Pc Functions Geometric Series

Geometric Series and Basic Properties of Series Formula: first term a If q  1, S   aq   . 1  q 1  ratio of terms n 0 

n

Series Definition

A series is an infinite sum of the form 

S   an  a1  a2  a3  K n 1

The finite sum

m

Sm   an  a1  a2  K  am

is

n 1

the mth partial sum of the series S. The series S converges if the sequence (Sm) converges and has a finite limit. If this is the case, then we say that the sum of the series S is the limit of the sequence (Sm). Notation



S   an  a1  a2  a3  K  lim Sm n 1

m 

Geometric Series Definition

A series S is geometric if the ratio of its two subsequent terms is constant.

A geometric series is of the form 

S  a  aq  aq  K   aq n . 2

n 0

Example

The following series are geometric  1 1  1 2  1   K   2   2 4 k 0  2 

k



1  2  4  8  K    2 

1  0.1  0.01  0.001  K    0.1 k 0

k 0

k

k

Partial Sums of Geometric Series Consider the partial sums S   aq of a geometric series m

m

n

n 0



S  a  aq  aq  K   aq n . 2

n 0

Observe m

Sm   aq n  a  aq  aq 2  K aq m 1  aq m n 0 m

qSm   aq n 1  aq  aq 2  K  aq m 1  aq m  aq m 1 n 0

Subtrackting the equations one gets Sm  qSm  a  aq m 1.

These terms cancel when subtracting the equations

If q  1, one gets 1  q m 1 Sm  a . 1 q

Sum of a Geometric Series It is clear the the partial sums 1  q m 1 Sm  a 1 q have a finite limit if and only if q  1. In this case the 



geometric series S   aq converges and S   aq n  n

n 0

Formula

n 0



If q  1, S   aq n  n 0

a . 1 q

a . 1 q

In the above formula the term “a” is the first term of the geometric series, and the term “q” is the ratio between two subsequent terms.

Example 1



1  0.1  0.01  K    0.1  n 0

n

1 10  1- 0.1 9

Observe that 1+0.1+0.001+… = 1.111… =10/9.

Examples 2

Study the convergence of the series Write

Solution



1 .  k 1 k  k  1

1 1 1   . k  k  1 k k  1

M 1 1   1      k  1  k 1 k  k  1 k 1  k M

We use the Partial Fraction Decomposition.

These terms cancel each other.

1  1 1   1 1  1 1  1 1  1         L         1 2  2 3  3 4  M  1 M   M M  1



 1 Conclude

1  M 1

 M   1.

The series converges and



1  1.  k 1 k  k  1

Sequences of the Terms of Series If  a converges, then lim a  0. Theorem 

k

k 1

Proof

k 

Consider SM  If



a k 1

k

k

M

a . k 1

k

converges, then lim SM  S is finite. M 

Observe that an  S n  Sn 1.

We get

lim an  lim  Sn  Sn 1   lim Sn  lim Sn 1  S  S  0.

n 

Corollary

n 

n 

n 

If lim an  0 or if the limit does not exist, the series n 



a k 1

k

diverges.

Examples 1

Show that the series

 1 diverges.  k



 k sin 

k=1

Solution

 1 k  1    1  0. The series diverges since lim k sin    lim k  k  1  k k k If k is a 2   1 k 1 large even 2 Show that the series  diverges. 2 number, this k 1 k=1 sin 





is close to 1

Solution

1 2  1 k  1 k k The series diverges since lim  lim  1 k  k  1 k2  1 1 2 k If k is a large odd number, this is does not exist. k



2



close to -1. Hence no limit.

1

Properties of Series Theorem 



 a and  b both converge. Let c be a constant. The series  ca ,   a  b  and   a  b  converge and  ca  c  ca ,

Assume that

k

 k 1

k 1



k

k 1

k

k 1



k

k





k

k 1 

 a k 1

k



 a k 1

k

k 1

k

k 1

k

k





k 1

k 1





k 1

k 1

 bk    ak   bk  bk    ak   bk

This result follows immediately from the properties of the limits.

Properties of Series The series

Observation



 a

k

k 1

if the series If the series

a

k



 a k 1

k

k

and



b k 1

k

diverge.

diverges and bk  ak for all k , then all terms of

 bk  are 0. Hence the series converges.

   1  2n  1  2        n   3n  3 n=1  n 1 n 1  3  

Example

a k 1



k 1

the series



 bk  may converge even

n

1 2 1 3  3  3  2   1 2 2 2 1 1 3 3