Divergence and Integral Tests

The Essentials

The divergence test states that if the following limit is a non-zero number or doesn't exist, then the series diverges:

\[ \lim_{n\to\infty}a_n \]

Note that it doesn't work the other way around.

The integral test is used when all terms of the series are positive and there exists a function \( f \) that is decreasing, continuous, and \( a_n=f(n) \). If these conditions are met, the series and the integral of the function from one to infinity either both converge or both diverge. Note that if they converge, the integral is not necessarily equal to the convergence.

This series is called a p-series:

\[ \sum_{n=1}^\infty \frac{1}{n^p} \]

A p-series will diverge if p is less than or equal to 1 and diverge if p is greater than 1.

After adding up a finite number of terms in a series \( S_N \), the remainder of the terms can be calculated if all terms of the series are positive and there exists a function \( f \) that is decreasing, continuous, and \( a_n=f(n) \). If these conditions are met, the remainder \( R_N \) can be estimated where \( R_N \) is the infinite sum of the series minus \( S_N \):

\[ \int_{N+1}^\infty f(x)dx

Example

For the series, determine the least value of N such that the sum \( S_N \) will estimate the series within 0.001:

\[ \sum_{n=1}^{\infty} \frac{1}{n^4} \]

The remainder is

\[ R_N < \int_N^\infty \frac{1}{x^4}dx \]
\[ R_N < \lim_{b\to\infty}\left( -\frac{1}{3}x^{-3}\Big|_N^b\right) \]
\[ R_N < \lim_{b\to\infty}\left(-\frac{1}{3b^3}+\frac{1}{3N^3}\right) \]
\[ R_N < \frac{1}{3N^3} \]

We want to remainder to be less than 0.001:

\[ \frac{1}{3N^3} < 0.001 \]
\[ \frac{1}{0.001} < 3N^3 \]
\[ 6.93 < N \]

We will round up so that the remainder is within the desired amount. Seven terms in the series (or \( S_7 \)) will be an accurate estimate of the series within 0.001.

Practice

  1. Use the divergence test to tell whether the series diverges or the test is inconclusive:
    \[ \sum_{n=1}^{\infty}e^{n^2} \]
  2. Use the integral test to tell whether the series diverges or converges:
    \[ \sum_{n=1}^\infty \frac{1}{n^4} \]
  3. Use the p-series test to tell whether the series diverges or converges:
    \[ \sum_{n=1}^\infty \frac{1}{\sqrt[3]{n^4}} \]
  4. Find the number of terms N that is required for the estimate \( S_N \) is accurate to 0.0001:
    \[ \sum_{n=1}^{\infty} \frac{1}{1+n^2} \]

Solution:

  1. The series diverges.
  2. The series converges.
  3. The series converges.
  4. N = 1000

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