Sequence (Calculus)

A sequence is an ordered list of real numbers ($a_1,a_2,a_3,\dots$) usually denoted by parenthesis $(a_n)$ or curly brackets $\{a_n\}$.

A sequence can also be rewritten as a function: $f:\mathbb{N}\to \mathbb{R},f(n)=a_n$ and plotted.

Limits of Sequences §

Since sequences only work when n is a natural number, Limit (Calculus) are usually only interesting as $n\to \infty$. The definition of a limit of a sequence is:

$$\underset{n\to \infty }{\lim }{a}_n=L$$

• If $L$ exists (i.e. is finite, real, etc) then the sequence converges.
• Else, the sequence is said to diverge.

Note, a sequence can converge to infinity, but this does not actually mean ${\lim }_{n\to \infty }{a}_n=\infty$. Instead, it is an abbreviation, in real analysis.

Limit Laws On Sequences §

The same Limit Laws that work for normal functions work for sequences:

Continuity On Sequences §

Continuity (Calculus) works basically the same way as it does on a normal function, assuming n is always natural:

Theorem - Sequences & Functions §

Let $f(x)$ be a real-valued function and $(a_n)$ be a sequence of real numbers such that $f(n)=a_n$:

$$\underset{x\to \infty }{\lim }f(x)=L\to \underset{n\to \infty }{\lim }{a}_n=L$$

This theorem means that if a sequence has a valid function that produces the same output values for $n$, then all the same techniques used on Limit (Calculus) of functions (sandwich theorem, L’Hôpital’s rule, etc.) can be applied on sequences.

Not that (as shown via the implies symbol), this does not happen both ways:

$$\underset{n\to \infty }{\lim }{a}_n=L\nrightarrow \underset{x\to \infty }{\lim }f(x)=L$$

Limits of sequences do not always hold on functions

Take the sequence $a_n=\sin (2\pi n)$, which always equates zero for $n\in \mathbb{N}$ Hence, the following limit is true for $a_n$ :

$$\underset{n\to \infty }{\lim }\sin (2\pi n)=0$$

However the same does not apply for the function version: $f(x)=\sin (2\pi x)$ as the value can oscillate between -1 and 1. I.e:

$$\underset{n\to \infty }{\lim }\sin (2\pi n)=0\nrightarrow \underset{x\to \infty }{\lim }\sin (2\pi x)=0$$