Sequence $\{a_n\} _{n=1}^{\infty}$

$= \{a_1, a_2, a_3 \ldots . a_n^n \ldots 1\}$

we mean a function

$f: \mathbb{N} \longrightarrow \mathbb{R}$

consider the sequence

$\{a_ n\}_{n=1}^{\infty}$

Consider $\{a_ n\}_ {n=1}^{\infty}; \ a_n=\sqrt{n}$

Consider $\{b_ n\}_ {n=1}^{\infty} \ b_n=\frac{1}{n}$

Write first five terms of the sequence given by the formula,

$a_n=n(n+2)$

Solution:

$\begin{aligned} & a_1=1(1+2)=3 \\ & a_2=2(2+2)=2 \cdot 4=8 \\ & a_3=3(3+2)=3 \cdot 5=15 \\ & a_4=4(4+2)=4 \cdot 6=24 \\ & a_5=5(5+2)=5 \cdot 7=35\end{aligned}$

$\{3,8,15,24,35, \cdots , \cdots, n(n+2), \cdots \}$

Find first 4 terms of the sequence given by $a_n=n \cdot \frac{n^2+5}{4}$

Solution:

$\begin{aligned} & a_1=1 \cdot \frac{1^2+5}{4}=1 \cdot \frac{6}{4} =\frac{3}{2} \\ & \begin{aligned} a_2 & =2 \cdot \frac{2^2+5}{4}=2 \cdot \frac{4+5}{4} \\ & =2 \cdot \frac{9}{4}=\frac{9}{2}\end{aligned} \\ & a_3=3 \cdot \frac{3^2+5}{4}\end{aligned}$

$a_3=3 \cdot \frac{9+5}{4}$

$\ =3 \cdot \frac{14}{4}$

$\ =3 \cdot \frac{7}{2} =\frac{21}{2}$

$a_4 =4 \cdot \frac{4^2+5}{4}$

$\ =4 \cdot \frac{16+5}{4}$

$\ =4 \cdot \frac{21}{4}$

$=21$

$\frac{3}{2}, \frac{9}{2}, \frac{21}{2}, 21$

Find $9^{\text {th }}$ term of the sequence $\{a_n\} _{n=1}^{\infty}$

$a_n=(-1)^{n-1} \cdot n^3$

Solution:

$a_9=(-1)^{9-1} \cdot 9^3$

$\begin{aligned} & =(-1)^8 \cdot 9^3 \ & =81 \times 9 \ & =729\end{aligned}$

First three terms of the sequence $\{a_n\} _{n=1}^{\infty}$

$a_ n = a_{n-1} - 1, n > 2$

$a_1=a_2=2$

Solution:

$\begin{aligned} & a_1=2, a_2=2 \text { given } \\ & a_3=a_{3-1}-1=a_2-1 \\ & =2-1 \\ & =1 \\ & \end{aligned}$

Find first 4 terms of the sequence $\{a_n\} _{n=1}^{\infty}$, where

$\begin{aligned}& a_1=3 \\ & a_n=3 a_{n-1} \quad n \geqslant 2 \end{aligned}$

Solution:

$\begin{aligned} & a_1=3 \quad \text { given } \\ & a_2=3 a_1=3 \cdot 3=3^2 \\ & a_3=3 a_2=3 \cdot 3^2=3^3 \\ & a_4=3 a_3=3 \cdot 3^3=3^4\end{aligned}$

• $\quad \quad \longrightarrow$ "closed form" expression for $a_n$.

Consider the sequence $\{a_n\}_{n=i}^{\infty}$ where $\quad a_n=\frac{1}{n}$

$1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}$

{$0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \cdots \frac{n-1}{n}, \cdots$}

{$0, \frac{1}{2}, \frac{2}{3}, \cdots 1-\frac{1}{n}, \cdots.$}

$\{\sqrt{n}\}_{n=1}^{\infty}$

$\{1, \sqrt{2}, \sqrt{3}, \cdots \sqrt{n}, \cdots \}$

$\begin{aligned} & a_{10001}=\sqrt{10001} \\ &>100 . \\ & a_m>k\end{aligned}$

$\{1,-1,1,-1, \cdots(-1)^{n+1}, \cdots\}$

Consider the sequence $\{a_n \}_{n=i}^{\infty}$

where $a_n = \frac{1}{n}$

1, $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$,

${0, \frac{1}{2}, \frac{2}{3}, \cdots . .1- \frac{1}{n}},.$.

A sequence $\{a_n \} _{n=1}^{\infty}$ is said to be convergent if as " $n$ " increases $a_n$ 's become sofficiently close to a number $"L".$

$L$ is called limit of the sequence.

$\operatorname{Lt}_ {n \rightarrow \infty} a_n=L$

Consider the

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

$a_n=\frac{5}{n^2}$

$\{a_ n\}_{n=1}^{\infty}$

$5, \frac{5}{2^2}, \frac{5}{3^2}, \frac{5}{4^2} \cdots$

$\operatorname{Lt}_{n \rightarrow \infty} \frac{5}{n^2}=0$

$a_n=\frac{4-7 n^6}{n^6+3}$

$a_n=\frac{n^6\left(\frac{4}{n^6}-7\right)}{n^6\left(1+\frac{3}{n^6}\right)}$

$\begin{aligned} a_n & =\frac{\frac{4}{n^6}-7}{1+\frac{3}{n^6}} \\ & \operatorname{Lt}_{n \rightarrow \infty} a_n & =\frac{-7}{1}=-7\end{aligned}$