Summary
IPartial Sums of SequencesAThe sigma notationBDefinitions and VocabularyCQuick ShortcutsIIIntroduction to convergent and divergent seriesADefinition and vocabularyBConvergence and divergence of infinite geometric seriesPartial Sums of Sequences
The sigma notation
Sigma notation
The sigma notation is a notation which means "sum up". Given a sequence \left\{ a_{1}, a_{2}, ...,a_{n}\right\}, the sum of the terms of this sequence is :
\sum_{k=1}^{n}a_{k}=a_{1}+a_{2}+a_{3}+\dots+a_{n}
\sum_{k=1}^{4}\left(3k\right)=\left(3\cdot1\right)+\left(3\cdot2\right)+\left(3\cdot3\right)+\left(3\cdot4\right)=30
\sum_{k=7}^{9}\left(k^{2}+1\right)=\left(7^{2}+1\right)+\left(8^{2}+1\right)+\left(9^{2}+1\right)=197
\sum_{p=0}^{6}\left(\sqrt{p}\right)=\sqrt{0}+\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}\approx10.83
\sum_{k=1}^{5}\left(7\right)=\left(7\right)+\left(7\right)+\left(7\right)+\left(7\right)+\left(7\right)=35
Sigma notation is also called summation notation.
Definitions and Vocabulary
If \left\{ a_{1}, a_{2}, a_{3},...,a_{n} \right\} and \left\{ b_{1}, b_{2}, b_{3},...,b_{n} \right\} are two sequences, then:
\sum_{k=1}^{n}\left( a_{k}+b_{k} \right)=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}
If \left\{ a_{1}, a_{2}, a_{3},...,a_{n} \right\} and \left\{ b_{1}, b_{2}, b_{3},...,b_{n} \right\} are two sequences, then:
\sum_{k=1}^{n}\left( a_{k}-b_{k} \right)=\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{n}b_{k}
If \left\{ a_{1}, a_{2}, a_{3},...,a_{n} \right\} is a sequence and if \alpha is a real number, then:
\sum_{k=1}^{n}\left(\alpha\cdot a_{k}\right)=\alpha\cdot\sum_{k=1}^{n}a_{k}
\sum_{k=1}^{n}\left( 5k^{2}+3k-7 \right)=\sum_{k=1}^{n}\left( 5k^{2} \right)+\sum_{k=1}^{n}\left( 3k \right)-\sum_{k=1}^{n}\left(7 \right)=5\sum_{k=1}^{n}\left( k^{2} \right)+3\sum_{k=1}^{n}\left( k \right)-\sum_{k=1}^{n}\left(7 \right)
Quick Shortcuts
The sum of the first n natural numbers is:
1+2+3+\cdots+n=\sum_{k=1}^{n}\left(k\right)= \dfrac{n\left(n+1\right)}{2}
1+2+3+...+250=\sum_{k=1}^{250}\left(k\right)=\dfrac{250\cdot\left(250+1\right)}{2}=\dfrac{250\cdot251}{2}=31{,}375
The sum of the first n squares of natural numbers is:
1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\sum_{k=1}^{n}\left(k^{2}\right)=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}
1^{2}+2^{2}+3^{2}+\cdots+15^{2}=\sum_{k=1}^{15}\left(k^{2}\right)=\dfrac{15\left(15+1\right)\left(2\cdot15+1\right)}{6}=1{,}240
The sum of the first n cubes of natural numbers is:
1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\sum_{k=1}^{n}\left(k^{3}\right)=\left[\dfrac{n\cdot\left(n+1\right)}{2}\right]^{2}
1^{3}+2^{3}+3^{3}+\cdots+20^{3}=\sum_{k=1}^{20}\left(k^{3}\right)=\left[\dfrac{20\cdot\left(20+1\right)}{2}\right]^{2}=44{,}100
If \left\{ a_{1}, a_{2}, ...,a_{n}\right\} is a finite arithmetic sequence with a common difference d, then the sum of the terms of the sequence is:
\sum_{k=1}^{n}a_{k}=a_{1}+a_{2}+a_{3}+\cdots+a_{n}=n\cdot\left( \dfrac{a_{1}+a_{n}}{2} \right)
Consider an arithmetic sequence with a_{1}=3 and a_{20}=98.
Calculate the sum of the first 20 terms of the sequence:
a_{1}+a_{2}+a_{3}+\cdots+a_{20}=20\cdot\left( \dfrac{a_{1}+a_{20}}{2} \right)=20\cdot\left( \dfrac{3+98}{2} \right)=1{,}010
If \left\{ a_{1}, a_{2}, ...,a_{n}\right\} is a finite arithmetic sequence with a common difference d, then the sum of the terms of the sequence is:
\sum_{k=1}^{n}a_{k}=a_{1}+a_{2}+a_{3}+\cdots+a_{n}=\dfrac{n}{2}\cdot\left[2a_{1}+\left(n-1\right)d\right]
Consider an arithmetic sequence with a_{1}=15 and common difference d=10.
Calculate the sum of the first 100 terms of the sequence:
a_{1}+a_{2}+a_{3}+\cdots+a_{100}=\dfrac{100}{2}\cdot\left[2\cdot\left(15\right)+99\cdot\left(10\right)\right]=51{,}000
The sequence of the first n natural numbers \left\{ 1, 2, 3, 4, ..., n \right\} is an arithmetic sequence with first term a_{1}=1 and common difference d=1.
In order to calculate the sum of this sequence, use the formula for the sum of the first terms of an arithmetic sequence:
1+2+3+\cdots+n=\dfrac{n}{2}\cdot\left[2a_{1}+\left(n-1\right)d\right]=\dfrac{n}{2}\cdot\left[2\left(1\right)+\left(n-1\right)\left(1\right)\right]=\dfrac{n}{2}\cdot\left[n+1\right]
This proves the formula that was given for the sum of the first n natural numbers.
Let \left\{ b_{1}, b_{2}, ...,b_{n}\right\} be a finite geometric sequence with a common ratio q\neq1. The sum of the terms of the sequence is:
\sum_{k=1}^{n}b_{k}=b_{1}+b_{2}+b_{3}+\cdots+b_{n}=b_{1}\cdot\dfrac{1-q^{n}}{1-q}
Consider a geometric sequence with b_{1}=2 and common ratio q=3.
We calculate the sum of the first 10 terms of the sequence:
b_{1}+b_{2}+b_{3}+\cdots+b_{10}=2\cdot\dfrac{1-3^{10}}{1-3}=59{,}048
Introduction to convergent and divergent series
Definition and vocabulary
Partial Sums
The partial sum is the sum of a particular number of terms of a given series or sequence. If \left( a_{n} \right) is a given sequence, then the partial sum of the first N terms can be written as:
S_N=\sum_{i=1}^{N}a_{i}=a_{1}+a_{2}+a_{3}+\cdots+a_{N}
Given the sequence:
a_{n}=\dfrac{n}{n+1}
The partial sum of the first term is:
S_{1}=a_{1}=\dfrac{1}{2}
The partial sum of the first two terms is:
S_{2}=a_{1}+a_{2}=\dfrac{1}{2}+\dfrac{2}{3}
The partial sum of the first three terms is:
S_{3}=a_{1}+a_{2}+a_{3}=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}
When we try to add an infinite number of terms of a sequence, we are dealing with an infinite series. As we determine what the terms of a sequence will be, we can calculate the sum of an infinite number of terms which sometimes can be a finite number.
If we try to calculate:
\dfrac{1}{2}+\dfrac{1}{2^{2}}+\dfrac{1}{2^{3}}+\cdots+\dfrac{1}{2^{n}}+\cdots
We observe that:
- \dfrac{1}{2}=0.5
- \dfrac{1}{2}+\dfrac{1}{4}=0.75
- \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}=0.875
- \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}=0.9\ 375
- \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}=0.96\ 875
Therefore, we can assume intuitively that this infinite sum equals 1.
Infinite series
Let \left\{ a_{n} \right\} be a sequence. An infinite series is an expression of the form:
a_{1}+a_{2}+a_{3}+\cdots=\sum_{n=1}^{\infty}a_{n}
Convergent series
An infinite series \sum_{n=1}^{\infty}a_{n} is convergent to the sum S, if its partial sums are convergent to S :
\lim\limits_{N \to\infty }S_{N}=S
In this case, we write:
\sum_{n=1}^{\infty}a_{n}=S
Consider the series:
\sum_{n=1}^{\infty}\dfrac{1}{2^{n}}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...
For this series:
S_{N}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2^{N}}
Calculate S_{N} using the sum of a geometric sequence formula:
S_{N}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2^{N}}=\dfrac{1}{2}\cdot\dfrac{1-\dfrac{1}{2^{N}}}{1-\dfrac{1}{2}}=1-\dfrac{1}{2^{N}}
We have:
\lim\limits_{N \to\infty }S_{N}=\lim\limits_{N \to\infty }\left(1-\dfrac{1}{2^{N}}\right)=1
Therefore, the series converge and:
` \sum_{n=1}^{\infty}\dfrac{1}{2^{n}}=1
Divergent series
Let \{a_n\} be a sequence. Then we say that the series \sum_{n=1}^\infty a_n is divergent if the limit \lim\limits_{N\to \infty}\sum_{n=1}^N a_n is infinite or does not exist.
Consider the series:
\sum_{n=1}^{\infty}{2^{n}}={2}+{4}+{8}+{16}+\cdots
For this series:
S_{N}={2}+2^{2}+2^{3}+ \cdots+{2^{N}}
Calculate S_{N} using the sum of a geometric sequence formula:
S_{N}={2}+2^{2}+2^{3}+ \cdots+{2^{N}}={2}\cdot\dfrac{1-{2^{N}}}{1-{2}}=2\cdot\left(2^{N}-1\right)
We have:
\lim\limits_{N \to\infty }S_{N}=\lim\limits_{N \to\infty }2\cdot\left(2^{N}-1\right)=\infty
Thus the series is divergent to infinity.
Consider the series:
\sum_{n=1}^\infty\left(-1\right)^n=-1+1-1+1-\cdots
Then:
S_N=\begin{cases} 0& \mbox{if } N\mbox{ is even} \cr \cr -1& \mbox{ if } N\mbox{ is odd} \end{cases}
Then the sequence S_N is the sequence which oscillates between -1 and 0. Therefore, the series \sum_{n=1}^\infty\left(-1\right)^n is a divergent series.
Consider the series:
\sum_{n=1}^\infty \dfrac{1}{n}=1+\dfrac{1}{2}+\cdots + \dfrac{1}{n}+\cdots
The above series is a famous example of a divergent series and is called the harmonic series.
Convergence and divergence of infinite geometric series
Geometric series
A geometric series with common ratio q\neq0 is a series defined by a geometric sequence c\cdot q^{n}, where c\neq 0. If the series begins at n=0, then:
S=\sum_{n=0}^{\infty}c\cdot q^{n}=c+cq+cq^{2}+cq^{3}+cq^{4}+\cdots
The geometric series with a constant factor c=7 and ratio q=3 is:
\sum_{n=0}^{\infty}7\cdot 3^{n}=7+7\cdot3+7\cdot3^{2}+7\cdot3^{3}+7\cdot3^{4}+7\cdot3^{5}+\cdots
If \left| q \right|\lt1, then the geometric series is convergent and:
S=\sum_{n=0}^{\infty}c\cdot q^{n}=\dfrac{c}{1-q}
Consider the following series:
\sum_{n=0}^{\infty}5\left(\dfrac{2}{3}\right)^{n}
Since \dfrac{2}{3}\lt1, we know that the series is convergent, and:
\sum_{n=0}^{\infty}5\left(\dfrac{2}{3}\right)^{n}=\dfrac{5}{1-\dfrac{2}{3}}=15
If \left| q \right| \geq1, then the geometric series is divergent.
Consider the following geometric series:
\sum_{n=0}^{\infty}5\cdot\left(-2\right)^{n}=5-5\cdot2+5\cdot2^{2}-5\cdot2^{3}+5\cdot2^{4}-5\cdot2^{5}+\cdots
We have:
\left| q \right|=\left| -2 \right|=2 and 2\geqslant1
Therefore, the series is divergent.
If a geometric series begins at k and if \left| q \right|\lt1, then the geometric series is convergent to \dfrac{cq^{k}}{1-q} :
S=\sum_{n=k}^{\infty}c\cdot q^{n}=cq^{k}+cq^{k+1}+cq^{k+2}+\cdots=\dfrac{cq^{k}}{1-q}
Consider the following series:
\sum_{n=2}^{\infty}7\left(\dfrac{3}{5}\right)^{n}
Since \dfrac{3}{5}\lt1, we know that the series is convergent to S and:
S=\dfrac{7\cdot \left(\dfrac{3}{5}\right)^{2}}{1-\dfrac{3}{5}}=\dfrac{63}{10}
In the above theorem, if k=0, then:
S_{N}=\sum_{n=o}^{N}a_{n}
In this situation, the geometric series is convergent to:
\dfrac{cq^{0}}{1-q}=\dfrac{c}{1-q}
Which is the formula from the previous theorem.