Definition:Convergence

Sequence
Let $$\left \langle {x_k} \right \rangle$$ be a sequence in $\mathbb{R}$.

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

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

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 $$l$$."

Such a sequence is convergent.

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 $$N \in \mathbb{N}$$ but this is not strictly necessary and can make proofs more cumbersome.

Real Function
Let $$f$$ be a real function defined on an open interval $$\left({a \, . \, . \, b}\right)$$ except possibly at some $$c \in \left({a \, . \, . \, b}\right)$$.

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 $$c$$''.

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


 * Let $$f \left({x}\right)$$ 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 $$f \left({x}\right)$$ be such that:

$$\forall \epsilon > 0: \exists X: \forall x > X: \left|{f \left({x}\right) - l}\right| < \epsilon$$.

Then $$\lim_{x \to \infty} f \left({x}\right) = l$$.