Given a sequence $\{a _n\}$, the expression $a_1+a_2+a_3+\ldots \ldots$ is called a Series

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

If $\{a_n\}$ is finite the corresponding Series is finite. If $\{a_n\}_{n=1}^{\infty}$ is infinite then corresponding series is infinite.

$a_1+a_2+\cdots+a_n \ \ \rightarrow$ finite series.

$a_1+a_2+ \cdots \ \ \rightarrow$ infinite series

Consider the Sequence $\{a_ n\}_{n=1}^{\infty}.$

We construct a new sequence $\{S_ n\}_{n=1}^{\infty}$ as follows,

$S_1=a_1$

$S_2=a_1+a_2$

$S_3=a_1+a_2+a_3$

$S_ n=a_ 1+a_ 2+\cdots+a_ n$ $\quad \quad =\sum_ {i=1}^n a_i$

$\vdots$

The sequence ${S_n}$ is called sequence of Partial sums of the series $\sum_{n=1}^{\infty} a_n.$

$S_n \rightarrow n^{\text {th }}$ partial sum

$S_n=a_1+a_2+\cdots+a_n$

What will be happen if $n$ becomes larger and larger?

we observe convergence or divergence of $\left(S_n\right)_{n=i}^{\infty}$

For a given sequence $\{a_n\}$ and corresponding series $\sum_{n=1}^{\infty} a_n$, construct sequence $\left(S_n\right)_{n=1}^{\infty}$, the sequence of partial sum.

If $(S_n)^{\infty}$ is convergent, then we say that $\sum_{n=1}^{\infty} a_n$ is convergent.

If $\underset{n \rightarrow \infty}{\operatorname{Lt}} S_n=L$

then $\sum_{n=1}^{\infty} a_n=L$

Sequence - what happens as you progress towards end.

Series "Summable"

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

$\sum a_n$ arises from the sequence $\left(\frac{1}{2^n}\right)_{n=0}^{\infty}$

$S_1=a_1=1$

• $S_2=a_1+a_2=1+\frac{1}{2}=\frac{3}{2}$

• $S_3=a_1+a_2+a_3=1+\frac{1}{2}+\frac{1}{4} =\frac{3}{2}+\frac{1}{4}=\frac{7}{4}$
  $\vdots$

$S_n=1+\frac{1}{2}+\frac{1}{3}+\quad+\frac{1}{n}$

$\left(S_n\right)_{n=1}^{\infty}$

$\begin{aligned} S_3 & =1+\frac{1}{2}+\frac{1}{4} \\ & =2-\frac{1}{4}\end{aligned}$

$S_4=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=2-\frac{1}{8}$

$\vdots$

$S_n=\frac{2^n-1}{2^{n-1}}$

$=2-\frac{1}{2^{n-1}} \ \ n= 1,2,3,\cdot\cdot\cdot$

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+ \cdot\cdot\cdot$

The sequence of partial sum $\left(S_n\right)_{n=1}^{\infty}$

$S_n:=2-\frac{1}{2^{n-1}}$

$\forall n \in \mathbb{N}$

$\operatorname{Lt}_{n \rightarrow \infty} \quad S_n=2$

$\therefore 1+\frac{1}{2}+\frac{1}{4}+\cdots \ \ \text { is convergent }$

$\begin{array}{r}1+\frac{1}{2}+\frac{1}{4}+\cdots =2\end{array}$

$2,4,6,8,10, \ldots 2 n,$

$5,10,15,20,25, \ldots$

$4,3,2,1,0,-1\ldots$

A sequence in which difference between two consecutive terms remains the same called an Arithmetic Sequence or (AP)

A sequence $\{a_ n \}_{n=1}^{\infty}$ is called arithmetic- sequence or AP if,

$a_{n+1}=a_n+d \quad \forall n \geqslant 1$

where "$d$" is a real number.

Here " $d$ " is called common difference.

If we add a constant to each term in an AP, the resulting Sequence is again an AP

$\left(a_n\right)_{n=1}^{\infty}$

$a_1, a_2, a_3, \ldots$

$\left(b_n\right)_{n=1}^{\infty} \quad b_n=a_n+d^{\prime}$

$a_1+d^{\prime}, a_2+d^{\prime}, a_3+d^{\prime} \cdots$