22. Algebra of Limits - part one

11 - Algebra of Limits - part 1

Algebra of Limits - Part 1

Let $(a_n)$ and $(b_n)$ be sequences with

$$a_n\to L\text{ and }{b}_n\to M\text{ as }n\to \infty$$

Let $c$ be a constant

1. Constant: If $a_n=c$, so $(a_n)$ is a constant sequence then

$$a_n\to c\text{ as }n\to \infty$$

2. Scalar Multiplication: Sequence $(c{a}_n)$ converges with

$$c{a}_n\to cL\text{ as }n\to \infty$$

3. Addition: Sequence $(a_n+b_n)$ converges with

$$a_n+b_n\to L+M\text{ as }n\to \infty$$

4. Subtraction: Sequence $(a_n-b_n)$ converges with

$$a_n-b_n\to L-M\text{ as }n\to \infty$$

5. Modulus: Sequence $(|a_n|)$ converges with

$$|a_n|\to |L|\text{ as }n\to \infty$$

Proof

1. Obvious from defintion
2. If $c=0$ then done by $(1)$, so assume that $c\ne 0$
   Take $ϵ>0$
   Since $a_n\to L$, there is $N$ such that if $n\le N$ then $|a_n-L|<ϵ$
   Now if $n\ge N$ then $|c{a}_n-cL|=|c||a_n-L|<|c|ϵ$
   Hence $(c{a}_n)$ converges to $cL$
3. Take $ϵ>0$
   Since $a_n\to L$, there is $N_1$ such that if $n\le N_1$ then $|a_n-L|<ϵ$
   Since $b_n\to M$, there is $N_2$ such that if $n\le N_2$ then $|b_n-M|<ϵ$
   Let $N=\max \{N_1,N_2\}$
   If $n\ge N$ then

$$|a_n-L|<ϵ\text{ and }|b_n-M|<ϵ$$

So

$$\begin{array}{rl}|(a_n+b_n)-(L+M)|& \le |a_n-L|+|b_n-M|(\text{by triangle inequality})\\ & <2ϵ\end{array}$$

4. By $(2)$ and $(3)$
5. Proved before - Modulus Sequence Converges

Link to original