Sequences and series

This is a quick summary of sequences and series for a refresh purpose only.

We will explore various situations so we can get up and running in problems involving series.

When we wake up in the morning, if we write down on a hourly basis all events happening during the day, we end up having a sequence of these events.

In a sequence we will enumerate 1,2,3,4, …..

We can even graph the sequence as it is a function.

We can summarize:

A sequence can be defined as a function having a domain in the set of positive numbers.

In the following sequence:

$\frac{1}{2}$ , $\frac{1}{4}$ , $\frac{1}{6}$ , $\frac{1}{8}$ ,….

We can see numbers in an ordered list. We call them $terms$ of our sequence.

Taking $f(n)=\frac{1}{2n}$ and giving $n$ values 1,2,3,4,….. We get the sequence shown above.

$f(1)=\frac{1}{2}$

$f(2)=\frac{1}{4}$

……………………..

$f(n)=\frac{1}{2n}$

The formula for the $nth$ term as above in $f(n)=\frac{1}{2n}$ is called the $General \quad term$ of the sequence.

The factoral notation:

if $n \geq 0$ is an integer, the factorial notation $n!$ is:

$0!=1$

$1!=1$

$n!=n(n-1)(n-2)....2\times1$ for $n\geq2$

Please note that:

$n!=n(n-1)!$

Examples

$3!=3\times2 \times1=6$

$5!=5\times 4 \times 3!=20\times 6=120$

Recursive formula:

We may be presented a situation where we have to find the next value based on previous value and a rule given to us. This rule is called the recursive formula.

Example:

$U_{1}=4$

$U_{n}=nU_{n-1}$

We have to know the previous term in order to find the next term.

In this example:

$U_{2}=2U_{1}=8$

$U_{3}=3U_{2}=3\times 8=24$

$U_{4}=4U_{3}=4\times 24=96$

See the picture?

Fibonacci Sequence:

$U_{1}=1$

$U_{2}=1$

$U_{n}=U_{n-2}+U_{n-1}$

We get the following:

$1,1,2,3,5,8,13 \cdots$

Summation or Sigma notation:

We cannot discuss sequences without using the powerful sigma notation:

The following sequence:

$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+\cdots+x_{n}$

Can be written using the sigma notation.

If $k$ is a variable going from $1$ to $n$ , we can say that this is the sum of $x_{k}$ when $k$ goes from $1$ to $n$

We can write:

$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+\cdots+x_{n}=\sum_{k=1}^{n}x_{k}$

Example:

$\sum_{k=1}^{n}x_{k}^{2}=1^{2}+1^{2}+2^{2}+3^{2}+4^{2}+\cdots+n^{2}$

In these formulas, $k$ is the $Index$ .

Properties of Sequences:

If $a^{n}$ and $b_{n}$ are seuqences and $c$ is a real number, we have the following THEOREM:

$\sum_{k=1}^{n}(ca_{k})=ca_{1}+ca_{2}+ca_{3}+ca_{4}+\cdots+ca_{n}=c\sum_{k=1}^{n}a_{k}$

$\sum_{k=1}^{n}(a_{k}+b_{k})=\sum_{k=1}^{n}a_{k}+\sum_{k=1}^{n}b_{k}$

$\sum_{k=1}^{n}(a_{k}-b_{k})=\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{n}b_{k}$

$\sum_{k=j+1}^{n}a_{k}=\sum_{k=1}^{n}a_{k}-\sum_{k=1}^{j}a_{k}$

where $0<j<n$

Some important sequences and series:

Calculating

$\sum_{k=1}^{n}k$

We have:

$(k+1)^{2}=k^{2}+2k.1+1$

$(k+1)^{2}-k^{2}=2k.1+1$

For various values of $k$

$2^{2}-1^{2}=2.1+1$

$3^{2}-2^{2}=2.2+1$

$4^{2}-3^{2}=2.3+1$

……………………………

$(n+1)^{2}-n^{2}=2.n+1$

——————————————————————–

$(n+1)^{2}-1=2(1+2+3+4+\cdots +n)+n$

$(n+1)^{2}-1=2\sum_{k=1}^{n}k+n$

$n^{2}+2n+1-1=2\sum_{k=1}^{n}k+n$

$n^{2}+2n=2\sum_{k=1}^{n}k+n$

$\sum_{k=1}^{n}k=\frac{n^{2}+2n-n}{2}$

$\sum_{k=1}^{n}k=\frac{n^{2}+n}{2}$

$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

This is the first relation:

$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$

We can prove

$\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$

Using:

$(k+1)^{3}-k^{3}=3k^{2}+3k+1$

$\sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$

Or, one more exponent:

$\sum_{k=1}^{n}k^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$

Arithmetic sequences:

We have already seen that a sequence is a list of numbers in a particular order.

Each number is called a term.In a arthmetic sequence, each term after the first is found by adding a constant value called common difference to the previous term.

The first term is noted $a_{1}$

The common difference is noted $d$ .

Looking at the natural numbers: $1,2,3,4,5\cdots n$

In this sequence, $a_{1}=1$ and $a_{n}=n$ with the common difference $d=1$
If we develop:

$a_{1}=a_{1}$

$a_{2}=a_{1}+d$

$a_{3}=a_{2}+d=a_{1}+d+d=a_{1}+2\cdot d$

$a_{4}=a_{3}+d=a_{2}+d=a_{1}+3\cdot d$

………………………………

$a_{n}=a_{1}+(n-1)d$

We can now see that the $nth$ term can be noted:When $n$ is a positive integer:

$a_{n}=a_{1}+(n-1)d$

EXAMPLE 1:

Given the sequence:

$100,96,92,\cdots$ , find the $7th$ term.

$a_{1}=100 <img src="https://www.mouctar.org/wp-content/ql-cache/quicklatex.com-a6b1deab9a3f709012df6c34e40776ec_l3.png" height="93" width="584" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[d=96-100=-4$ $a_{n}=a_{1}+(n-1)d$ $a_{7}=100+(7-1)(-4)=100-24=76$ EXAMPLE 2: Given the sequence: $5,13,21,\cdots$, find the $nth$ term. $a_{1}=5\]" title="Rendered by QuickLaTeX.com"/>d=13-5=21-13=8$

$a_{n}=a_{1}+(n-1)d$

$a_{n}=5+(n-1)(8)=5+8n-8=8n-3$

Arithmetic means is simply the terms between any non consecutive terms of an arithmetic sequence.

Example:

33 and 35 are two arithmetic means between 31 and 37 in the arithmetic sequence $31,33,35,37, \cdots$

Arithmetic series:

Arithmetic series are merely the sum of the terms of an arithmetic sequence.

Arithmetic Sequence: $14,10,6,2$

Arithmetic Series: $14+10+6+2$

We have already seen some arithmetic series.

For the first $n$ terms of an Arithmetic series:

The sum $S_{n}$ is:

$S_{n}=\frac{n(a_{1}+a_{n})}{2}$

Proof:

$S_{n}=1+2+3+4+\cdots+(n-2)+(n-1)+n$

$S_{n}=n+(n-1)+(n-2)+\cdots+4+3+2+1$

————————————

$2S_{n}=(n+1)\times n$

$s_{n}=\frac{n(n+1)}{2}$ with $a_{1}=1$ and $a_{n}=n$

Given:

$a_{1}=24$

$a_{n}=114$

$s_{n}=414$

Find $a_{3}$ and $a_{5}$

Solution:

$S_{n}=\frac{n(a_{1}+a_{n})}{2}$

$414=\frac{n(24+114)}{2}$

$414=n(12+57)$

$69n=414\Rightarrow n=\frac{414}{69}=6$

Now for the common difference: $a_{n}=a_{1}+(n-1)d$

$114=24+(6-1)d$

$5d=114-24$

$5d=90$

$d=\frac{90}{5}$

$d=18$

Now we can find any term:

$a_{3}=24+2\times18=60$

$a_{5}=24+4\times 18=96$

We can actually write the full sequence:

$24,42,60,78,96,114$

Geometric sequences:

In a geometric sequence, each term after the first is found bu multiplying the previous term by what we call the common ratio $r$ .

We note:

$a_{n}=r \cdot a_{n-1}$

The common ratio:

$r=\frac{a_{n}}{a_{n-1}}$

The same way we found the $nth$ term for the arithmetic sequence, the can find the one for the geometric sequence:

$a_{n}=a_{1} \cdot r^{n-1}$ with $n$ any positive integer.

Given:

$2,8,32,128,\cdots$

Suggest an equation for $a_{n}$

Solution:

$r=\frac{8}{2}=\frac{128}{32}=4$

Equation:

$a_{n}=2\times 4^{n-1}$

Given:

$405,135,45, \cdots$

Find the next two terms

Solution

$r=\frac{45}{135}=\frac{1}{3}$

The next two terms:

$45\times \frac{1}{3}=15$

$15\times \frac{1}{3}=5$

Now we have:

$405,135,45, 15,5 \cdots$

Geometric Means

The missing terms between two nonconsecutive terms of a geometric sequence are called geometric means.

Geometric series

Geometric series is the sum of the terms of a geometric sequence.

$S_{n}=a_{1}+a_{1}r+a_{1}r^{2}+a_{1}r^{3}+ \cdots +a_{1}r^{n-1}$

$rS_{n}=a_{1}r+a_{1}r^{2}+a_{1}r^{3}+ \cdots +a_{1}r^{n-1}+a_{1}r^{n}$

If we subtract:

$(1-r)S_{n}=a_{1}-a_{1}r^{n}$ All the rest cancels out.

Finally we can write: For $n$ terms: with $r\neq 1$

$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$

Calculate:

$\sum_{n=1}^{6}5\cdot 2^{n-1}$

In here the first term is:

$a_{1}=5$

$a_{6}=160$

$r=2$

$S_{6}=\frac{5(1-2^{6})}{1-2}=315$

Question:

Find the first term of a geometric series with $S_6=2548$ and a common ratio of 3.

Solution:

$S_{6}=\frac{a_{1}(1-r^{6})}{1-r}$

$S_{6}=\frac{a_{1}(1-3^{6})}{1-3}$

$2548=\frac{a_{1}(1-3^{6})}{1-3}$

$2548\times (-2)=a_{1}(-728)$

$a_{1}=\frac{2548\times (-2)}{-728}$

$a_{1}=7$

Infinite Geometric Series

Series of the following form are called infinite series:

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+ \cdots$

We can only get partial sums here.

However, when $-1<r<1$ , when $n$ becomes very large we have:

$S_{n}=\frac{a_{1}(1-r^{n})}{1-r}=\frac{a_{1}}{1-r}-\frac{a_{1}r^{n}}{1-r}$

The second term Will be very close to $0$ and has a limit of $0$ as $n$ approaches $\infty$

We have the final formula:

$S_{n}=\frac{a_{1}}{1-r}$

With $-1<r<1$

Given:

$Z=\sum_{n=1}^{\infty}24(-\frac{1}{5})^{n-1}$

Find $Z$ :

We can see that:

$a_{1}=24$

$r=-\frac{1}{5}$

$Z=\frac{24}{1-(-\frac{1}{5})}$

Finally: $Z=20$

Please note that some series with alternating signs do not converge.

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Sequences and series

Sequences and series

The factoral notation:

Properties of Sequences:

Some important sequences and series:

Arithmetic sequences:

Arithmetic series:

Geometric sequences:

Geometric Means

Geometric series

Infinite Geometric Series

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