Let
Is positive and continuous on the interval ? Yes No
The derivative,
Is decreasing on the interval ? Yes No
Evaluate the improper integral:
Does the infinite series converge or diverge? Converge Diverge
We determine the convergence or divergence of an infinite series using a related improper integral.
The rationale for the integral test is that both the improper integral and the the sum of an infinite series can be interpreted as the area under a graph. Moreover, these areas will be comparable. Hence, we can use the logic of the comparison theorem of improper integrals: “less than convergent is convergent and greater than divergent is divergent”. The two figures below show that if the function satisfies the hypotheses of the integral test and if , then we can use the behavior of the integral to make conclusions about the behavior of the series.
Let
Is positive and continuous on the interval ? Yes No
The derivative,
Is decreasing on the interval ? Yes No
Evaluate the improper integral:
Does the infinite series converge or diverge? Converge Diverge
Let
Is positive and continuous on the interval ?Yes No
The derivative,
Is decreasing on the interval ? Yes No
Evaluate the improper integral:
Does the infinite series converge or diverge?Converge Diverge
Let
Is positive and continuous on the interval ?Yes No
The derivative,
Is decreasing on the interval ? Yes No
Evaluate the improper integral:
Does the infinite series converge or diverge? Converge Diverge
Let
Is positive and continuous on the interval ?Yes No
The derivative,
Is decreasing on the interval ? Yes No
Evaluate the improper integral (hint, use integration by parts):
Does the infinite series converge or diverge? Converge Diverge
We have seen in the section on improper integrals that the -integral
converges if and diverges if . Since the function is continuous, positive, and decreasing on the interval , the integral test can be applied to conclude that the associated -series has the same behavior. This is stated in the following theorem.