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Key concepts

  • Given the infinite series
    n = 1 a n = a 1 + a 2 + a 3 +

    and the corresponding sequence of partial sums { S k } where
    S k = n = 1 k a n = a 1 + a 2 + a 3 + + a k ,

    the series converges if and only if the sequence { S k } converges.
  • The geometric series n = 1 a r n 1 converges if | r | < 1 and diverges if | r | 1 . For | r | < 1 ,
    n = 1 a r n 1 = a 1 r .
  • The harmonic series
    n = 1 1 n = 1 + 1 2 + 1 3 +

    diverges.
  • A series of the form n = 1 [ b n b n + 1 ] = [ b 1 b 2 ] + [ b 2 b 3 ] + [ b 3 b 4 ] + + [ b n b n + 1 ] +
    is a telescoping series. The k th partial sum of this series is given by S k = b 1 b k + 1 . The series will converge if and only if lim k b k + 1 exists. In that case,
    n = 1 [ b n b n + 1 ] = b 1 lim k ( b k + 1 ) .

Key equations

  • Harmonic series
    n = 1 1 n = 1 + 1 2 + 1 3 + 1 4 +
  • Sum of a geometric series
    n = 1 a r n 1 = a 1 r for | r | < 1

Using sigma notation, write the following expressions as infinite series.

1 + 1 2 + 1 3 + 1 4 +

n = 1 1 n

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1 1 2 + 1 3 1 4 + ...

n = 1 ( −1 ) n 1 n

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sin 1 + sin 1 / 2 + sin 1 / 3 + sin 1 / 4 +

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Compute the first four partial sums S 1 ,… , S 4 for the series having n th term a n starting with n = 1 as follows.

a n = sin ( n π / 2 )

1 , 1 , 0 , 0

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In the following exercises, compute the general term a n of the series with the given partial sum S n . If the sequence of partial sums converges, find its limit S .

S n = 1 1 n , n 2

a n = S n S n 1 = 1 n 1 1 n . Series converges to S = 1 .

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S n = n ( n + 1 ) 2 , n 1

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S n = n , n 2

a n = S n S n 1 = n n 1 = 1 n 1 + n . Series diverges because partial sums are unbounded.

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S n = 2 ( n + 2 ) / 2 n , n 1

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For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

n = 1 n n + 2

S 1 = 1 / 3 , S 2 = 1 / 3 + 2 / 4 > 1 / 3 + 1 / 3 = 2 / 3 , S 3 = 1 / 3 + 2 / 4 + 3 / 5 > 3 · ( 1 / 3 ) = 1 . In general S k > k / 3 . Series diverges.

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n = 1 ( 1 ( −1 ) n ) )

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n = 1 1 ( n + 1 ) ( n + 2 ) ( Hint: Use a partial fraction decomposition like that for n = 1 1 n ( n + 1 ) . )

S 1 = 1 / ( 2.3 ) = 1 / 6 = 2 / 3 1 / 2 , S 2 = 1 / ( 2.3 ) + 1 / ( 3.4 ) = 2 / 12 + 1 / 12 = 1 / 4 = 3 / 4 1 / 2 , S 3 = 1 / ( 2.3 ) + 1 / ( 3.4 ) + 1 / ( 4.5 ) = 10 / 60 + 5 / 60 + 3 / 60 = 3 / 10 = 4 / 5 1 / 2 , S 4 = 1 / ( 2.3 ) + 1 / ( 3.4 ) + 1 / ( 4.5 ) + 1 / ( 5.6 ) = 10 / 60 + 5 / 60 + 3 / 60 + 2 / 60 = 1 / 3 = 5 / 6 1 / 2 .

The pattern is S k = ( k + 1 ) / ( k + 2 ) 1 / 2 and the series converges to 1 / 2 .

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n = 1 1 2 n + 1 ( Hint: Follow the reasoning for n = 1 1 n . )

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Suppose that n = 1 a n = 1 , that n = 1 b n = −1 , that a 1 = 2 , and b 1 = −3 . Find the sum of the indicated series.

n = 1 ( a n + b n )

0

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n = 1 ( a n 2 b n )

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n = 2 ( a n b n )

−3

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n = 1 ( 3 a n + 1 4 b n + 1 )

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State whether the given series converges and explain why.

n = 1 1 n + 1000 ( Hint: Rewrite using a change of index.)

diverges, n = 1001 1 n

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n = 1 1 n + 10 80 ( Hint: Rewrite using a change of index.)

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1 + 1 10 + 1 100 + 1 1000 +

convergent geometric series, r = 1 / 10 < 1

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1 + e π + e 2 π 2 + e 3 π 3 +

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1 + π e + π 2 e 4 + π 3 e 6 + π 4 e 8 +

convergent geometric series, r = π / e 2 < 1

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1 π 3 + π 2 9 π 3 27 +

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For a n as follows, write the sum as a geometric series of the form n = 1 a r n . State whether the series converges and if it does, find the value of a n .

a 1 = −1 and a n / a n + 1 = −5 for n 1 .

n = 1 5 · ( −1 / 5 ) n , converges to −5 / 6

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a 1 = 2 and a n / a n + 1 = 1 / 2 for n 1 .

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a 1 = 10 and a n / a n + 1 = 10 for n 1 .

n = 1 100 · ( 1 / 10 ) n , converges to 100 / 9

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a 1 = 1 / 10 and a n / a n + 1 = −10 for n 1 .

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Use the identity 1 1 y = n = 0 y n to express the function as a geometric series in the indicated term.

x 1 + x in x

x n = 0 ( x ) n = n = 1 ( −1 ) n 1 x n

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1 1 + sin 2 x in sin x

n = 0 ( −1 ) n sin 2 n ( x )

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Evaluate the following telescoping series or state whether the series diverges.

n = 1 2 1 / n 2 1 / ( n + 1 )

S k = 2 2 1 / ( k + 1 ) 1 as k .

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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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