Definition:Convergence

Topological Space
Let $$T = \left({A, \vartheta}\right)$$ be a topological space.

Let $$\left \langle {x_k} \right \rangle$$ be a sequence in $T$.

Then $$\left \langle {x_k} \right \rangle$$ converges to the limit $$l \in T$$ if, for any open set $$U \subseteq T$$ such that $$l \in U$$, $$\exists N \in \R: n > N \Longrightarrow x_n \in U$$.

Such a sequence is convergent.

Metric Space
Let $$\left({X, d}\right)$$ be a metric space.

Let $$\left \langle {x_k} \right \rangle$$ be a sequence in $X$.

Then $$\left \langle {x_k} \right \rangle$$ converges to the limit $$l$$ iff:


 * $$\forall \epsilon > 0: \exists N \in \R: n > N \Longrightarrow d \left({x_n, l}\right) < \epsilon$$

Or equivalently, using the definition of neighborhood:


 * $$\forall \epsilon > 0: \exists N \in \R: n > N \Longrightarrow x_n \in N_\epsilon \left({l}\right)$$

We can write "$$x_n \to l$$ as $$n \to \infty$$", or $$\lim_{n \to \infty} x_n \to l$$.

This is voiced "As $$n$$ tends to infinity, $$x_n$$ tends to (the limit) $$l$$."

It can be seen that by the definition of open set in a metric space, this definition is equivalent to that for convergence in a topological space.

Standard Number Fields
When $$X$$ is one of the standard number fields $$\Q, \R, \C$$, and the metric $$d$$ is the usual (Euclidean) metric, the condition on convergence becomes:

$$\left \langle {x_k} \right \rangle$$ converges to the limit $$l$$ iff:


 * $$\forall \epsilon > 0: \exists N \in \R: n > N \Longrightarrow \left|{x_n - l}\right| < \epsilon$$

where $$\left|{x}\right|$$ is the modulus of $$x$$.

The validity of this definition derives from the fact that:
 * the Rational Numbers form Metric Space;
 * the Real Number Line is Metric Space;
 * the Complex Plane is Metric Space.

Divergent Sequence
A sequence which is not convergent is divergent.

Comment
The sequence $$x_1, x_2, x_3 \ldots$$ can be thought of as a set of approximations to the number $$l$$, in which the higher the $$n$$ the better the approximation.

The distance $$\left|{x_n - l}\right|$$ between $$x_n$$ and $$l$$ can then be thought of as the error arising from approximating $$l$$ by $$x_n$$.

Note the way the definition is constructed.

"Given any value of $$\epsilon$$, however small, we can always find a value of $$N$$ such that ..."

If you pick a smaller value of $$\epsilon$$, then (in general) you would have to pick a larger value of $$N$$ - but the implication is that, if the sequence is convergent, you will always be able to do this.

Note also that $$N$$ depends on $$\epsilon$$. That is, for each value of $$\epsilon$$ we (probably) need to use a different value of $$N$$.

Note
Some sources insist that $$N \in \N$$ but this is not strictly necessary and can make proofs more cumbersome.

Series
Let $$S$$ be one of the standard number fields $$\Q, \R, \C$$.

Let $$\sum_{n=1}^\infty a_n$$ be a series in $S$.

Let $$\left \langle {s_N} \right \rangle$$ be the sequence of partial sums of $$\sum_{n=1}^\infty a_n$$.

It follows that $$\left \langle {s_N} \right \rangle$$ can be treated as a sequence in the metric space $$S$$.

If $$s_N \to s$$ as $$n \to \infty$$, the series converges to the sum $$s$$.

Divergent Series
A series which is not convergent is divergent.

Convergence of a Function on a Metric Space
Let $$M_1 = \left({A_1, d_1}\right)$$ and $$M_2 = \left({A_2, d_2}\right)$$ be metric spaces.

Let $$c$$ be a limit point of $$M_1$$.

Let $$f: A_1 \to A_2$$ be a mapping from $$A_1$$ to $$A_2$$ defined everywhere on $$A_1$$ except possibly at $$c$$.

Let $$f \left({x}\right)$$ tend to the limit $$L$$ as $$x$$ tends to $$c$$.

Then $$f$$ converges to the limit $$L$$ as $$x$$ tends to $$c$$.

Convergence of Real and Complex Functions
As:
 * The real number line $$\R$$ under the usual metric forms a metric space;
 * The complex plane $$\C$$ under the usual metric forms a metric space;

the definition holds for real and complex functions.

Divergent Function
There are multiple ways that a function can be divergent. Here are some samples:


 * Let $$f: \R \to \R$$ be such that:


 * $$\forall H > 0: \exists \delta > 0: f \left({x}\right) > H$$ provided $$c < x < c + \delta$$

Then (using the language of limits), $$f \left({x}\right) \to +\infty$$ as $$x \to c^+$$.


 * Let $$f: \R \to \R$$ be such that:

$$f \left({x}\right) = \begin{cases} 0 & : x \in \Q \\ 1 & : x \notin \Q \end{cases}$$

Then $$x$$ converges to neither $$0$$ nor $$1$$ and hence is divergent (although, it needs to be noted, not to infinity).