Sequence

$f : S \longrightarrow {R}$

• $S \subseteq \mathbb{N} \cup \{0\}$
• $\{a_n\}_{n=1}^{\infty} \quad a_n=f(n)$

$(a_n)_{n=1}^{\infty} \text { or }$

• $\{a_1, a_2, a_3, \ldots a_n \cdot \ldots \}$

Closed form expression

• $\{a_ n\}_ {n=1}^{\infty} \quad a_n=\frac{1}{n^2} \forall n \geqslant N$

• e.g. Fibonacci sequence

1) $(\frac{1}{n^2})_{n=1}^{\infty}$

$1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16} \cdots \frac{1}{n^2}, \cdots$

$\underset{n\rightarrow\infty}{Lim} \frac{1}{n^2}=0$

2) $\{(-1)^n\}_{n-2}^{\infty}$

$1, -1, 1, -1, ........$

3) $\{n^2\}_{n=1}^{\infty}$

$a_1, a_2, a_3 -\text { real no's }$

$a_1+a_2+a_3 = a_2+a_1+a_3 = a_3+a_2+a_1$

For instance sum of 247, 198 and 2

$198 + 2 = 200$

$198+2+247 = 200 + 247 = 447$

Decimal expansion

$\frac{10}{3} = 3.333...........$

$3 + \frac{3}{10} + \frac{3}{10^2} + \frac{3}{10^3} + \cdots = \frac{10}{3}$

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

$n^{th} \$ Summand: $\frac{1}{2^n}$

Observation

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

$1+2+3+4+ \cdots$ cannot be finite.

$1 + (\frac{-1}{2}) + ( \frac{1}{3}) + (-\frac{1}{4}) + \cdots$

$[1+(\frac{-1}{2})] + [\frac{1}{3} + (-\frac{1}{4})] + \cdots$

Suppose we rearrange the terms of given series as follows:

$(-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+1)$

$(-\frac{1}{10}-\frac{1}{12}-\frac{1}{14}-\frac{1}{16}-\frac{1}{18} + \frac{1}{3})$

$1+(-\frac{1}{2})+(\frac{1}{3})+(-\frac{1}{4}) +\cdots$

$[1+(-\frac{1}{2})]+[\frac{1}{3}+(-\frac{1}{4})] + \cdots$

$a_1, a_2 \cdots a_n$

$a_1+a_2+a_3 \cdots + a_n$

Sigma notation, $\sum$

$a_1+a_2+ \cdots +a_n = \underset {i=1} {\overset{n} \sum} a_i$

• $i-\text { index of summation }$
• $a_1+a_2+ \cdots + a_n$

$= \underset{i=1}{\overset{n}\sum} a_i = \underset{j=1}{\overset{n}\sum} a_j = \underset{r=1}{\overset{n}\sum} a_r$

Consider a sequence $\{a_r\}_{r=1}^{\infty}$

$a_m, a_{m+1} \cdots a_n$

i.e.

$a_m + a_{m+1} + \cdots + a_n = \underset{i=m}{\overset {n}\sum} a_i$

Find $\ \underset{j=1} {\overset{5} \sum} j^2$

Solution:

$\underset{j=1} {\overset{5} \sum} j^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2$

$= 1+4+9+16+25$

$= 55$

Find $\underset{r=1}{\overset{8} \sum} (-1)^r$

Solution:

$\underset{r=1}{\overset{8} \sum} (-1)^r = (-1+1)+(-1+1)+(-1+1)+(-1+1)$

$\quad \quad = 0$

Given a sequence $\{a_n\}$, the expression, $a_1+a_2+a_3+ \cdots$

is called a series.

associated with the sequence $\{a_n\}.$

Sequence -ordered list of numbers

Series - sum