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n = 1 ( 1 2 ) n 1 = 1 + 1 2 + 1 4 + 1 8 +

is a geometric series with initial term a = 1 and ratio r = 1 / 2 .

In general, when does a geometric series converge? Consider the geometric series

n = 1 a r n 1

when a > 0 . Its sequence of partial sums { S k } is given by

S k = n = 1 k a r n 1 = a + a r + a r 2 + + a r k 1 .

Consider the case when r = 1 . In that case,

S k = a + a ( 1 ) + a ( 1 ) 2 + + a ( 1 ) k 1 = a k .

Since a > 0 , we know a k as k . Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for r = 1 . For r 1 , to find the limit of { S k } , multiply [link] by 1 r . Doing so, we see that

( 1 r ) S k = a ( 1 r ) ( 1 + r + r 2 + r 3 + + r k 1 ) = a [ ( 1 + r + r 2 + r 3 + + r k 1 ) ( r + r 2 + r 3 + + r k ) ] = a ( 1 r k ) .

All the other terms cancel out.

Therefore,

S k = a ( 1 r k ) 1 r for r 1 .

From our discussion in the previous section, we know that the geometric sequence r k 0 if | r | < 1 and that r k diverges if | r | > 1 or r = ± 1 . Therefore, for | r | < 1 , S k a / ( 1 r ) and we have

n = 1 a r n 1 = a 1 r if | r | < 1 .

If | r | 1 , S k diverges, and therefore

n = 1 a r n 1 diverges if | r | 1 .

Definition

A geometric series is a series of the form

n = 1 a r n 1 = a + a r + a r 2 + a r 3 + .

If | r | < 1 , the series converges, and

n = 1 a r n 1 = a 1 r for | r | < 1 .

If | r | 1 , the series diverges.

Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than n = 1 or the exponent involves a linear expression for n other than n 1 . As long as we can rewrite the series in the form given by [link] , it is a geometric series. For example, consider the series

n = 0 ( 2 3 ) n + 2 .

To see that this is a geometric series, we write out the first several terms:

n = 0 ( 2 3 ) n + 2 = ( 2 3 ) 2 + ( 2 3 ) 3 + ( 2 3 ) 4 + = 4 9 + 4 9 · ( 2 3 ) + 4 9 · ( 2 3 ) 2 + .

We see that the initial term is a = 4 / 9 and the ratio is r = 2 / 3 . Therefore, the series can be written as

n = 1 4 9 · ( 2 3 ) n 1 .

Since r = 2 / 3 < 1 , this series converges, and its sum is given by

n = 1 4 9 · ( 2 3 ) n 1 = 4 / 9 1 2 / 3 = 4 3 .

Determining convergence or divergence of a geometric series

Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.

  1. n = 1 ( −3 ) n + 1 4 n 1
  2. n = 1 e 2 n
  1. Writing out the first several terms in the series, we have
    n = 1 ( −3 ) n + 1 4 n 1 = ( −3 ) 2 4 0 + ( −3 ) 3 4 + ( −3 ) 4 4 2 + = ( −3 ) 2 + ( −3 ) 2 · ( −3 4 ) + ( −3 ) 2 · ( −3 4 ) 2 + = 9 + 9 · ( −3 4 ) + 9 · ( −3 4 ) 2 + .

    The initial term a = −3 and the ratio r = −3 / 4 . Since | r | = 3 / 4 < 1 , the series converges to
    9 1 ( −3 / 4 ) = 9 7 / 4 = 36 7 .
  2. Writing this series as
    e 2 n = 1 ( e 2 ) n 1

    we can see that this is a geometric series where r = e 2 > 1 . Therefore, the series diverges.
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Determine whether the series n = 1 ( −2 5 ) n 1 converges or diverges. If it converges, find its sum.

5 / 7

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We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.

Writing repeating decimals as fractions of integers

Use a geometric series to write 3. 26 as a fraction of integers.

Since 3. 26 = 3.262626 , first we write

3.262626 = 3 + 26 100 + 26 1000 + 26 100,000 + = 3 + 26 10 2 + 26 10 4 + 26 10 6 + .

Ignoring the term 3, the rest of this expression is a geometric series with initial term a = 26 / 10 2 and ratio r = 1 / 10 2 . Therefore, the sum of this series is

26 / 10 2 1 ( 1 / 10 2 ) = 26 / 10 2 99 / 10 2 = 26 99 .

Thus,

3.262626 = 3 + 26 99 = 323 99 .
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Write 5.2 7 as a fraction of integers.

475 / 90

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Practice Key Terms 7

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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