A Brief Analysis of the Geometric Series UPLOAD VERSION

7/29/2019 A Brief Analysis of the Geometric Series UPLOAD VERSION

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1 A Brief Analysis Of The Geometric Series

A Brief

Analysis Of

The Geometric

SeriesType 1 Mathematical

investigation

MD. Rakibur Rahman


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Introduction

A geometric series is essentially a sequence of real* numbers, where

each term after the first is found by multiplying the first term (t1) by a non-

zero constant known as the common ratio (r).

For example: the sequence 2, 8, 32, 128, is a geometric sequence,

where the first term or t1 is 2, while the common ratio (r) is 4.

While the name first term is self explanatory, the common ratio is

the non-zero constant by which each receding term is multiplied by to obtain

the next. In our example, the common ratio is:

nd

2

st

1

2 term 8= = 4

1 term 2

t

t

Thus, in a more general manner, if the nthterm is tn, then the previous

term is tn-1, then the common ratio or r:

1

n

n

tr

t

Now, the definition of a term of the geometric sequence is the first

term (t1 or a) multiplied by the common ratio to the power of the term

number minus one. To elaborate, for example: if the 1 st term is 5, the common

ratio is 2, then the 4th term will be:

4 1 35 4 = 5 4 = 320

Again, in a more general manner, if the first term is a, the common

ratio is r, then the nth term or tn:

1= nnt a r

This leads us to conclude that the general geometric sequence must be

formulated as shown below:

2 3 1, , , , ... , na a r a r a r a r

*Real numbers are defined as numbers that can be expressed using the real number system, and

thus excludes all imaginary and complex numbers

Sum of a geometric series


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One of the most important, as well as useful applications/

functions of the geometric series is the sum of the series. If a geometric series

is defined as:

2 3 1, , , , ... , na a r a r a r a r

Then the sum of the series or Snof the first n number of terms is as follows:

2 3 1+ + ... + nnS a a r a r a r a r

For example, 2, 8, 32, 128 is a geometric series. The sum of the first 4 terms of

the series is:

4 2 8 32 128 = 170S

Therefore, the sum of the first 4 terms of the given series is 170.

The sum of a geometric series is one of the most useful

applications of algebra used in our financial world. Interests, loans and

compounding function calculations would not be possible without the use of

geometric sequences. Patterns across natural artifacts occur while following

the rules of a geometric sequence. In physics, rate of change of many functions

occur with respect to a geometric sequence. Chemical reactions at the ionic

levels occur according to a geometric approach. Biological micro-organisms

multiply and breed with respect to a geometric sequence. The importance of a

geometric sequence and the sum is essential to almost any field of education.

Derivation of the sum formula


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Now, it is relatively simple to calculate the sum of the first 4

numbers, considering the fact only a few terms need to be taken into account

while calculating. However, when the sum of, for example, first 201 terms

need to be calculated, a more effective as well as efficient approach needs to

be considered. Algebra can be used to manipulate the general sequence to

obtain a more feasible equation.

The sum of general geometric sequence is defined as:

2 3 1+ + ... + nnS a a r a r a r a r

Now, the sum this geometric sequence can also be defined as:

1

1

nb

n

b

S a r

Here, the sum of the equation1ba r is found from when b = 1, which gives us

a, to when b = n, which gives us 1nar .

Therefore,

1 2 3 1

1

1 2 3 1

1

= + + ... +

(1 ... ) ---- equation 1

nb n

b

nb n

b

a r a a r a r a r a r

a r a r r r r

Multiplying both sides by r in the second previous statement we get,

1 2 3 4

1

= + + ... +n

b n

b

r a r a r ar a r a r a r

Now, subtracting (n x a) from both sides gives us nnumber of as on each side,

therefore an a for each term in the series:

1 2 3 4

1

1 2 3 4

1

- ( ) = -a+ -a+ -a ... +

- ( ) ( 1) ( 1) ( 1) ( 1) ... ( 1)

nb n

b

nb n

b

r a r n a a r a ar a r a r a r a

r a r n a a r a r a r a r a r

(continued on next page)

(continued from last page)

Now, factoring a on the right side we get,


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1 2 3 4

1

1 2 3 4

1

1 2 3 4

1

1 2 3 4

1

1 2 3 4

1

- ( ) ( 1 1 1 1 ... 1)

- ( ) ( ... )

( ... ) ( )

( ... )

( ... ) --

nb n

b

nb n

b

n

b n

b

nb n

b

nb n

b

r a r n a a r r r r r

r a r n a a r r r r r n

r a r a r r r r r n n a

r a r a r r r r r n n

r a r a r r r r r -- equa tion 2

Now, subtracting equation 2 from equation 1 we get,

1 1 2 3 1 2 3 4

1 1

1 1 2 2 3 3 1 1

1 1

(1 ... ) - ( ... )

(1 ... ... )

n nb b n n

b b

n nb b n n n

b b

a r r a r a r r r r a r r r r r

a r r a r a r r r r r r r r r

We notice that all of the terms between 1 and rncancels out, thus,

1 1

1 1

1

1

1

1

1

1

(1 )

(1-r) = (1 )

(1 )

(1 )

( 1)

( 1)

n nb b n

b b

nb n

b

nnb

b

nnb

b

a r r a r a r

a r a r

a ra r

r

a ra r

r

Now we remember that

1

1

nb

n

b

S a r (by definition)

Thereby giving us,

( 1)

( 1)

n

n

a rS

r

In order to test the new equation, let us consider the

geometric series:


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2, 4, 8, 16, 32 and 64

Now, the sum of the first 6 terms of the series by adding the terms manually is:

62 4 8 16 32 64 126S

Now, the first term or a in this series is 2, while the common ratio is 2, and the

number of terms is 6. Using the formula:

6

6

( 1)

( 1)

2(2 1)126

2 1

n

n

a rS

r

S

The sum of the 6 terms given by the derived formula and the manual

calculation is equal.

Therefore, the derived formula is correct and can be used to

calculate the sum of a geometric series.

Convergent geometric series


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The simplest definition of a convergent geometric series is: A

geometric series whose sum of multiple terms converges, or approaches

towards a certain and real value. Now, convergent geometric series are unique

and therefore, have unique characteristics. There are certain circumstances

under which the sum of a geometric series tends to approach a certain value,

or limit. These circumstances are imposed upon the geometric series by two

contributor, the smaller being the first term or a, while the largest contributor

is the common ratio, or r.

The value of the first term does not have any significant effect on the

converging characteristics of the series. For our example, we will pick a to be

equal to 2.

The value of the common ratio however, greatly contributes to the

convergence of the geometric series.

Fig. 1) Generic number line determined for convergence testing

In the number line of fig. 1, values ranging from and including x, -2,

-1, -0.5, 1

x

, 01

x

, 0.5, 1, 2 and x has been chosen. For the most obvious

reasons, 0 is one of the primary choices made, as it separates positive values

from negative ones. 1 and -1 are also chosen here, because multiplication by

the prior does not change the value of the term, while the latter only causes

sign change. 0.5 and -0.5 are two of the most generic and common fractions or

decimal values used in mathematics, while 1

xand

1

xrepresent general

fractions, where { x > 1|x N}, or x is greater than 1, while being an element

of the natural number system. This allows to test all values up to 0 0}.

If the values of the Sn in table 2 are carefully observed, a pattern can be

observed. The Sn values 2, 4, 6, 8 represent an arithmetic sequence. In this

sequence, the first term (a) is 2, while the common difference (d) is:

2 1 4 2 2d t t

Therefore in this case, the Sn for the geometric sequence represents the tn for

the arithmetic sequence given above.

As we know for an arithmetic sequence:

At t1 = a and common ratio = r,

( 1)nt a n d

In our case, a = 2 and d = 2;

2 ( 1)2

2 2 22

n

n

n

t n

t nt n

Term numbertnfor r = 1

(arn-1)Sn(changes as n increases)

1 2 2

2 2 4

3 2 6

4 2 8

n 2 2n

Table 2) Obtained values oftn and Sn for first term of 2 and common ratio of 1

Here, as n increases, 2n or Sn also increases. Thus, as n , 2nas well. Therefore, at r = 1, the geometric sequence sum is divergent.

a = 2, r = 0.5

Following the method outlined on page 7, we get the values in table 3.

Term number tnfor r = 0.5 Sn(changes as n increases)


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(arn-1)

( 1)

( 1)

na r

r

1 2 2

2 1 3

3 0.5 3.5

4 0.25 3.75n 22 n

2(2) 4n

Table 3) Obtained values oftn and Sn for first term of 2 and common ratio of

0.5

If we analyze the equation

2(2) 4n , we observe that as n ,

2(2) n 0. Therefore, as

2(2) n 0,

2(2) 4n 4 . Therefore, as the

sum of the geometric series is converging to a certain value, namely 4, we can

conclude that as r = 0.5, the geometric series is convergent.

a = 2, r = 0

Following the method outlined on page 7, we get the values in table 4. .

Term numbertnfor r = 0

(arn-1)

Sn(changes as n increases)

( 1)

( 1)

na r

r

1 2 2

2 0 2

3 0 2

4 0 2

n 0 2

Table 4) Obtained values oftn and Sn for first term of 2 and common ratio of 0

We observe from the above table that as n, Sn=2.

Therefore, the sum of the geometric series converges to the value of

the first term, when the common ratio is 0. In other words, a common ratio of

0 results in a convergent geometric series.

a = 2, r = -2


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Following the method outlined on page 7 for r = 2, we get the values in

table 5.

Term numbertnfor r = -2

(arn-1)


( 1)

( 1)

na r

r

1 2 2

2 -4 -2

3 8 6

4 -16 -10

n ( 2)n 2 2

( )( 2)3 3

n

Table 5) Obtained values oftn and Sn for first term of 2 and common ratio of -2

In2 2

( )( 2)3 3

n , as n , (-2) , therefore

2 2( )( 2)

3 3

n

. Thus, as the common ratio of a geometric series is -2, the sum of the

series is not convergent (i.e. divergent).

a = 2, r = -1

Following the method outlined on page 8 for r =1, we get the values in

table 6 for r = -1 and a = 2.

Term numbertnfor r = -1


1 2 2

2 -2 0

3 2 2

4 -2 0n 2 ( 1)n 2 if n is odd, 0 if n is even

Table 6) Obtained values oftn and Sn for first term of 2 and common ratio of -1

We can see from the Sn values that as n increases, when n is an odd

number the sum is 2, while when n is odd the sum is 0. Thus, the sum oscillates

between 0 and 2. Therefore, it does not converge or diverge, and is therefore

neither convergent nor divergent.

a = 2, r = -0.5

Following the method outlined on page 9, we get the values in table 7.

Term numbertnfor r = -0.5



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( 1)

( 1)

na r

r

1 2 2

2 -1 1

3 0.5 1.5

4 -0.25 1.25

n1

4( )2

n

4 1 4( )

3 2 3

n

Table 7) Obtained values oftn and Sn for first term of 2 and common ratio of -

0.5

If we analyze the equation4 1 4

( )3 2 3

n , we see that as n,

1( )

2

n as well. Therefore, this leads to the conclusion that

4 1( )

3 2

n

, and thus the sum of the series converges to4

3at larger values of n.

Therefore, at r = -0.5, the geometric series is convergent.

a = 2, r = x

Following the method outlined on page 7, we get the values in table 8

for a = 2 and r = x.

Term number

tnfor r = x

(arn-1)


( 1)( 1)

n

a rr

1 2 2

2 2x 2+2x

3 2x 2+2x+2x

4 2x3 2+2x+2x2+2x3

n 12xn 2 2

1

nx

x

Table 8) Obtained values oftn and Sn for first term of 2 and common ratio of x

Here, we see that as the value of n approaches larger and larger

values,2 2

1

nx

x

, so does the value of xn. This is because { x > 1|x N}, and

therefore as the value of the exponent increases, so does the value of the

positive natural integer which is greater than 1. Therefore, as n ,


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2 2

1

nx

x

, for x, given { x > 1|x N}. Therefore, the series is not

convergent for given values of x at { x > 1|x N}.

a = 2, r = -x

Following the method outlined on page 7, we get the values in table 9

for a = 2 and r = -x.

Term numbertnfor r = -x

(arn-1)


( 1)

( 1)

na r

r

1 2 2

2 -2x 2-2x

3 2x 2-2x+2x

4 -2x

2-2x+2x -2x

n 12( )nx 2( ) 2

1

nx

x

Table 9) Obtained values oftn and Sn for first term of 2 and common ratio of -x

Analysis of the equation2( ) 2

1

nx

x

shows us that as n, (-x)n .

This is because when n is odd, the value of (-x)n becomes -xn, and after being

divided by the negative denominator, the sum approaches positive infinity. The

scenario is the opposite for when n is even, and thus the sum approachesnegative infinity. However, this is insignificant, because for neither case the

geometric series converges to a certain value. Therefore, for a common ratio

ofx, the geometric series is divergent*.

*Divergent stands for not convergent, or notconverging towards a certain value.

a = 2, r =1

x

Following the method outline in the previous pages, we obtain the

values in table 10.

Term number tnfor r =1

x Sn(changes as n increases)


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(arn-1)

( 1)

( 1)

na r

r

1 2 2

22

x2+

2

x

32

2

x 2+

2

x+

2

2

x

43

2

x 2+

2

x+

2

2

x+

3

2

x

n 11

2( )n

x

12( ) 2

11

n

x

x

Table 10) Obtained values oftn and Sn for first term of 2 and common ratio of

1

x

In the equation,

12( ) 2

11

n

x

x

, as n , the value of (1

x)n0. This is because

of the fact that { x > 1|x N}. Thus, as (1

x)n0,

12( )n

x0, and therefore

12( ) 2

11

n

x

x

2 2

1 11

x

or x

x

. Therefore, as the common ratio of a

geometric series is1

x, the series will converge to

2

1

x

x

as n approaches

infinity, given: { x > 1|x N}.

a = 2, r = -1

x

Following the method outline in the previous pages, we obtain the

values in table 11.


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Term numbertnfor r = -

1

x

(arn-1)


( 1)

( 1)

na r

r

1 2 2

2-

2

x 2-

2

x

32

2

x 2-

2

x+

2

2

x

4 -3

2

x 2-

2

x+

2

2

x-

3

2

x

n112( )n

x

12( ) 2

11

n

x

x

Table 11) Obtained values oftn and Sn for first term of 2 and common ratio of -

1

x

In the equation:

12( ) 2

11

n

x

x

, as n ,1

( )n

x 0. This is because

as x is greater than 1 and is a natural number and is in the denominator, as its

exponents value increases, then value of1

x decreases. As the exponent

approaches ,1

x approaches zero. Therefore, the value of

12( )n

x

approaches zero. This also implies that

12( ) 2

11

n

x

x

2

1

x

x, as n

.Therefore, when1

x is the common ratio of a geometric series, the series

tends to converge to 21

xx

, given { x > 1|x N}.

Results

Now, to sum up our results, table 12 is constructed below.


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Common ratio (r) Nature of geometric series2 divergent

1 divergent

0.5 convergent

0 convergent

-2 divergent

-1 oscillatory-0.5 convergent

x divergent

-x divergent

1

x convergent

1

x convergent

Table 12) Analyzed common ratios and their respective geometric series nature

We notice from the above table that only 0, 0.5, -0.5,1

x and

1

xproduce geometric series which are convergent. All other values of common

ratio produce divergent series.

Now, it had been previously specified that x > 1. This range

ensures that the value of1

xproduces a value which is less than 1. We notice

from our chart that when r is equal to 1, the series in no longer convergent,

but for values of r < 1; the series is convergent (i.e. at1

x, 0, 0.5, -0.5 and

1

x ) up to -1< r. Again, we notice that as r is equal to -1, the series is not

convergent, but for values of r > -1, the series is convergent. This leads us to

conclude the range of the acceptable r values on a number line (in fig. 2), in

order to obtain a convergent geometric series.

-x -2 -1 -0.5

1

x

0 1

x0.51 2 x


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Fig. 2) The range of usable common ratios in order to obtain a geometric series

From the above figure and analyzed data we can therefore conclude

that, in order for a convergent geometric series to form

-1 < r < 1

Or, the common ratio is greater than -1 and less than 1.

Derivation of infinite sum formula

Now, we understand that in order to calculate the sum of an infinitely

continuous geometric series, the series must be convergent. Also, we know

that in order for any geometric series to be convergent, the common ratio must

be greater than -1 and less than 1.

Here, by definition:

0

b

b

S a r

In other words, the sum of an infinite series is the sum of all values

from t1 to tb or tn (here b is used instead of n for future algebraic manipulation),

where the value of b approaches infinity.

Now,

0

0

lim

b

b

nb

nb

S a r

S a r

Slight algebraic manipulation shows us that the prior equation is equal

to the latter, because in both the sum of the equation approaches infinity.

Only in the second one, the value of b approaches n, while the value of n

approaches infinity (making both equations equal).

0

limn

b

nb

S a r

Now by definition, we know that:


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0

nb

b

a r

= nS

Now,

If

( 1)( 1)

n

n a rSr

Then,

0

nb

b

a r

=( 1)

( 1)

na r

r

Thus substituting this in0

limn

b

nb

S a r

we get,

(ac co rding to p roperties of limits)

( 1)lim

( 1)

lim - lim( 1) ( 1)

lim lim( 1) ( 1)

n

n

n

n n

n

n n

a rS

r

a a rS

r r

a r aS

r r

Now, we know that -1 < r < 1; thus when n , rn0. Thus, as rn 0,

( 1)

na r

r 0.

Thus,

(as the value of n doe s not have a ny effect o n this equa tion, the limit ca n be remo ved)

= 0 - lim( 1)

=-( 1)

= where -1


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Verification of formula

Now, from the previous section we understand that at common ratio = -

0.5, the geometric series is convergent, where the sum of the geometric series

converges to4

3(as evaluated algebraically on page 12). Now, using our

derived formula at =1

aS

r

, let us verify the geometric series where a = 2, r =

-0.5.

Here,

2=

1 0.5

4=

3

S

S

Thus, as both algebraic manipulation and our derived formula gives us

the sum of infinity as4

3, we can verify the formula to be valid and usable in

future cases.

Interpretation of transformation for given equations

Visualization of an idea or fact is how human beings analyze critical

data as if it were simple. This is the function of a graph. Simply stated, a graph

is generally an equation plotted on multiple axes in order to represent theequation visually, for critical analysis. Although this may be done using algebra

and/or limits, it is much easier to understand the equation visually first, and

then evaluate or confirm using algebra. We will analyze some equations now

and later interpret to how this relates to a geometric series.

Equation set 1


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In the equation set 1:

1

2

3

2 -- equation 1

2 1 --equa tion 2

3(2 1) --equat ion 3

x

x

x

y

y

y

Equation 1 undergoes a set of transformations in order to become equation 3.

Fig. 3) shows the three of these equations plotted on the same set of axis. Note

that these equations are drawn to represent the equations only, and not the

geometric series which they are meant to represent later on.

A ti-83 Plus graphing calculator may be used for plotting these graphs

for ease of analysis. The following range of variables may be used:

x: [0,9, 1]

y: [0, 12 , 1]

A screenshot* of the window is given as below:

*An artificially computer generated mod of the Ti-83+ was used via ROM dump


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Now, the following graphs are plotted on y1, y2, y3 as shown:

Now, we get the graph as given below (graph also included in fig. 3):


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In general sense, a graphical transformation is when a set of changes

are made to an original equation in order to obtain a different and unique

equation. Let us consider the equation set 1 as listed above to discuss the

transformations applied there.

Referring to fig. 3), we see that Equation 1, which is 1 2xy , is an

exponential curve which intersects the y-axis through the point (0,1). Now, the

graph is moved down (vertically translated) by 1 unit on the y-axis in order to

obtain equation 2. We can see this algebraically as well as equation 2 is

2 2 1xy , which is essentially 1 subtracted from equation 1. As we can

observe in the figure, equation 2 intersects the y axis at (0,0), which is 1 unit

below the intersection of equation 1.

Equation 3 is essentially equation 2 with a vertical stretch factor of 3. In

Lehmans terms, equation 3 has been stretched 3 units on a vertical aspect

from equation 2. This can be observed visually in figure 3, as well as

mathematically, as equation 3 is 3 3(2 1)xy , which is basically 3 times

2 2 1xy .

On a visual terrain:

1 2xy 2 2 1

xy 3 3(2 1)xy

Equation set 2

Equation set 2 utilizes the same set of skills used in equation set 1.

Equation set 2:

1

2

3

1( ) -- eq ua tion 1

2

1( ) 1 --equa tion 2

2

13[( ) 1] --eq ua tion 3

2

x

x

x

y

y

y

Vertically translated by -1 unit Vertical stretch factor of 3 applied


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For equation set 2, the range of x and y are as follows:

x: [0, 9, 1)

y: [-3, 1, 1]

The window (which directly corresponds with the range) is as follows:

The graph obtained (also illustrated in fig. 4) is as follows:

Referring to fig. 4, we observe that as x values approaches infinity, the y

values of the equations y1, y2, y3 respectively approach 0, -1 and -3.


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From equation 1 or 11

( )2

xy , the equation 21

( ) 12

xy was moved down

1 unit or vertically translated 1 unit down. The vertical asymptote for equation

1, which is y = 0, was transformed into y = -1 for equation 2, which was also

vertically translated 1 unit down.

From equation 2 to 3, 21

( ) 12

xy or equation 2 was

transformed with a vertical stretch factor of 3 with respect to the y-axis. Inthis case, the vertical asymptote was transformed from y =-1 to y = -3, which is

a vertical transformation with a stretch factor of 3 as well.

1

1( )2

xy 21

( ) 12

xy 31

3[( ) 1]2

xy



25/41


Equation set 3

Although equation set 3 looks very similar to equation set 2, it is

extremely different when we graph it.

Equation set 3 is defined as follows:

1

2

3

1( )

2

1( ) 1

2

13[( ) 1]

2

x

x

x

y

y

y

We will be using the same windows as equation set 3 to graph this equation set.

After applying the values inside the calculator, we obtain the following graph:

This of course, from a direct perspective, signifies nothing but a

blank graph. However, if we look at it in an algebraic approach initially, and

then a graphical, we will understand the problem.

Our function consists of an exponential graph with a negative

base. Let us take 11

( )2

xy as an example. If for example, x was equal to

natural numbers 0, 1 and 2, the corresponding y values would be 1, -0.5 and

0.25 respectively, using simple substitution method.

Now, let us consider non-natural numbers such as1

2. Using the

real numbers system, 11

( )2

xy cannot be evaluated at x =1

2, since it

consists of taking the second root of a negative number, which cannot be

accomplished without the use of complex numbers.


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This means that using a real number system, equation set 3

cannot be graphed continuously, which results in a discontinuous function. Of

course, this can be graphed continuously using polar co-ordinates on a three

dimensional plane, however since this increases the complexity of this report

and is not included in the International Baccalaureate Standard Level

curriculum, we will not be going into details here.

Therefore we can conclude that, using the real number co-

ordination graphing system, equation set 3, which essentially consists of

exponential equations with negative bases, is only defined at:

x:[x N]*

This is verified by using the trace function of the ti-83+ and

placing x values which only consist of the natural number system, thus giving us

responding y co-ordinates.

Now, graphing the defined points on the same window as

equation set 2 in fig. 5) shows us that the points on the graphs are transformedaccording to the following measures:

1

1(- )

2

xy 2

1(- ) 1

2

xy

3

13[(- ) 1]

2

xy

The equation set 3 has the same values for vertical asymptotes as set 2.

*Hint: Incidentally, the sum of a geometric series is also defined at n:[n N]; but we will be goinginto details later!



27/41


Equation set 4

Equation set 4 is, as discussed above, only defined at x:[x N].

Equation set 4 is defined as below:

1

2

3

( 2)

( 2) 1

3[( 2) 1]

x

x

x

y

y

y

Again, graphing on the ti-83+ does not provide any useful visual

information. Therefore, we will be using the same graphing method as used in

fig. 5); i.e. by tracing the graph at x N, and then plotting the points in fig. 6).

Our two variables x and y will be defined as:

x: [x N|0, 3, 1]

y: [-12, 12, 1]


28/41


The equation set 4 has been transformed by the following

measures, as observations from figure 6 and algebraic characteristics dictate:

1 (-2)xy 2 (-2) 1

xy 3 3((-2) 1)xy

The given equations do not approach or converge to a certain

value, but extend towards positive and negative infinity on both axes.

Therefore, they do not have any vertical or horizontal asymptotes.



29/41


Rewriting final equation of set 1 in terms of Sum of geometric series formula

The final equation given in equation set 1 is 3y , where:

3 3(2 1)x

y

Now, our purpose is to rewrite this equation in terms of the equation derived

for the sum of the geometric series, or:

( 1)

( 1)

n

n

a rS

r

Now,

3

3

3(2 1)

13(2 1)

2 1

x

x

y

y

This resembles the arbitrary equation

( 1)

( 1)

n

n

a rS

r


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Where, a = 3, r = 2, x = n, nS = 3y .

This allows us to conclude that equation 3 of equation set 1

represents the sum of a geometric series, where the variables are defined as a

= 3, r = 2, x = n, nS = 3y .

Now as we know, a geometric series is defined generally as follows:

2 3 1, , , , ... , na a r a r a r a r

In this case, we know that a = 3 and r = 2,

Thus the series represented by equation set 1s final equation is:

2 13,3 (2), 3 (2) ,..., 3 (2)n



3

13[( ) 1]

2

xy

Using algebraic manipulation,

3

3

13[( ) 1]

2

1-0.5{3[( ) 1]}

20.5

x

x

y

y

3

1-1.5[( ) 1]}

21

12

x

y


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In terms of

( 1)

( 1)

n

n

a rS

r

Where a = -1.5, r =1

2, x = n, nS = 3y

This allows us to conclude that equation 3 of equation set 2 represents the sum

of a geometric series, where the variables are defined as

a = -1.5, r =1

2, x = n, nS = 3y .


2 3 1, , , , ... , na a r a r a r a r

In this case, we know that a = -1.5 and r = 12

,


2 11 1 11.5, 1.5 ( ), -1.5 ( ) ,..., -1.5 ( )2 2 2

n



3

13[(- ) 1]

2

xy

Now,

3

3

3

13[(- ) 1]

2

1-1.5{3[(- ) 1]}

2 1.5

1-4.5[(- ) 1]}

21

12

x

x

x

y

y

y


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Which is in terms of

( 1)

( 1)

n

n

a rS

r

3

1Where a = -4.5, r = - , x = n and y

2n

S


2 3 1, , , , ... , na a r a r a r a r

In this case, we know that a = -4.5 and r =1

2 ,


2 11 1 14.5, 4.5 (- ), -4.5 (- ) ,..., -4.5 (- )

2 2 2

n



3 3((-2) 1)xy

Here,

3

3

3

3((-2) 1)

-3{3((-2) 1)}

3

-9((-2) 1)}

2 1

x

x

x

y

y

y

Which is in the form

( 1)

( 1)

n

n

a rS

r,

3Where, a 9, r 2,x = n and y

nS



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2 3 1, , , , ... , na a r a r a r a r

In this case, we know that a = -9 and r = -2,


2 19, 9 (-2), -9 (-2) ,..., -9 (-2)n

Convergence and divergence of final equation in the 4 sets

To define convergence of a graph, it is essentially when as the x

co-ordinates increases or decreases infinitely, the y co-ordinates reach a

certain, real number value. For example, in the graph y = x2, as x approaches

positive infinity, the y-values also approach positive infinity. Therefore, y = x2

is not convergent. While on the other hand, in the equation1

yx

, as x

approaches positive infinity, y approaches 0. Therefore,1

yx

is converging

towards y = 0.

Convergence of final equation in set 1

The final equation in equation set 1 is defined as:

3 3(2 1)xy

Which can be rewritten as,

3

3(2 1)

2 1

x

y


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Thereby representing the geometric series

2 13,3 (2), 3 (2) ,..., 3 (2)n

Now if we look at 3 3(2 1)xy algebraically, as x approaches ,

3

3

3

lim lim 3(2 1)

lim lim 3(2) lim( 3)

lim not def ined, or y also approaches infinity

x

x x

x

x x x

x

y

y

y

We can observe this in fig. 3), where we see that as x-approaches infinity, y-

also approaches infinity.

Now, in terms of the geometric sum formula, let us consider the

sequence:

2 13,3 (2), 3 (2) ,..., 3 (2)n

( 1)

( 1)

n

n

a rS

r

Here a = 3, r =2

3(2 1)

(2 1)

,

lim lim 3(2 1)

lim do es not exist, or ap proac hes po sitive infinity.

n

n

n

nn n

nn

S

Now

S

S

In addition, if we try to apply our infinite geometric sum

formula:

= where -1


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Therefore, in the equation 3 3(2 1)x

y as well as the geometric series

represented by the equation, the y-values and the sum of geometric series

value is not convergent.

Convergence of final equation in set 2

3

13[( ) 1]

2

xy is the final equation in set 2, and represents the

geometric series,2 11 1 11.5, 1.5 ( ), -1.5 ( ) ,..., -1.5 ( )

2 2 2

n

Observations from fig. 4) show us that as x approaches positive infinity,

the y-values approach -3.

Algebraically,

3

3

3

3

3

13[( ) 1]

2

1lim lim 3[( ) 1]

2

1lim 3lim ( ) lim 3

2

lim 0 3

lim 3

x

x

x x

x

x x x

x

x

y

y

y

y

y

Thus, as x approaches positive infinity in 313[( ) 1]2

xy , the y

values approach -3.

Using the geometric series2 11 1 11.5, 1.5 ( ), -1.5 ( ) ,..., -1.5 ( )

2 2 2

n ,


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( 1)

( 1)

n

n

a rS

r

Where a = -1.5, r =1

2

1-1.5(( ) 1)

2

1( 1)2

13( ) 3

2

1lim 3lim( ) lim 3

2

lim 0 3

lim 3

n

n

n

n

n

nn n n

n

n

nn

S

S

S

S

S

Now, using the derived formula:

= where -1


37/41


3

13[(- ) 1]

2

xy is the final equation in set 3, and we observe

that it is a discontinuous function, only defined at x:[x N]. This is also the

same scenario with its geometric series, where:

2 11 1 14.5, 4.5 (- ), -4.5 (- ) ,..., -4.5 (- )2 2 2

n , and n:[n N)

Algebraically,

3

3

3

3

13[(- ) 1]

2

1lim 3lim(- ) lim 3

2

lim 0 3

lim 3

x

x

x x x

x

x

y

y

y

y

Thus, as x approaches positive infinity, the y-values approach -3.

If we look at fig. 5) we observe that as the x co-ordinates

approach positive infinity, the y-values approach -3 as well.

In the geometric series,

2 11 1 14.5, 4.5 (- ), -4.5 (- ) ,..., -4.5 (- )2 2 2

n

The sum of the series represents the equation 31

3[(- ) 1]2

xy

Where,

( 1)

( 1)

n

n

a rS

r

and a = -4.5, r=

1-2


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1-4.5[(- ) 1]

2

1( 1)

2

1

lim lim{3[(- ) 1]}2

1lim 3lim (- ) lim 3

2

lim 3

n

n

n

nn n

n

nn n n

nn

S

S

S

S

Therefore, as n approaches positive infinity, the sum of the geometric

series approaches -3.

If we examine the series using our infinite sum formula of,

= where -1


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Algebraically,

3

3

3

3((-2) 1)

lim 3lim( 2) lim 3

lim does no t exist , o r approaches

x

x

x x x

x

y

y

y

Using the geometric sum formula,

( 1)

( 1)

n

n

a rS

r

where a =-9 and r = -2,

-9((-2) 1)

( 2 1)

3(-2) 3

lim 3 lim (-2) lim 3

lim DNE or a pp roa ches

n

n

n

n

nn

n n n

nn

S

S

S

S

If we apply our infinite sum formula here,

= where -1


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2 11 1 14.5, 4.5 (- ), -4.5 (- ) ,..., -4.5 (- )2 2 2

n

Now, the initial observation made from the above equation, as well as

the other equation sets is that:

The base of the exponent, i.e.1

-2

in this case, represents the

common ratio value in its respective geometric series.

This can be verified using other equations in the sets as well.

The second observation that is made from this report is that,

The value of the first term, or a, is equal to the vertical stretch

factor, i.e. 3 in this case, multiplied by the common ratio subtracted by 1.

In simpler terms: a = (vertical stretch factor) (common ratio -1)

One can also conclude that,

a or t1 = (vertical stretch factor) (base of exponent -1) [as base of

exponent = common ratio]

The third, and most important observation made:

If the base of the exponent is less than 1 or greater than -1, then

the equation, as well as its sum of infinity of the geometric series, will

converge to a real value. On the other hand, if the base of the exponent is

greater than 1 or less than -1, then the equation as well as its sum of

infinity will be divergent.

In general terms,

If zx is the given exponent in y = k(zx-1), Then:


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if -1

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A Brief Analysis of the Geometric Series UPLOAD VERSION