Sequence

A sequence of real numbers is a function $f:\mathbb{N}\to \mathbb{R}$ we denote it as $(a_n{)}_{n\ge 0}$ with $a_n=a(n)$ we can define a sequence either recursive or exclusive

Operations

Let $(a_n),(b_n)$ be two convergent sequence with ${\lim }_{n\to \infty }{a}_n=a$ and ${\lim }_{n\to \infty }{b}_n=b$ Then:

• ${\lim }_{n\to \infty }(a_n\pm b_n)=a\pm b$
• ${\lim }_{n\to \infty }(a_n{b}_n)=ab$
• If $b_n\ne 0\forall n\ge 0\wedge b\ne 0$ then ${\lim }_{n\to \infty }\left(\frac{a_n}{b_n}\right)=\frac{a}{b}$
• If there is $K>1$ such that $a_n>b_n\forall n\ge K$ then $a>b$

Definition: Convergence

A sequence $(a_n{)}_{n\ge 1}$ converges to a limit $l\in \mathbb{R}$ if for any arbitrarily small distance $\epsilon$, we can find a point in the sequence beyond which all subsequent terms are within that distance of the limit.

Theorem

A sequence $(a_n{)}_{n\ge 1}$ converges to a limit $l\in \mathbb{R}$ if $\forall \epsilon >0\exists N\in \mathbb{N}\forall n>N(|a_n-l|<\epsilon )$

The limit of a convergent sequence is unique, can be prove by contradiction

We can find similarities to the definition of convergence in a topology space

Every convergent sequence is a Cauchy sequence. In Real numbers, the converse also holds because of completeness.

How to prove convergence

To prove a convergence property, we usually focus on finding a suitable $N$ for a given $\epsilon$ without explicitly dealing with the set of indices

Example

$a_n=\frac{n}{n+1}$ and we try to prove ${\lim }_{n\to \infty }{a}_n=1$ $|a_n-1|=\left|\frac{n}{n+1}-1\right|=\left|-\frac{1}{n+1}\right|=\frac{1}{n+1}⟹\epsilon >\frac{1}{N+1}$ Given $\epsilon >0$ there exists $N$ such that $N>\frac{1}{\epsilon }-1$ (by Archimedean principle). Then for all $n>N$, we have $n+1>N+1>\frac{1}{\epsilon }$, so $\frac{1}{n+1}<\epsilon$. Thus $|a_n-1|<\epsilon$ for all $n>N$. This proves that $a_n$ converges to $1$

Divergence

• positive: $\forall M>0\exists N>0\forall n>N:a_n>M$
• negative: $\forall M>0\exists N>0\forall n>N:a_n<-M$