# Section 7.1

:::{prf:definition} Sequence
:label: defSeq

A finite sequence is a function that has a set of natural numbers
of the form $\{1,2,3,...,n\}$ as its domain. An infinity sequence
has the set of natural numbers as its domain.

The general term of the sequence is usually denoted $a_{n}$.

:::

::::{prf:example}
:label: seqExam1

Write the first five terms of the sequence $a_{n}=5n-2$.

:::{dropdown} Solution:

In this case the domain is $\{1,2,3,4,5\}$ and 
\begin{align*}
a_{1} & =5(1)-2=3\\
a_{2} & =5(2)-2=8\\
a_{3} & =5(3)-2=13\\
a_{4} & =5(4)-2=18\\
a_{5} & =5(5)-2=23
\end{align*}

So the sequence $\{a_{i}\}_{i=1}^{5}=\{3,8,13,18,23\}$.

:::
::::

:::{prf:definition} Convergence and Divergence
:label: defConDiv

If the terms of an infinite sequence get closer and closer to some
real number, the sequence is said to be convergent and to converge
to that real number. 

A sequence that does not converge to any number is divergent.

:::

The sequence $a_{n}=\frac{1}{n}$ is a convergent sequence and converges to $0$.

The sequence $b_{n}=n^{2}$ is divergent and continue to get larger and larger without bound.

::::{prf:example}
:label: findSeqExam1

If $a_{1}=2$ and $a_{n}=3\cdot a_{n-1}-1$ for $n\ge1$, then find the first five terms of the sequence $a_{n}$.

:::{dropdown} Solution:

We have

\begin{align*}
a_{1} & =2\\
a_{2} & =3a_{2-1}-1\\
 & =3a_{1}-1\\
 & =3(2)-1\\
 & =5\\
a_{3} & =3(5)-1\\
 & =14\\
a_{4} & =3(14)-1\\
 & =41\\
a_{5} & =3(41)-1\\
 & =122
\end{align*}

:::
::::

:::{prf:definition} Series
:label: defSeries

A finite series is an expression of the form

$$
S_{n}=a_{1}+a_{2}+\dots+a_{n}=\sum_{i=1}^{n}a_{i}.
$$

An infinite series is an expression of the form

$$
S_{\infty}=a_{1}+a_{2}+\dots=\sum_{i=1}^{\infty}a_{i}.
$$

We say the letter $i$ is the index of the summation.

:::

::::{prf:example}
:label: seriesExam1

Evaluate 

$$
\sum_{i=1}^{5}(i^{2}+i+1).
$$

:::{dropdown} Solution:

First we define $a_{i}=i^{2}+i+1$ and we want to find the first $5$ terms of the sequence. 

\begin{align*}
a_{1} & =(1)^{2}+1+1=3\\
a_{2} & =(2)^{2}+(2)+1=7\\
a_{3} & =(3)^{2}+(3)+1=13\\
a_{4} & =(4)^{2}+(4)+1=21\\
a_{5} & =(5)^{2}+(5)+1=31
\end{align*}

Next, we evaluate

\begin{align*}
\sum_{i=1}^{5}(i^{2}+i+1) & =\sum_{i=1}^{5}a_{i}\\
 & =a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\\
 & =3+7+13+21+31\\
 & =75
\end{align*}

Therefore, 

$$
\sum_{i=1}^{5}(i^{2}+i+1)=75.
$$

:::
::::

::::{prf:example}
:label: seriesExam2

Evaluate $\sum_{i=1}^{3}f(x_{i})\Delta x$ where $f(x)=\frac{1}{x}$;
$x_{1}=1$ , $x_{2}=3$, $x_{3}=5$, and $\Delta x=2$.

:::{dropdown} Solution:

First, we will define $a_{i}=f(x_{i})\Delta x$ and we will find the first 3 terms of the sequence

\begin{align*}
a_{1} & =f(x_{1})\Delta x\\
 & =f(1)(2)\\
 & =\frac{1}{1}(2)\\
 & =2\\
a_{2} & =f(x_{2})\Delta x\\
 & =f(3)(2)\\
 & =\frac{1}{3}\cdot2\\
 & =\frac{2}{3}\\
a_{3} & =f(x_{3})\Delta x\\
 & =f(5)(2)\\
 & =\frac{1}{5}\cdot2\\
 & =\frac{2}{5}\\
\sum_{i=1}^{3}f(x_{i})\Delta x & =f(x_{1})\Delta x+f(x_{2})\Delta x+f(x_{3})\Delta x\\
 & =2+\frac{2}{3}+\frac{2}{5}\\
 & =\frac{46}{15}
\end{align*}

Therefore, we have

$$
\sum_{i=1}^{3}f(x_{i})\Delta x=\frac{46}{15}.
$$

:::
::::

:::{prf:property} Summation Properties
:label: sumProper

If $a_{1}$, $a_{2}$, $a_{3}$, ..., $a_{n}$ and $b_{1}$, $b_{2}$,
$b_{3}$, ..., $b_{n}$ are two sequences, and $c$ is a constant, then for every positive integer $n$, the following holds true.

1. $\sum_{i=1}^{n}c=cn$
1. $\sum_{i=1}^{n}ca_{i}=c\sum_{i=1}^{n}a_{i}$
1. $\sum_{i=1}^{n}(a_{i}+b_{i})=\sum_{i=1}^{n}a_{i}+\sum_{i=1}^{n}b_{i}$
1. $\sum_{i=1}^{n}(a_{i}-b_{i})=\sum_{i=1}^{n}a_{i}-\sum_{i=1}^{n}b_{i}$
1. $\sum_{i=1}^{n}i=\dfrac{n(n+1)}{2}$
1. $\sum_{i=1}^{n}i^{2}=\dfrac{n(n+1)(2n+1)}{6}$
1. $\sum_{i=1}^{n}i^{3}=\dfrac{n^{2}(n+1)^{2}}{4}$
:::

::::{prf:example}
:label: sumPropExam1

Evaluate 

$$
\sum_{i=1}^{10}(3i^{2}+5)
$$

:::{dropdown} Solution:

Here we will use the properties of summation
\begin{align*}
\sum_{i=1}^{10}(3i^{2}+5) & =\sum_{i=1}^{10}3i^{2}+\sum_{i=1}^{10}5\\
 & =3\sum_{i=1}^{10}i^{2}+\sum_{i=1}^{10}5\\
 & =3\cdot\frac{n(n+1)(2n+1)}{6}|_{n=10}+5n|_{n=10}\\
 & =3\cdot\frac{(10)((10)+1)(2(10)+1)}{6}+5(10)\\
 & =3\cdot385+50\\
 & =1205
\end{align*}

Therefore, 

$$
\sum_{i=1}^{10}(3i^{2}+5)=1205.
$$

:::
::::