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Although convergence of N f ( x ) d x implies convergence of the related series n = 1 a n , it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,

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

is a geometric series with initial term a = 1 / e and ratio r = 1 / e , which converges to

1 / e 1 ( 1 / e ) = 1 / e ( e 1 ) / e = 1 e 1 .

However, the related integral 1 ( 1 / e ) x d x satisfies

1 ( 1 e ) x d x = 1 e x d x = lim b 1 b e x d x = lim b e x | 1 b = lim b [ e b + e −1 ] = 1 e .

Using the integral test

For each of the following series, use the integral test to determine whether the series converges or diverges.

  1. n = 1 1 / n 3
  2. n = 1 1 / 2 n 1
  1. Compare
    n = 1 1 n 3 and 1 1 x 3 d x .

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

    Thus the integral 1 1 / x 3 d x converges, and therefore so does the series
    n = 1 1 n 3 .
  2. Compare
    n = 1 1 2 n 1 and 1 1 2 x 1 d x .

    Since
    1 1 2 x 1 d x = lim b 1 b 1 2 x 1 d x = lim b 2 x 1 | 1 b = lim b [ 2 b 1 1 ] = ,

    the integral 1 1 / 2 x 1 d x diverges, and therefore
    n = 1 1 2 n 1

    diverges.
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Use the integral test to determine whether the series n = 1 n 3 n 2 + 1 converges or diverges.

The series diverges.

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The p -series

The harmonic series n = 1 1 / n and the series n = 1 1 / n 2 are both examples of a type of series called a p -series.

Definition

For any real number p , the series

n = 1 1 n p

is called a p -series    .

We know the p -series converges if p = 2 and diverges if p = 1 . What about other values of p ? In general, it is difficult, if not impossible, to compute the exact value of most p -series. However, we can use the tests presented thus far to prove whether a p -series converges or diverges.

If p < 0 , then 1 / n p , and if p = 0 , then 1 / n p 1 . Therefore, by the divergence test,

n = 1 1 / n p diverges if p 0 .

If p > 0 , then f ( x ) = 1 / x p is a positive, continuous, decreasing function. Therefore, for p > 0 , we use the integral test, comparing

n = 1 1 n p and 1 1 x p d x .

We have already considered the case when p = 1 . Here we consider the case when p > 0 , p 1 . For this case,

1 1 x p d x = lim b 1 b 1 x p d x = lim b 1 1 p x 1 p | 1 b = lim b 1 1 p [ b 1 p 1 ] .

Because

b 1 p 0 if p > 1 and b 1 p if p < 1 ,

we conclude that

1 1 x p d x = { 1 p 1 if p > 1 if p < 1 .

Therefore, n = 1 1 / n p converges if p > 1 and diverges if 0 < p < 1 .

In summary,

n = 1 1 n p { converges if p > 1 diverges if p 1 .

Testing for convergence of p -series

For each of the following series, determine whether it converges or diverges.

  1. n = 1 1 n 4
  2. n = 1 1 n 2 / 3
  1. This is a p -series with p = 4 > 1 , so the series converges.
  2. Since p = 2 / 3 < 1 , the series diverges.
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Does the series n = 1 1 n 5 / 4 converge or diverge?

The series converges.

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Estimating the value of a series

Suppose we know that a series n = 1 a n converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum n = 1 N a n where N is any positive integer. The question we address here is, for a convergent series n = 1 a n , how good is the approximation n = 1 N a n ? More specifically, if we let

R N = n = 1 a n n = 1 N a n

be the remainder when the sum of an infinite series is approximated by the N th partial sum, how large is R N ? For some types of series, we are able to use the ideas from the integral test to estimate R N .

Remainder estimate from the integral test

Suppose n = 1 a n is a convergent series with positive terms. Suppose there exists a function f satisfying the following three conditions:

  1. f is continuous,
  2. f is decreasing, and
  3. f ( n ) = a n for all integers n 1 .

Let S N be the N th partial sum of n = 1 a n . For all positive integers N ,

S N + N + 1 f ( x ) d x < n = 1 a n < S N + N f ( x ) d x .

In other words, the remainder R N = n = 1 a n S N = n = N + 1 a n satisfies the following estimate:

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

This is known as the remainder estimate    .

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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