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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: Copyright © Elizabeth Stapel 1999-2009 All Rights Reserved

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:

• Evaluate the following:

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:

So the value of the summation is 2 097 150

• Evaluate S₁₀ for 250, 100, 40, 16,....

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:

• Find a_n 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:

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

Then:

• 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:

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:

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:

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

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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