Sequences and Series

Sequences

Definition : A sequence is (for us) an infinite list of numbers which can be ordered and in which there is a first term.

Notation: = , ... , , ...

e.g. , ... , , ... e.g. = 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, ....

e.g. (famous) the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, ... = { } where = 1 and for .

Here are the first 20 terms of the Fibonacci sequence:

> with(combinat,fibonacci):seq(fibonacci(n),n=1..20);

Warning, new definition for fibonacci

Here follows an interesting sequence of matrices:

> with(linalg):f:=matrix(2,2,[1,1,1,0]);seq(evalm(f^n),n=1..10);

Definition : = L means that the terms of can be made arbitrarily close to L by taking n sufficiently large.

Note: If n is an integer and and = L then = L . It follows that the usual limit laws that we learned for functions also apply to sequences.

Definition : If exists (and = L) then { }is said to converge to L . Otherwise { } is divergent or diverges.

Theorem : If and when n is an integer, then .

e.g. since = 0, the sequence { } converges to 0 but { } and { } both diverge (and for different reasons, why?)

Limit Laws : All the limit laws from Cal I apply to sequences (see table on page 696 ). For example :

If and are convergent sequences then

The squeeze theorem also applies and says that if and for and = L , then = L .

As a consequence of the squeeze theorem is an important litle fact:

If then .

Definition : is increasing if for all . It is decreasing if ... It is monotonic if it is either increasing or decreasing.

Definition : is bounded above if there is a number M such that for all . Similar definition for bounded below .

Theorem : Every bounded, monotonic sequence is convergent.

Series

Definition : If is a sequence then its associated series is = = + ....

If is a series then we associate with it a sequence of partial sums, , where .

=> , , , and so on...

e.g. Consider and its sequence of partial sums.

From the table you should be fairly convinced that = 1 and it is overwhelmingly tempting to conclude that . And that's what we will conclude and it inspires the following definition.

Definition : The series is convergent if and only if its associated sequence of partial sums converges. If converges to L then we say that = L.

e.g. the Geometric Series

= + .... is a geometric series with first term a and common ratio r . According to Maple, we have that (see blackboard for proof):

> Sum(a*r^(n-1),n=1..infinity)=sum(a*r^(n-1),n=1..infinity);

*** The above formula is only true if . ***

e.g. 1) Find 2) Find + ...

1)

> S:=Sum(1/(2^n),n = 1 .. infinity)=sum(1/(2^n),n = 1 .. infinity);

2)

> Sum(-2*(-3/4)^n,n=0..infinity)=sum(-2*(-3/4)^n,n=0..infinity);

3) Write 1.42323232323..... as a ratio of integers.

Note: 1.42323232323... = 1.4 + + + + ... = + + + + ...

> 1.4+Sum(23/10*1/100^n,n=1..infinity)=14/10+sum(23/10*1/100^n,n=1..infinity);

> evalf(1409/990,75);

Theorem : If is convergent then .

Note: the converse of the above theorem is not true. For example, the Harmonic Series is not covergent (see blackboard).

Test for Divergence : If then diverges.

e.g.

> an:=2/(n^2+3*n+2);bn:=1/3^n;

> convert(an,parfrac,n);

> A:=Sum(an,n=1..infinity);B:=Sum(bn,n=1..infinity);value(A);value(B);

> Sum(an+bn,n=1..infinity)=sum(an+bn,n=1..infinity);

3 quick e.g.'s. Determine the donvergence/divergence of:

1.

> Sum((-3)^(n-1)/4^n,n=1..infinity)=sum((-3)^(n-1)/4^n,n=1..infinity);

2.

> Sum((3^n+2^n)/(6^n),n = 1 .. infinity)=sum((3^n+2^n)/(6^n),n = 1 .. infinity);

3.

> Sum(ln(n/(n+1)),n=1..infinity)=sum(ln(n/(n+1)),n=1..infinity);Limit(ln(x/(x+1)),x=infinity)=limit(ln(x/(x+1)),x=infinity);evalf(seq(sum(ln(n/(n+1)),n=1..i),i=1..10),4);evalf(seq(sum(ln(n/(n+1)),n=1..i),i=90..100),4);evalf(seq(sum(ln(n/(n+1)),n=1..i),i=990..1000),4);

=> It looks like the series diverges. It does. Prove it!

e.g. Find x so that S = converges.

Note: S is a Geometric Series with first term and common ratio . It will converge <=>

> plot([abs(1/x),1],x,y=0..4,thickness=2,colour=[black,red]);

> solve(abs(1/x)=1,x);

=> converges for and . The sum will then be

> simplify((1/x)/(1-(1/x)));

The Integral Test

e.g. Show that the Harmonic Series diverges.

> with(student):

> leftbox(1/x,x=1..16,15,title=`y = 1/x`);

Looking at the picture above it should be clear that + .... > and we know from earlier work with improper integrals that = . It follows that the hamonic series, + ..., is divergent.

e.g. Show that is convergent.

> rightbox(1/(x^2+1),x=0..10,10,title=`y = 1/(x^2+1)`);

Calculate the areas of the rectangles in the picture and we get = + ... < . Therefore, it follows that the sequence of partial sums, , of the series is bounded above by (see below) and clearly is monotonic increasing. We stated a theorem about sequences that therefore guarantees the convergence of (and we know that its sum is something less than 1.3 ).

> I1:=1/2+Int(1/(x^2+1),x=1..infinity);value(I1);

The Integral Test

The two preceeding examples motivate the integral test:

If f is a continuous, positive, decreasing function and then

(i) If is convergent, then is convergent

(ii) If is divergent, then is divergent

Notes: 1) It is not necessary for the series to start at n = 1. It is only necessary to make sure that the "starting numbers" match. e.g. to study we would use the intgral .

2) It is not necessary that f be always decreasing but only that it be monotonic decreasing for all x beyond some number N (i.e. f must be ultimately decreasing).

The p-series: .

If p < 1 then we know that and if p = 0 then = 1. In both cases, the series diverges by the term test for divergence. If p > 0 then we can see that is continuous and decreasing on and we can use the integral test on .

It follows that the p -series is convergent if p > 1 and divergent if .

The plots of three different "p-functions" below is informative.

> s1:=1/x^(1.5):s2:=1/x:s3:=1/sqrt(x):

> plot([s1,s2,s3],x=0..50,colour=[black,red,magenta],thickness=2,title=`y = 1/x^1.5, 1/x and 1/x^(1/2)`);

Determine the behaviour of .

> f:=x*exp(1)^(-x);plot(f,x=0..6,thickness=2);

Clearly, is positive for x > 0 and the picture makes it look like f is decreasing on . To prove decreasing you can look at the derivative and see that it is indeed dcreasing for all x > 1.

> fp:=diff(f,x);solve(fp<0);

> Int(x*exp(1)^(-x),x=1..infinity)=int(x*exp(1)^(-x),x=1..infinity);

Maple says that the integral converges so we conclude that so does the series.

> value(Sum(n*exp(1)^(-n),n = 1 .. infinity));

The Comparison Tests

The Comparison Test

If and are both positive term series, then:

(i) if is convergent and for all n then is also convergent.

(ii) if is divergent and for all n then is also divergent.

Determine the convergence/divergence of: a) b) c)

> Sum(1/(n^2+n-1),n = 1 .. infinity)=sum(1/(n^2+n-1),n = 1 .. infinity);

> evalf(%);

> plot([1/(x^2+x-1),1/x^2],x=0..4,y=0..1.5,discont=true,colour=[black,red],thickness=2);

> Sum(5/(2+3^n),n=1..infinity);evalf(%);

> plot([5/(2+3^n),5/3^n],n=0..5,y=0..5,colour=[black,magenta],thickness=2);

> Sum(ln(n)/n,n=1..infinity);evalf(%);

> plot([ln(n)/n,1/n],n=0..6,y=-2..2,colour=[black,red],thickness=2);

> evalf(seq(sum(ln(i)/i,i=1..n),n=1..50),3);

=> diverges by the comparison test (compare to a p-series with p = 1).

The Limit Comparison Test

Suppose that and are both positive term series and that where and . Then either both series converge or both series diverge.

e.g.

> restart:an:=sin(1/n);bn:=1/n;Limit(an/bn,n=infinity)=limit(an/bn,n=infinity);