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sachik

Member

Registered: 2008-02-28

Posts: 9

Geometric Series convergion and divergion

Could someone help me solve this... I already solved parts a) and b) ( the work is below) but i need to solve part c) but im not really sure how to do that.

1.
a)
Values of r
1. r=2   
2.r=3   
3.r=4   
4.r=5   
5.r=6   

Sum 

             10 Terms    50 Terms                   100 Terms
1.    2046    2.25 x 10^15    2.53 x 10^30
2.    59048    7.17 x 10^23    5.15 x 10^47
2.    699050    8.45 x 10^29    1.07 x 10^60
4.    4882812    4.44 x 10^34    3.94 x 10^69
5.    24186470    3.23 x 10^38    2.61 x 10^77

Work

1.    Sn=2(1-2^10 )/(1-2) = 2046
    Sn=2(1-2^50 )/(1-2) = 2.25 x 10^15
    Sn=2(1-2^100 )/(1-2) = 2.53 x 10^30

2.    Sn=2(1-3^10 )/(1-3) = 59048
    Sn=2(1-3^50 )/(1-3) = 7.17 x 10^23
    Sn=2(1-3^100 )/(1-3) = 5.15 x 10^47

3.    Sn=2(1-4^10 )/(1-4) = 699050
    Sn=2(1-4^50 )/(1-4) = 8.45 x 10^29
    Sn=2(1-4^100 )/(1-4) = 1.07 x 10^60

4.    Sn=2(1-5^10 )/(1-5) = 4882812

    Sn=2(1-5^50 )/(1-5) = 4.44 x 10^34

                  Sn=2(1-5^100 )/(1-5) =3.94 x 10^69

5.    Sn=2(1-6^10 )/(1-6) = 24186470
    Sn=2(1-6^50 )/(1-6) = 3.23 x 10^38
                   Sn=2(1-6^100 )/(1-6)= 2.61 x 10^77

b)
Values of r
1. r=.9
2.r=.8
3.r=.7
4.r=.6
5.r=.5

Sum
               10 Terms               50 Terms         100 Terms
1.    13.03             19.89         19.99
2.    8.92             9.999857275        9.999999998
3.    6.4783498            6.6666665        6.666666667
4.    4.969766912      5       5
5.    3.99609375    4         4

Work

1.    Sn=2(1-〖.9〗^10 )/(1-.9) = 13.03
    Sn=2(1-〖.9〗^50 )/(1-.9) = 19.89
    Sn=2(1-〖.9〗^100 )/(1-.9) =19.99

2.    Sn=2(1-〖.8〗^10 )/(1-.8) = 8.9262581
    Sn=2(1-.8)/(1-.8) = 9.999857275
    Sn=2(1-〖.8〗^100 )/(1-.8) = 9.999999998

3.    Sn=2(1-〖.7〗^10 )/(1-.7) = 6.4783498   
                Sn=2(1-.7)/(1-.7) = 6.6666665
    Sn=2(1-〖.7〗^100 )/(1-.7) = 6.666666667

4.    Sn=2(1-〖.6〗^10 )/(1-.6) = 4.969766912

    Sn=2(1-〖.6〗^50 )/(1-.6) = 5
                  Sn=2(1-〖.6〗^100 )/(1-.6) =5

5.    Sn=2(1-〖.5〗^10 )/(1-.5) = 3.99609375
    Sn=2(1-〖.5〗^50 )/(1-.5) = 4
                   Sn=2(1-〖.5〗^100 )/(1-.5)= 4

a)  Chose 5 different values for r such that |r|> 1. For each of these geometric series, let the first term g1 = 2. Determine the sum of each of these geometric series for 10 terms, 50 terms, and 100 terms. Organize your data into a chart. (10 points)

b) Chose 5 different values for r such that
|r| <1 For each of these geometric series, let the first term g1 = 2. Determine the sum of each of these geometric series for 10 terms, 50 terms, and 100 terms. Organize your data into a chart. (10 points)

c) The word converge means to approach or draw near to a particular value. For
example, as x gets very large, converges to zero. The word diverge means
does not converge. For example as x gets very large, x^2 diverges; as x gets large, x^2 does not approach any particular value. Make a conjecture about the conditions under which a geometric series will converge. Test your conjecture using g1 (not equals) 2 and some of the values of r. Think about the formulas for geometric series. Write a mathematical argument to support your conjecture. (30 points)

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Geometric Series convergion and divergion

From Part b) you might begin to realise that if |r|<1, the terms will converge to zero eventually. You could use this as your conjecture for part c).

When testing your conjecture, just experiment around abit. See how the terms behave for different values of r.

The formula for a geometric series is

What do you think will happen when

and |r|<1?

sachik

Member

Registered: 2008-02-28

Posts: 9

Re: Geometric Series convergion and divergion

the sum would keep getting smaller and smaller....right?

technically it would never hit zero, because n is infinite, but it would be pretty darn close to zero

sachik

Member

Registered: 2008-02-28

Posts: 9

Re: Geometric Series convergion and divergion

what mathematical arguement would i be able to use to support my conjecture

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,343

Re: Geometric Series convergion and divergion

If |r|>1, you get a divergent series with every term greater than the preceding term.
In such cases, the sum of n terms as n tends to infinity cannot be determined. The sum tends to infinity as n tends to infinity.

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It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

sachik

Member

Registered: 2008-02-28

Posts: 9

Re: Geometric Series convergion and divergion

isnt the formula for a sum of a geometric series Sn= a1(1-r^n)/1-r

because yours is a little different

sachik

Member

Registered: 2008-02-28

Posts: 9

Re: Geometric Series convergion and divergion

so would i be able to use this as a mathematical arguement to support conjecture?

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,343

Re: Geometric Series convergion and divergion

isnt the formula for a sum of a geometric series Sn= a1(1-r^n)/1-r

This formula is applicable when |r|<1.
When |r|>1, the formula given in the earlier post would hold good.

---

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Geometric Series convergion and divergion

Technially both can be used in all cases, it's just that which the easier one is varies depending on |r|.

When |r|<1, sachik's formula has positive numerator and denominator, making it slightly easier than the other one. When |r|>1, the reverse is true.

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Why did the vector cross the road?
It wanted to be normal.