An Infinite sequence is a function f(n) whose domain is an
infinite subset of whole numbers.
We usually write a sequence {a_{n}} . where f(n) = a_{n} The following 4 animations are the graphs of the sequences
{(.5)^{n}} {n/(n+1)}
{e^{n}/n!} and {(1)^{n}}
You'll notice that sequences are discontinuous everywhere but never
the less play an important role in Mathematics as we will see when we consider Infinite Series.
an = (.5)^n
an = n/ (n+1)
an = e^n/n!
an = (1)^n
As you could tell from the animations
the 1st and 3d converge to 0, the second converges to 1 and the 4th diverges.
But do we mean by convergence exactly?
Informally by lim a_{n} = L we mean for any ε there is a point beyond which L ε <
a_{n} < L + ε. for every n.
Formally : given any ε>0
there exists a number M such that L ε < a_{n} < L + ε. whenever n>
M
Or as is usually stated given any ε>0 there exists a number M such that 
an  L  < ε
whenever n > M.
The following animation shows this.
Animation Limit Demonstration
How do we compute limits?
Thee are 2 very
important theorems
1. Given {a_{n}} if there is a continuous differentiable function f(x) such that
f(n) = an
then lim a_{n} = lim f(x) therefore we can use our results on differentiable functions
and L'Hopital's Rule.
Examples 1 and 2 fit into this category. But what about the 3d? the Gamma Function
notwithstanding we need the following
Thm If {a_{n}} is increasing and bounded
above or decreasing and bounded below it converges.
The first 3 animations make this fairly obvious,However for
a Fromal Proof click the link below
Proof for increasing and bounced sequences
Alternating Sequences
Another important theorem
Suppose b_{n} = (1)^{n} a_{n} where an >0.
Then if lim a_{n}
= 0 then lim b_{n }= 0
If lim a_{n} ≠ 0 then { b_{n}} diverges.
Even if lim a_{n} = L The alternating sequence (1)^{n} a_{n }diverges as
the subssequence of even terms converges to L and the subssequence of odd terms converges to  L therefore the sequence diverges.
The next 2 animations show the divergent case and the 3d shows convergence of an alternating sequence.
(1)^n (n/(n+1))
(1)^n * n
Converegent (1)^n (.5)^n
