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10.4 Integral Test for Convergence

Syllabus
2020
Topic
10.4
Level
—

Match a Positive Decreasing Series to an Improper Integral

Suppose an=f(n)a_n=f(n) and, for all x≥Nx\ge N, the function ff is continuous, positive, and decreasing. Then the terms can be compared with adjacent strips under the graph, so the infinite series and corresponding improper integral either both converge or both diverge.

\sum_{n=N}^{\infty}a_n\text{ converges}\iff\int_N^{\infty}f(x),dx\text{ converges},\qquad a_n=f(n)

  1. Choose f(x)f(x) so that f(n)=anf(n)=a_n.
  2. State where ff is continuous, positive, and decreasing.
  3. Rewrite ∫N∞f(x) dx\int_N^{\infty}f(x)\,dx as a limit and evaluate it.
  4. Transfer only the convergence or divergence conclusion to the series.

Example: for ∑n=1∞1/(n2+1)\sum_{n=1}^{\infty}1/(n^2+1), let f(x)=1/(x2+1)f(x)=1/(x^2+1). For x≥1x\ge1, ff is continuous and positive, and f′(x)=−2x/(x2+1)2<0f'(x)=-2x/(x^2+1)^2<0, so it is decreasing. Also,
∫1∞dxx2+1=lim⁡b→∞[arctan⁡x]1b=π2−π4=π4.\int_1^{\infty}\frac{dx}{x^2+1}=\lim_{b\to\infty}[\arctan x]_1^b=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}.
The improper integral converges, so the series converges by the integral test.

The value π/4\pi/4 in the example is the integral's value, not the sum of the series. The integral test determines convergence behavior; it does not normally calculate the exact series sum. If positivity or decreasing behavior fails, this test has not been justified.