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Geometric Series (page 5 of 5)

Sections: Terminology and notation, Basic examples, Arithmetic and geometric sequences, Arithmetic series, Finite and infinite geometric series


You can take the sum of a finite number of terms of a geometric sequence. And, for reasons you'll study in calculus, you can take the sum of an infinite geometric sequence, but only in the special circumstance that the common ratio r is between 1 and 1; that is, you have to have r | < 1.

For a geometric sequence with first term a1 = a and common ratio r, the sum of the first n terms is given by:

sum, from i = 1 to n, of a-sub-i  is equal to (a) [ (1 - r^n) / (1 - r) ]

Note: Your book may have a slightly different form of the partial-sum formula above. For instance, the "a" may be multiplied through the numerator, the factors in the fraction might be reversed, or the summation may start at i = 0 and have a power of n + 1 on the numerator. All of these forms are equivalent, and the formulation above may be derived from polynomial long division.

In the special case that r | < 1, the infinite sum exists and has the following value:

    sum, from i = 1 to infinity, of a-sub-i  is equal to  a/(1 - r) Copyright Elizabeth Stapel 2006-2011 All Rights Reserved


  • Evaluate the following:
    • sum, from i = 1 to 20, of [3(-2)^i]

    The first few terms are 6, 12, 24, so this is a geometric series with common ratio r = 2. (I can also tell that this must be a geometric series because of the form given for each term: as the index increases, each term will be multiplied by an additional factor of 2.) The first term of the sequence is a = 6. Plugging into the summation formula, I get:

      computation of summation value

    So the value of the summation is 2 097 150

  • Evaluate S10 for 250, 100, 40, 16,....
  • The notation "S10" means that I need to find the sum of the first ten terms. The first term is a = 250. Dividing pairs of terms, I get 100 250 = 2/5, 40 100 = 2/5, etc, so the terms being added form a geometric sequence with common ratio r = 2/5. When I plug in the values of the first term and the common ratio, the summation formula gives me:

     

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      calculation of S_10

      S_10 = 6509734 / 15625

Note: If you try to do the above computations in your calculator, it may very well return the decimal approximation of 416.62297... instead of the fractional (and exact) answer. As you can see in the screen-capture to the right, entering the values in fractional form and using the "convert to fraction" command still results in just a decimal approximation to the answer. But (warning!) the decimal approximation will almost certain be regarded as a "wrong" answer! Take the time to find the fractional form!

 

screen-capture from calculator

  • Find an if S4 = 26/27 and r = 1/3.
  • They've given me the sum of the first four terms, S4, and the value of the common ratio r. Since there is a common ratio, I know this must be a geometric series. Plugging into the geometric-series-sum formula, I get:

      S_4 = a[(1 - r^4)/(1 - r)], so 26/27 = a(40/27)

    Multiplying on both sides by 27/40 to solve for the first term a = a1, I get:

      (26/27)(27/40) = a = 13/20

    Then:

      a-sub-n = (13/20) (1/3)^(n - 1)

  • Show, by use of a geometric series, that 0.3333... is equal to 1/3.
  • There's a trick to this. I first have to break the repeating decimal into separate terms:

      0.333... = 0.3 + 0.03 + 0.003 + 0.0003 + ...

    This shows the repeating pattern of the non-terminating (never-ending) decimal explicitly: For each term, I have a decimal point, followed by a steadily-increasing number of zeroes, and then ending with a "3". This can be written in fractional form, and then converted into geometric-series form:

      0.333... = 3/10 + 3/100 + 3/1000 + ... = sum [ (3/10) (1/10)^(n-1) ]

    Then 0.333... is an infinite geometric series with a = 3/10 and r = 1/10. Since r | < 1, I can use the formula for summing infinite geometric series:

      0.333... = (3/10) [1/(1 - (1/10))] = (3/10)(10/9) = 3/9 = 1/3

Using the summation formula to show that the geometric series "expansion" of 0.333... has a value of one-third is the "showing" that the exercise asked for. You can use this method to convert any repeating decimal to its fractional form:

  • By use of a geometric series, convert 1.363636... to fractional form.
  • First I'll break this into its constituent parts, so I can find the pattern:

      1.363636.. = 1 + 0.36 + 0.0036 + 0.000036 + ...

    There are two digits that repeat, so the fractions are a little bit different. But this is still a geometric series:

      1.363636... = 1 + 0.36 + 0.0036 + ... = 1 + sum [(9/25)(1/100)^(n-1)]

    Then this is the leading "1", plus a geometric series having a = 9/25 and r = 1/100. Then the sum is:

      1.363636... = 1 + (9/25)(1/(1 - 1/100)) = 1 + 4/11 = 15/11

Note: This technique can also be used to convert any repeating decimal into fractional form, and also can be used to prove that 0.999... = 1.

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Cite this article as:

Stapel, Elizabeth. "Geometric Series." Purplemath. Available from
    http://www.purplemath.com/modules/series5.htm. Accessed
 

 



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