18. Limits - modulus and inequalities

Modulus on a sequence

Let $(a_n)$ be a convergent sequence then $(|a_n|)$ also converges
Moreover if $a_n\to L$ as $n\to \infty$ then

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

Proof

Say $a_n\to L$ as $n\to \infty$
Take $ϵ>0$ then
There is $N$ such that if $n\ge N$ then $|a_n-L|<ϵ$
If $n\ge N$ then by the Reverse Triangle Inequality then

$$||a_n|-|L||\le |a_n-L|<ϵ$$

So $(|a_n|)$ converges and $|a_n|\to |L|$ as $n\to \infty$

Can also be proved using Sandwiching Lemma

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Limits preserve weak inequalities

Let $(a_n)$ and $(b_n)$ be real sequences, and assume that

$$a_n\to L\text{ and }{b}_n\to M\text{ as }n\to \infty \text{ and }{a}_n\le b_n\text{ for all }n$$

Then

$$L\le M$$

Proof

Suppose, for a contradiction, that it is not the case that $L\le M$ so

$$L>M\text{ (by trichotomy)}$$

Let $ϵ=\frac{1}{2}(L-M)>0$
Since $a_n\to L$ as $n\to \infty$, there is $N_1$ such that

$$\text{if }n\ge N_1\text{ then }|a_n-L|<ϵ$$

Since $b_n\to M$ as $n\to \infty$, there is $N_2$ such that

$$\text{if }n\ge N_1\text{ then }|a_n-M|<ϵ$$

For $n\ge \max \{N_1,N_2\}$, we have

$$a_n>L-ϵ\text{ and }{b}_n<M+ϵ$$

Hence

$$\begin{array}{rl}L-ϵ& <{\alpha }_n\le b_n<M+ϵ\\ L-M& <2ϵ=L-M\end{array}$$

This is a contradirction hence $L=M$

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Sandwiching

Let $(a_n),(b_n)\text{ and }(c_n)$ be real sequences with

$$a_n\le b_n\le c_n\text{ for all }n\ge 1$$

If

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

Then

$$b_n\to L\text{ as }n\to \infty$$

Proof

Take $ϵ>0$
Since $a_n\to L$ as $n\to \infty$, there is $N_1$ such that

$$\text{if }n\ge N_1\text{ then }|a_n-L|<ϵ$$

Since $c_n\to L$ as $n\to \infty$, there is $N_2$ such that

$$\text{if }n\ge N_2\text{ then }|c_n-L|<ϵ$$

Then for $n\ge \max \{N_1,N_2\}$ we have

$$L-ϵ\le a_n\le b_n\le c_n\le L+ϵ$$

Hence

$$|b_n-L|<ϵ$$

So $b_n\to L$ as $n\to \infty$

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