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We illustrate [link] in [link] . In particular, by representing the remainder R N = a N + 1 + a N + 2 + a N + 3 + as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by N f ( x ) d x and bounded below by N + 1 f ( x ) d x . In other words,

R N = a N + 1 + a N + 2 + a N + 3 + > N + 1 f ( x ) d x

and

R N = a N + 1 + a N + 2 + a N + 3 + < N f ( x ) d x .

We conclude that

N + 1 f ( x ) d x < R N < N f ( x ) d x .

Since

n = 1 a n = S N + R N ,

where S N is the N th partial sum, we conclude that

S N + N + 1 f ( x ) d x < n = 1 a n < S N + N f ( x ) d x .
This shows two graphs side by side of the same decreasing concave up function y = f(x) that approaches the x axis in quadrant 1. Rectangles are drawn with a base of 1 over the intervals N through N + 4. The heights of the rectangles in the first graph are determined by the value of the function at the right endpoints of the bases, and those in the second graph are determined by the value at the left endpoints of the bases. The areas of the rectangles are marked: a_(N + 1), a_(N + 2), through a_(N + 4).
Given a continuous, positive, decreasing function f and a sequence of positive terms a n such that a n = f ( n ) for all positive integers n , (a) the areas a N + 1 + a N + 2 + a N + 3 + < N f ( x ) d x , or (b) the areas a N + 1 + a N + 2 + a N + 3 + > N + 1 f ( x ) d x . Therefore, the integral is either an overestimate or an underestimate of the error.

Estimating the value of a series

Consider the series n = 1 1 / n 3 .

  1. Calculate S 10 = n = 1 10 1 / n 3 and estimate the error.
  2. Determine the least value of N necessary such that S N will estimate n = 1 1 / n 3 to within 0.001 .
  1. Using a calculating utility, we have
    S 10 = 1 + 1 2 3 + 1 3 3 + 1 4 3 + + 1 10 3 1.19753 .

    By the remainder estimate, we know
    R N < N 1 x 3 d x .

    We have
    10 1 x 3 d x = lim b 10 b 1 x 3 d x = lim b [ 1 2 x 2 ] N b = lim b [ 1 2 b 2 + 1 2 N 2 ] = 1 2 N 2 .

    Therefore, the error is R 10 < 1 / 2 ( 10 ) 2 = 0.005 .
  2. Find N such that R N < 0.001 . In part a. we showed that R N < 1 / 2 N 2 . Therefore, the remainder R N < 0.001 as long as 1 / 2 N 2 < 0.001 . That is, we need 2 N 2 > 1000 . Solving this inequality for N , we see that we need N > 22.36 . To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is N = 23 .
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For n = 1 1 n 4 , calculate S 5 and estimate the error R 5 .

S 5 1.09035 , R 5 < 0.00267

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Key concepts

  • If lim n a n 0 , then the series n = 1 a n diverges.
  • If lim n a n = 0 , the series n = 1 a n may converge or diverge.
  • If n = 1 a n is a series with positive terms a n and f is a continuous, decreasing function such that f ( n ) = a n for all positive integers n , then
    n = 1 a n and 1 f ( x ) d x

    either both converge or both diverge. Furthermore, if n = 1 a n converges, then the N th partial sum approximation S N is accurate up to an error R N where N + 1 f ( x ) d x < R N < N f ( x ) d x .
  • The p -series n = 1 1 / n p converges if p > 1 and diverges if p 1 .

Key equations

  • Divergence test
    If a n 0 as n , n = 1 a n diverges .
  • p -series
    n = 1 1 n p { converges if p > 1 diverges if p 1
  • Remainder estimate from the integral test
    N + 1 f ( x ) d x < R N < N f ( x ) d x

For each of the following sequences, if the divergence test applies, either state that lim n a n does not exist or find lim n a n . If the divergence test does not apply, state why.

a n = n 5 n 2 3

lim n a n = 0 . Divergence test does not apply.

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a n = ( 2 n + 1 ) ( n 1 ) ( n + 1 ) 2

lim n a n = 2 . Series diverges.

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a n = ( 2 n + 1 ) 2 n ( 3 n 2 + 1 ) n

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a n = 2 n 3 n / 2

lim n a n = (does not exist). Series diverges.

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a n = 2 n + 3 n 10 n / 2

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a n = e −2 / n

lim n a n = 1 . Series diverges.

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a n = tan n

lim n a n does not exist. Series diverges.

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a n = 1 cos 2 ( 1 / n ) sin 2 ( 2 / n )

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a n = ( 1 1 n ) 2 n

lim n a n = 1 / e 2 . Series diverges.

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a n = ( ln n ) 2 n

lim n a n = 0 . Divergence test does not apply.

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State whether the given p -series converges.

n = 1 1 n n

Series converges, p > 1 .

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n = 1 1 n 4 3

Series converges, p = 4 / 3 > 1 .

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Practice Key Terms 4

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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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