- #1
Dr. Seafood
- 121
- 0
Call {a1, a2, a3, ...} = {an} a "convergent sequence" if
[tex]\exists L \in \mathbb{R} : \quad \forall \epsilon > 0 \quad \exists N \in \mathbb{N} : (\forall n > N \quad (n > N \implies |a_n - L| < \epsilon))[/tex]
in which case we write [itex]\lim_{n \rightarrow \infty} a_n = \lim a_n = L[/itex]. Of course this is the usual definition for sequences of reals. Consider the sequence
[tex]\{{n \over {n + 1}}\} = \{{1 \over 2}, {2 \over 3}, {3 \over 4}, ... \}[/tex]
It appears the numbers approach 1, intuitively. We'll try to show by definition that [itex]\lim a_n = 1[/itex]. Given [itex]\epsilon > 0[/itex], we'll try to find the corresponding positive integer N so that we satisfy the above definition. I get this far:
[tex]|{n \over n + 1} - 1| = |{-1 \over n + 1}| = |{1 \over n + 1}| < \epsilon \implies 1 < {\epsilon}(n + 1)[/tex]
By the Archimedean property, we can always find (n + 1) so that this is true, i.e. for any [itex]\epsilon > 0[/itex]. But this doesn't help me find the particular fixed integer N which corresponds to our choice of [itex]\epsilon[/itex]. How is this done?
[tex]\exists L \in \mathbb{R} : \quad \forall \epsilon > 0 \quad \exists N \in \mathbb{N} : (\forall n > N \quad (n > N \implies |a_n - L| < \epsilon))[/tex]
in which case we write [itex]\lim_{n \rightarrow \infty} a_n = \lim a_n = L[/itex]. Of course this is the usual definition for sequences of reals. Consider the sequence
[tex]\{{n \over {n + 1}}\} = \{{1 \over 2}, {2 \over 3}, {3 \over 4}, ... \}[/tex]
It appears the numbers approach 1, intuitively. We'll try to show by definition that [itex]\lim a_n = 1[/itex]. Given [itex]\epsilon > 0[/itex], we'll try to find the corresponding positive integer N so that we satisfy the above definition. I get this far:
[tex]|{n \over n + 1} - 1| = |{-1 \over n + 1}| = |{1 \over n + 1}| < \epsilon \implies 1 < {\epsilon}(n + 1)[/tex]
By the Archimedean property, we can always find (n + 1) so that this is true, i.e. for any [itex]\epsilon > 0[/itex]. But this doesn't help me find the particular fixed integer N which corresponds to our choice of [itex]\epsilon[/itex]. How is this done?