16 - Convergence of a Sequence

Convergence of a Sequence

Let $(a_n)$ be a real sequence
Let $L\in \mathbb{R}$ then

$(a_n)$ converges to $L$ as $n\to \infty$ if

$$\forall ϵ>0:\exists n\in \mathbb{N}\text{ s.t. }\forall n\ge N,|a_n-L|<ϵ$$

It can also be written as

$$(a_n)\to L\text{ as }n\to \infty \text{ or }\underset{n\to \infty }{\lim }{a}_n=L$$

Alternative definition for convergence $(a_n)$ converges to $L$ as $n\to \infty$ if

$$\forall ϵ>0:\exists n\in \mathbb{N}\text{ s.t. }\forall n>N,|a_n-L|\le ϵ$$

Tip - for when proving a sequence convergence $|a_n-L|<cϵ$ where $c$ is some constant that is good enough! This is because you can just scale everything back to get $ϵ$

If you can show that

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Convergence vs Divergence of a Sequence

Let $(a_n)$ be a real sequence then

1. $(a_n)$ converges, or is convergent, if there is

$$L\in \mathbb{R}\text{ s.t. }{a}_n\to L\text{ as }n\to \infty$$

2. $(a_n)$ diverges, or is divergent if $(a_n)$ doesn’t converge

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Tail of a Sequence

Let $(a_n)$ be a sequence then
Tail of $(a_n)$ is a sequence $(b_n)$, for some natural number $k$

$$b_n=a_{n+k}\text{ for }n\ge 1$$

TLDR: $(b_n)$ is obtained by deleting the first $k$ terms of $(a_n)$

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Tails Lemma lemma

Let $(a_n)$ be a sequence

1. If $(a_n)$ converges to a limit $L$ then every tail of $(a_n)$ converges and to $L$

2. If a tail $(b_n)=(a_{n+k})$ of $(a_n)$ converges, then $(a_n)$ converges

Proof

1. Take a tail of $(a_n)$ so for $k\ge 1$ and let $b_n=a_{n+k}$ for $n\ge 1$
   Assume that $(a_n)$ converges to a limit L
   Take $ϵ>0$
   Then there is $N\in \mathbb{N}$ such that

$$\text{if }n\le N\text{ then }|a_n-L|<ϵ$$

But if $n\ge N$ then $n+k\ge N$ hence

$$|a_{n+k}-L|<ϵ⟹|b_n-L|<ϵ$$

Hence $(b_n)$ converges and $b_n\to L$ as $n\to \infty$
2) Assume that $(b_n)=(a_{n+k})$ converges
Then there is $L\in \mathbb{R}$ such that

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

Take $ϵ>0$
Then there is $N$ such that if $m\ge N$ then

$$|b_m-L|<ϵ⟹|a_{m+k}-L|<ϵ$$

Now if $n\ge N+k$ then $n=m+k$ where $m\ge N$ so

$$|a_n-L|<ϵ$$

So $(a_n)$ converges and

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

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When showing convergence you don't have to find the smallest $n$

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You can also show it works for $0<ϵ<1$ which naturally extends for $ϵ>0$

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