Chapter 2 Flashcards
Definition: convergence
A sequence (a_n) is said to converge to a limit l∈ℝ if there exists an N ∈Naturals st for all n>N We have |a_n -l | < ε
Theorem 2.11: convergence of a sequence to a limit and another limit value
If a sequence converged to a limit then that limit is unique. More precisely if (a_n) if a sequence st lim a_n =l and lim a_n= l’ also (as n tends to Infinity) then l = l’
Proof:
Assume that l is not equal to l’.
Then by definition of convergence we have an M and an N for each limit WE USE ε/2 . By the triangle inequality,
|l-l’| = | l-l’ -a_n + a_n |
Less than or equal to
|a_n-l’| + |l-a_n| =. |a_n-l’| + |a_n-l|
Less than
ε/2 + ε/2 = ε
Since the limits aren’t equal 0 < |l-l’| < ε , holds for any ε. Taking ε = |l-l’| /2 gives the contradiction we require.
Definition: divergence
A sequence which fails to converge is said to diverge
(a_n) diverges to infinity if given any k>0 there exists an N in the naturals st for all n >N a_n > k
Respectively a_n < -k for negative infinity
Definition: bounded above, bounded, bounded below
A sequence (a_n) is said to be
Bounded above/bounded below/ bounded
If the set {a_n,n∈N } is
Bounded above/ bounded below/ bounded
THM: 2.12: convergence and boundedness of sequences
If a sequence (a_n) is convergent then it’s also bounded
PROOF:
1) we need to find a k >0 st |a_n| is less than or equal to K for all naturals. We know (a_n) converges to some l in the reals ….
By triangle inequality
|a_n| less than or equal to |a_n-l | + |l|
Less than ε + |l|
So FOR n>N we take k= ε +|l|
FOR n less than or equal to N: we have |a_n| less than or equal to max({ |a₁|, |a₂|,…,|a_n|})
Hence for all n in the naturals k= max({ |a₁|, |a₂|,…,|a_n|,ε +|l| })
THM 2.21 algebra of limits
Proof from book (too long)
Suppose that (a_n) and (b_n) are convergence sequences with the limits l and m respectively.
1) sequences whose nth term is a_n + b_n converges to l+m
2) sequence whose nth term is a_n • b_n converges to lm
3) if c is real then the sequence whose nth term is c•a_n converges to cl
4) if b_n ≉ 0 for all n and also m ≉0 then the sequence whose nth term is a_n / b_n converges to l/m
PROOF
LOOK IN NOTES AS TOO LONG
1) Given any ε>0 there exists an N∈N st if n>N, then |a_n -l | < ε/2 and there also exists N∈N st if n>M |b_n -m| < ε/2. Choose any n > max{N,M} and by the triangle inequality |a_n + b_n - (l +m)|= |a_n -l + b_n -m | less than or equal to | a_n -l| + |b_n -m| < ε/2 + ε/2= ε
2) similarly for n > max{N,M} |a_nb_n -lm|= |a_nb_n- lm +lb_n -lb_n| by triangle inequality is less than or equal to |b_n(a_n -l)| + | l( b_n -m) |= |b_n||a_n-l| + |l||b_n-m|
Assuming l not equal to 0, don’t bother with l=0.
Now as sequence (b_n) is convergent and thus by theorem 2.1.2 is bounded. Hence there is a real number K bigger than 0 st |b_n| ≤ K for all n. So ..
|a_nb_n - on | ≤ K|a_n - l | + |l||b_n - m| …
For any ε> 0 there exists M in N st |b_n -m| < ε/ 2|l| whenever n is bigger than M.
Combining we get for n bigger than Max{M,N},
|a_nb_n - lm| < K ε/2K + |l| ε/2|l| = ε/2 + ε/2 = ε
Probing in case where l not equal to 0 if equal 0 …
3) follows 2) where (b_n) is the constant sequence with nth term c
4) firstly we show that
THM 2.2.2: The sandwich rule
Suppose given three sequences (a_n), (b_n) and (c_n), so that for all n∈N we have a_n ≤b_n ≤ c_n. If (a_n) and (c_n) both converge to the same limit l, then (b_n) also converges to l.
Proof:
We must show that given any ε> 0 there exists an N∈N st for all n>N.
|b_n -l | < ε. Ie both b_n -l less than ε and l- b_n less than ε and There exists an M_1 ∈N st for all n bigger than M_1, b_n -n ≤ c_n -l ≤ |c_n -l| < ε. And also there exists an M_2∈N st got all n >M_2 l- b_n≤ l -a_n ≤ |l - a_n| = |a_n -l| < ε.
Taking N =max{M_1,M_2} the result follows.
Definition: monotonic increasing and decreasing
A sequence (a_n) is monotonic increasing if a_n+1 ≥ a_n for all n∈N.
It’s monotoníc decreasing if a_n+1 ≤ a_n for all n∈N
We can see that (a_n) is monotonic decreasing if and only if (-a_n) is monotoníc decreasing
THM 2.3.1
Monotone convergence theorem
1) if a sequence (a_n) Is bounded above and monotonic increasing then it is convergence and limit as n tends to infinity of a_n = sup_(n∈N)[ (a_n)]
2) if (a_n) is bounded below and monotonía decreasing then convergent and limit as n tends to infinity of a_n = inf_(n∈N)[ (a_n)]
Proof: since (a_n) is bounded above α= sup_(n∈N)(a_n) exists by the completeness property.
By proposition, given any ε>0 there exists an N∈N st a_N bigger than α-ε. As (a_n) is monotonic increasing: for all n bigger than N a_n ≥ a_(N+1) ≥ a_N bigger than α-ε, it follows that |α-a_n| = α -a_n
Similarly 2
Monotonía and converges or diverges to
Collorary: if the sequence (a_n) is monotonic increasing / monotonic decreasing then either it converges or diverges to +infinity/ -infinity
Proof:
Suppose (a_n) is monotonic decreasing, then either bounded below or it isn’t. If it is bounded below then it converges by previous theorem and if it isn’t then given any k>0 we can find N in the naturals such that a_n > K, o/w would have lower bound. Since monontonic decreasing have a_n < K for all nbigger than or equal to N sequence diverges to minus infinity
A sequence and subsequence
A sequence (y_n) is a subsequence of a sequence (x_n) if there is a strictly increasing function σ: N -> N st y_n = x_σ(n) for all n in the naturals.
Equivalently a subsequence of (x_n, n in N) is a sequence of the form (x_r, r in N) ie (x_n_1 , x_n_2, x_n_3,…) where n_1 < n_2 < n_3
Proposition convergence of the subsequence
If the sequence (a_n) converges to some limit l, then every subsequence of (a_n) also converges to l.
Proof:
Given ε>0 , there exists an N∈N st if n bigger than N, then |a_n -l | less than ε.
Let (a_n_r) be a subsequence of (a_n) and choose k∈N st n_k bigger than N. Then for all r bigger than K, |a_n_r - l| less than ε.
Monotone subsequence
THM 2.4.2: every sequence has a monotone subsequence
Proof:
Let (a_n) be a sequence and define set
C= {N∈N, a_m < a_N for all m>N}.
The set C is either bounded above or it isn’t. Suppose that C is bounded above: then C is finite.
Choose an n₁∈N as follows:
If C is the empty set then n₁ =1, if C is non empty then n₁= max(C) +1. In either case n₁ is not an element of C for all n bigger than or equal to n₁.
But
•n₁ not in C implies a_n₂ ≥ a_n₁ for some n₂ bigger than n₁
•n₂ not in C implies a_n₃ ≥ a_n₂ for some n₃ bigger than n₂
And then we use induction to completed the proof that (a_n_j ∈N) is monotonic increasing.
If C is unbounded then we can find an infinite sequence (n₁,n₂,n₃,…) in C with n₁less than n₂less than n₃…
•n₁∈C implies a_m less than a_n₁ for all m bigger than n₁ which implies a_n₂ is less than a_n₁
•n₂∈C implies a_m less than a_n₂ for all m bigger than n₂ which implies a_n₃ is less than a_n₂
And again we use induction to complete the proof that (a_n_k, k∈N) us monotonic decreasing.
Boleando- Weirstrass
Every bounded sequence has a convergent subsequence.
Proof:
Suppose (a_n) is bounded. By theorem 2.4.2 it has a monotone subsequence, (a_n_r), which is itself bounded.
Then (a_n_r) converges by theorem that states monotone sequence converges if bounded. (Thm2.3.1)
Cauchy sequence
A sequence (a_n) is said to be a Cauchy sequence if given any ε> 0 , there exists an N ∈N st for all m and n > N we have | a_m - a_n| < ε
Every convergent sequence is Cauchy
Every Cauchy sequence is bounded
Arbitrarily close terms but says nothing in convergence.
THM Cauchy sequence and converges …
If (a_n) is a Cauchy sequence in R, then it converges to a limit in R
Proof: by every convergent sequence being Cauchy
We have that (a_n) is bounded and hence by the Bolzano-Weirstrass theorem is has a subsequence (a_n_r) converging to some l in the reals. Proving that (a_n) converges to l:
Given any ε>0 there exists an M₁∈N st for all m,n bigger than M₁ we have |a_m -a_n| < ε/2 and there exists M₂ ∈N st n_r bigger than M₂.
So we have that |a_n_r -l| < ε/2.
Hence for all n bigger than M₁ and choosing r so that n_r is bigger than max{M₁,M₂} we have
|a_n -l| ≤ |a_n -a_n_r| + |a_n_r - l| < ε/2 + ε/2 = ε
Definition: complete subset and Cauchy
A non-empty subset A subset of R is said to be complete if every Cauchy Sequence taking values in A converges to a limit in A.
Ie R is complete by theorem for (a_n) Cauchy sequence in R always converges to a limit in R
Q is not complete eg approximating root 2. By rationals
Convergent sequences and Cauchy …
Every convergent sequence is Cauchy:
Proof
Problem 41
Cauchy and bounded …
Every Cauchy sequence is bounded:
Proof:
Problem 42
(a_n) and a_n
(a_n) is sequence and nth term is a_n
What else modulus tells us in the convergence of a sequence |a_n - l | < ε
|a_n - l | < ε if and only if l-ε < a_n < l + ε
We interpret Into equality and so we can show it converges if we show the rhs
Convergence also tells us that as n tends to infinity |a_n - l | gets very small
Proof of convergence given sequence:
We find a SPECIFIC N for ε, that is we found N from ε and thus taking this N, for all n above, can show that in fact |a_n - l | < ε
Notation vs definition of convergence
Notation is not the definition.
Notation would be lim as n tends to infinity of a_n =l etc not to do with |a_n - l | < ε
Example: prove that lim as n tends to infinity of r^n =0 whenever 0 ≤ r < 1
BY CASES:
•r=0: clearly for all n∈N r^n =0
• 0< r < 1: we have that 1/r > 1.
Let 1/r = 1 + h where h >0.
Then by BERNOUILLS INEQUALITY we have
1/r^n = ( 1 +h )^n ≥ 1 +nh
And thus r^n ≤ 1/ (1+nh).
We must show that, given any ε>0 there exists an N∈N st for all n > N |r^n| = r^n < ε. So we require an N st (1+Nh)^-1 less than ε which implies that N bigger than (1-ε) / εh by algebra.
By the ARCHIMEDEAN PROPERTY for real (1-ε) / εh there does exists such an N. Then for all n>N. r^n ≤ (1+mg)^-1 < 1/ (1+Nh) < ε
Example- find limit as n tends to infinity of (2n -7n^3)/(6n^2 +11n^3)
We divide both by the highest power in the denominator: n^3
Hence it becomes the limit as n tends to infinity of
2/n^2 -7 divided by 6/n + 11n
Which becomes -7/11
Example: find the limit as n tends to infinity of cos(n)/n
We use the sandwich rule. As for all n in the naturals cos(n is bounded bewueen -1 and 1. Then using the sandwich rule with -1/n and 1/n the limit is 0
Example- EXAM STYLE QUESTION
Consider the sequence (a_n) given by
a_1=0, a_(n+1) = (3a_n +1)/(a_n +3)
For all n bigger than 1.
- Use induction to show that a_n is bounded within [0,1] for all n in the naturals
- Show that (a_n) is monotonic increasing
- Explain why limit as n tends to infinity of a_n exists and find its value
- true for n=1 base case, inductive hyp gives a_n+1 bigger than 0
a_n+1 -1 shown to be less than or equal to 0 also. By Induction etc - for all n in the naturals we can show that a_n - a_n+1 is less than or equal to 0. (Numerator of a fraction is)
- by previous parts it converges by theorem 2.3.1 convergence of monotoníc bounded
By taking limits on both sides of recurrence relation limit as n tends to unfinished is 1
Example- The golden section as a limit Constructing by recursion
a_1=1 and a_(n+1) = sqrt ( 1+ a_n)
Calculating first few terms (6) looks like increasing and bounded above.
Proving this: looks like 2 is an upper bound. Prove this by contradiction. Suppose there exists an N such that a_n is less than or equal to 2 for all n less than or equal to N and that a_(N+1) is bigger than 2. From our calculations we know it’s at least bigger than 6. Substitutions in the recurrence in a_(N+1) is bigger than 2 and squaring gives that 1+ a_N is bigger than 4. Which gives a_N is bigger than 3. But this is a contradiction. increaskng. Is proven by squaring the recursion and subtracting equations for a_n and a_n-1 is show it’s bigger than 0. Must combine all equations to cancel and give numbers on denominator.
To how u has a limit use theorem and take limits of the recurrence relation and use algebra of limits. Two sols are found.
Example : e as a limit
e = limit as n tends to infinity of (1+ 1/n) to the power of n
Proving that the limit exists and that e is bounded between 2 and 3
We show minotnix increasing and bounded above .
Monotmonic increasing:
By the theorem of means, and using a_1= a_2=..a_n-1 = 1+ (1/n-1) and a_n =1.
Geometric mean (1+ (1/(n-1)))^((n-1)/(n))
Arithmetic mean is 1+ 1/n
Hence the theorem of means raised to the power of n gives equations for a_n and a_n+1 in inequality to show increasing.
To show bounded use the BINOMIAL theorem to expand and inequalities from this to show bounded above by using a geometric series compared
Guess the limit as n tends to infinity of the sequence whose nth term is
3/4 • ( 0.5 - (7/n^3)) and prove
Limit is 3/8:
We need to show that given an epsilon bigger than 0 we can find an N in the naturals so that if n is bigger than N
| a_n -l | = | -21/4n^3 | = 21/4n^3 is less than epsilon.
This occurs if and only if n is bigger than (21/4epsilon)^(1/3). BY THE ARCHIMIDEAN PROPERTY such an N exists and so for n bigger than N
21/4n^3 is less than 21/4N^3 is less than epsilon
And thus the limit is 3/8.
