-1)^n(n+1)}{3^{n}}\}_{n=1}^{\infty} \hspace{5em} a_{n} = \frac{(-1)^n(n+1)}{3^{n}}\hspace{5em}\{-\frac{2}{3},\frac{3}{9}, -\frac{4}{27}, \frac{5}{81},..., \frac{(-1)^n(n+1)}{3^{n}},... \}\)
\(\{\sqrt{n-3}\}_{n=3}^{\infty} \hspace{5em} a_{n} = \sqrt{n-3}, n\geq3\hspace{5em}\{0,1,\sqrt{2}, \sqrt{3},..., \sqrt{n-3},...\}\)
\(\{cos\frac{n\pi}{6}\}_{n=0}^{\infty} \hspace{5em} a_{n} = cos\frac{n\pi}{6}, n\geq 0 \hspace{5em}\{1, \frac{\sqrt{3}}{2},\frac{1}{2}, 0,...,cos\frac{n\pi}{6},...\}\)
2.2 Series
- A series is the sum of a number of terms of a sequence.
- When writing series, the shorthand \(\sum\) notation is used to represent the sum of a number of terms having a common form.
- The series \(f(1)+ f(2)+\dots+ f(n−1)+ f(n)\) can be written as \[\sum_{r=1}^nf(r).\]
2.2.1 Arithmetic sequences and series
A sequence in which each term after the first term is obtained from the preceding term by adding a fixed number (Common difference), is called an arithmetic sequence or Arithmetic Progression.
- The sequence defined by \[u_1=a \text{ and } u_n=u_{n-1} + d \text{ for } n\geq2\] gives \[a,a+d, a+2d,\dots\]
- The \(n\)th term (i.e. the solution) is given by \(u_n = a + (n −1)d\).
This is the arithmetic sequence with first term \(a\) and common difference \(d\)
The arithmetic series with \(n\) terms \[a+(a+d)+(a+2d)+\dots+ [a+(n-1)d]\] has sum \[S_n = \frac{n}{2}(\text{first term} + \text{ last term)}\] \[S_n = \frac{n}{2}(2a+(n-1)d)\]
2.2.2 Geometric sequences and series
- The sequence defined by \[u_1=a \text{ and } u_n=ru_{n-1} \text{ for } n\geq2\] gives \[a,ar, ar^2,\dots\]
- The \(n\)th term is given by \(u_n = ar^{n −1}\).
This is the geometric sequence with first term \(a\) and common ratio \(r\).
The geometric series with \(n\) terms \[a+ar+ar^2+\dots+ ar^{n-1}\] has sum \[S_n = \frac{a(1-r^n)}{1-r} \text{ or } \frac{a(r^n-1)}{r-1} \text{ for } r\neq1\]
Important results relating to the \(\sum\) notation
\[\sum_{r=1}^n\{f(r)+ g(r)\} = \{f(1)+ g(1)\}+ \{f(2)+ g(2)\}+\dots + \{f(n)+ g(n)\}\] \[= \{f(1)+ f(2)+ \dots+ f(n)\}+ \{g(1)+ g(2)+\dots +g(n)\}\] \[= \sum_{r=1}^nf(r)+ \sum_{r=1}^ng(r)\]
\[\sum_{r=1}^naf(r)= af(1)+ af(2)+ \dots+ af(n)\] where \(a\) is a constant. \[=a\{f(1)+ f(2)+ \dots+ f(n)\}\] \[=a\sum_{r=1}^nf(r)\]
\(\sum_{r=1}^nr=\frac{n(n+1)}{2}\)
\(\sum_{r=1}^nr^2=\frac{n}{6}(n+1)(2n+1)\)
\(\sum_{r=1}^nr^3=\frac{n^2}{4}(n+1)^2\) or \(\left[\frac{n(n+1)}{2}\right]^2\)
\(\sum_{r=1}^n1=(1+1+\dots+1) = n\)
2.2.3 Methods of proof
- Mathematical induction
- The difference method
2.2.3.1 Mathematical induction
- Let \(n\) be a natural number. Then the aim is to show that some statement \(P(n)\) involving \(n\) is true for any \(n\).
The following steps are used in Mathematical induction
- Let \(P(n)\) be a statement
- Show that the statement is true for \(P(1)\) and \(P(2)\). (i.e. \(P(n)\) is true for \(n=1\) and \(n=2.\))
- Assume that \(P(k)\) is true (i.e. \(P(n)\) is true for \(n=k\)).
- Show that \(P(k+1)\) follows from \(P(k).\)
2.2.3.2 The difference method
- The process of proof by induction is a powerful mathematical tool. However it has the disadvantage that, in order to employ the method it requires the answer.
- There are, however, direct methods of proof such as the method of differences, or the difference method.
- The difference method can be summarised as follows,
\[\sum_{r=1}^n \left\{f(r)-f(r-1) \right\} = f(n) - f(0)\]
where \(f\) is any function suitably defined on the non-negative integers.
- This is also know as the fundamental theorem of summation:
2.3 Infinite Sequences
- When dealing with infinite sequences, each term \(a_{n}\) will have a successor \(a_{n+1}\)
Definition: Convergence and Divergence
A sequence \(\{a_n\}\) has the limit \(L\) and we write
\[\lim_{n\rightarrow \infty}a_n=L \text{ or } a_n \rightarrow L \text{ as } n\rightarrow \infty\]
if we can make the term \(a_n\) as close to L as \(n\) becomes sufficiently large.
If \(\lim_{n\rightarrow \infty}a_n\) exists, we say the sequence converges (or is convergent).
If \(\lim_{n\rightarrow \infty}a_n\) does not exist, we say the sequence diverges (or is divergent).
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2.3.1 Limit Laws for Sequences
If \(\{a_n\}\) and \(\{b_n\}\) are convergent sequences and \(c\) is a constant, then
\(\lim_{n \to \infty}(a_n+ b_n)= \lim_{n \to \infty}a_n+ \lim_{n \to \infty}b_n\)
\(\lim_{n \to \infty}(a_n- b_n)= \lim_{n \to \infty}a_n- \lim_{n \to \infty}b_n\)
\(\lim_{n \to \infty}ca_n= c \;\lim_{n \to \infty}a_n\)
\(\lim_{n \to \infty}c= c\)
\(\lim_{n \to \infty}(a_nb_n)= \lim_{n \to \infty}a_n. \lim_{n \to \infty}b_n\)
\(\lim_{n \to \infty}\frac{a_n}{b_n}= \frac{\lim_{n \to \infty}a_n} {\lim_{n \to \infty}b_n} \text{ if } \lim_{n \to \infty}{b_n} \neq 0\)
\(\lim_{n \to \infty}(a_n)^p= [\lim_{n \to \infty}a_n]^p \text{ if } p>0 \text{ and } a_n>0\)
Squeeze Theorem for Sequences
If \(a_n \leq b_n \leq c_n\) for \(n\geq n_0\) and \(\lim_{n \to \infty}a_n= \lim_{n \to \infty}c_n= L,\) then \(\lim_{n \to \infty}b_n=L\)
Theorem:
\[\text{If }\;\; \lim_{n \to \infty}|a_n|=0, \text{ then } \;\;\lim_{n \to \infty}a_n=0.\]
Theorem
The sequence \(\{r^n\}\) is convergent if \(-1 <r\leq1\) and divergent for all other values of \(r\).
\[\begin{equation} \lim_{n \to \infty}r^n= \begin{cases} 0 & \text{if } -1<r<1\\ 1 & \text{if } r=1 \end{cases} \end{equation}\]
Definition
A sequence \(\{a_n\}\) is called increasing if \(a_n<a_{n+1}\) for all \(n \geq 1,\) that is, \(a_1<a_2<a_3<\dots.\) It is called decreasing if \(a_n>a_{n+1}\) for all \(n\geq 1.\) It is called monotonic if it is either increasing or decreasing.
Definition
If there exists a number \(m\) such that \(m \leq a_n\) for every \(n\) we say the sequence is bounded below. The number \(m\) is sometimes called a lower bound for the sequence.
If there exists a number \(M\) such that \(a_n \leq M\) for every \(n\) we say the sequence is bounded above. The number \(M\) is sometimes called an upper bound for the sequence.
If the sequence is both bounded below and bounded above we call the sequence bounded.
2.4 Infinite Series
- Consider the infinite sequence \(\{a_n\}_{n=1}^\infty\).
- Consider the partial sums
\[\begin{align*} s_1=a_1 \\ s_2=a_1+a_2 \\ s_3=a_1+a_2+a_3 \\ s_4=a_1+a_2+a_3+a_4 \\ &\vdots\\ s_n=a_1+a_2+a_3+a_4+\dots +a_n =\sum_{i=1}^na_i \end{align*}\]
These partial sums form a new sequence \(\{s_n\}_{n=1}^\infty,\) which may or may not have a limit.
Consider the limit of the sequence of partial sums, \(\{s_n\}_{n=1}^\infty.\)
\[\lim_{n\to\infty}s_n= \lim_{n\to\infty}\sum_{i=1}^na_i=\sum_{i=1}^\infty a_i\]
- This \(\sum_{i=1}^\infty a_i\) (\(=a_1+a_2+a_3+a_4+\dots +a_n \dots\)) is called an infinite series
Definition
Consider the series \(\sum_{n=1}^\infty a_n= =a_1+a_2+a_3+\dots.\) Let \(s_n\) denote its \(n\)th partial sum:
\[s_n= \sum_{i=1}^n a_i= a_1+a_2+\dots +a_n\]
If the sequence \(\{s_n\}\) is convergent and \(\lim_{n\to \infty}s_n=s\) exist as a real number, then the series \(\sum_{i=1}^\infty a_i\) is called convergent and we write,
\(a_1+a_2+\dots +a_n+\dots=s\) or \(\sum_{n=1}^\infty a_n=s\)
The number \(s\) is called the sum of the series. If the sequence of partial sums is divergent then the infinite series is also called divergent.
Geometric Series
- An important example of an infinite series is the geometric series
\[\sum_{n=1}^\infty ar^{n-1}=a+ar+ar^2+\dots\] is convergent if \(|r|<1\) and its sum is \[\sum_{n=1}^\infty ar^{n-1}= \frac{a}{1-r}\qquad\qquad|r|<1\]
If \(|r|\geq1,\) the geometric series is divergent.
Theorem
If the series \(\sum_{n=1}^\infty a_n\) is convergent, then \(\lim_{n\to \infty}a_n=0.\)
Proof
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Note 1
- Consider any series \(\sum a_n.\)
This associates two sequences
- the sequence \(\{s_n\}\) of its partial sums
- the sequence \(\{a_n\}\) of its terms.
If \(\sum a_n\) is convergent, then
- the limit of the sequence \(\{s_n\}\) is \(s\) (the sum of the series)
- the limit of the sequence \(\{a_n\}\) is 0.
Note 2 - The converse of the above theorem is not true in general.
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The Test for Divergence
If \(\lim_{n\to \infty}a_n\) does not exist or if \(\lim_{n\to \infty}a_n \neq0\), then the series \(\sum_{n=1}^\infty a_n\) is divergent.
Theorem
If \(\sum a_n\) and \(\sum b_n\) are convergent series, then
\(\sum_{n=1}^\infty ca_n= c\sum_{n=1}^\infty a_n \;\;\) \(c\) is a constant.
\(\sum_{n=1}^\infty (a_n+b_n)= \sum_{n=1}^\infty a_n+ \sum_{n=1}^\infty b_n\)
\(\sum_{n=1}^\infty (a_n-b_n)= \sum_{n=1}^\infty a_n- \sum_{n=1}^\infty b_n\)
Reading:
Stewart, J., Clegg, D. K., & Watson, S. (2020). ‘Infinite Sequences and Series’, Calculus: early transcendentals. Cengage Learning.
Acharjya, D. P. (2009). Fundamental approach to discrete mathematics. New Age International.
Tutorial
- Find a formula for the general term \(a_{n}\) of the sequence, assuming that the pattern of the first few terms continues.
- \(\left\{\frac{3}{5},-\frac{4}{25}, \frac{5}{125}, -\frac{6}{625}, \frac{7}{3125},...\right\}\)
- \(\{1,1,2,3,5,8,13,21,\dots\}\)
- \(\left\{\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \dots \right\}\)
- \(\left\{2,7,12,17, \dots \right\}\)
- \(\left\{1,-\frac{2}{3}, \frac{4}{9} , -\frac{8}{27}, \dots \right\}\)
- Show that \[\sum_{r=1}^n(6r^2+4r-1) = n(n+2)(2n+1)\]
- Show by method of induction
- \(\sum_{r=1}^nr=\frac{n(n+1)}{2}\)
- \(\sum_{r=1}^nr^2=\frac{n}{6}(n+1)(2n+1)\)
- \(\sum_{r=1}^nr^3=\left[\frac{n(n+1)}{2}\right]^2\)
- \(\sum_{r=1}^n\frac{1}{r(r+1)}=\frac{n}{n+1}\)
By considering \(n^3 − (n − 1)^3\) and similar expressions, find the formula for \(\sum_{r=1}^nr^2\) in terms of \(n\), assuming the results for \(\sum_{r=1}^nr\)
By considering \(n^5 − (n − 1)^5\) and similar expressions, find the formula for \(\sum_{r=1}^nr^4\) in terms of \(n\), assuming the results for \(\sum_{r=1}^nr\), \(\sum_{r=1}^nr^2\) and \(\sum_{r=1}^nr^3.\)
- Determine whether the sequence \(a_n= (-1)^n\) is convergent or divergent.
- Determine whether the sequence converges or diverges. If it converges, find the limit
- \(a_n= n(n-1)\)
- \(a_n= \frac{n}{n+1}\)
- \(a_n = \frac{(-1)^n}{n}\)
- \(a_n = \frac{4n^2+2}{n^2+n}\)
- \(a_n= \frac{2^n}{3^{n+1}}\)
- \(a_n= \frac{(-1)^{n-1}n}{n^2+1}\)
- Discuss the convergence of the sequence \(a_n=n!/n^n,\) where \(n!= 1\times2\times\dots\times n.\)
- Determine if the following sequences are monotonic and/ or bounded
- \(\{-n^2\}_{n=0}^\infty\)
- \(\{(-1)^{n+1}\}_{n=1}^\infty\)
- \(\left\{ \frac{2}{n^2}\right\}_{n=5}^\infty\)
- Determine if the following series converges or diverges. If it converges determine its value.
- \(\sum_{n=1}^\infty n\)
- \(\sum_{n=1}^\infty (-1)^n\)
- \(\sum_{n=1}^\infty \frac{1}{3^{n-1}}\)
- \(5-\frac{10}{3}+\frac{20}{9}- \frac{40}{27}+\dots\)
- \(\sum_{n=1}^\infty 2^{2n}3^{1-n}\)
- \(\sum_{n=1}^\infty x^n,\) where \(|x| <1\)
- \(\sum_{n=1}^\infty\frac{1}{n(n+1)}\)
- Show that harmonic series \[\sum_{n=1}^\infty \frac{1}{n} = 1+\frac{1}{2}+ \frac{1}{3} + \frac{1}{4}+\dots\] is divergent.
Show that the series \(\sum_{n=1}^\infty\frac{n^2}{5n^2+4}\) diverges.
Determine whether the series is convergent or divergent. If it is convergent, find its sum.
- \(\sum_{n=1}^\infty\left(\frac{3}{n(n+1)+ }+\frac{1}{2^n}\right)\)
- \(\sum_{n=1}^\infty\left(\frac{3}{n(n+3)+ }+\frac{5}{4^n}\right)\)
- \(\sum_{n=1}^\infty \frac{1}{n^2+3n+2}\)
- \(\sum_{n=1}^\infty \frac{1}{n^2+4n+3}\)
- \(\sum_{n=1}^\infty 9^{-n+2}4^{n+1}\)
- \(\sum_{n=1}^\infty \frac{(-4)^{3n}}{5^{n-1}}\)
- \(\sum_{n=1}^\infty \left(\frac{1}{n^2+4n+3}- 9^{-n+2}4^{n+1}\right)\)