Comparison Test for Integrals
If on : convergence of forces convergence of ; divergence of to forces divergence of .
Proof
Let ; this is increasing since , and bounded since
Since is monotone and bounded, exists. By the definition of supremum, for every there is such that for ,
hence, taking limits as , : the improper integral of converges to .
For (2), since , necessarily. Hence for every there is such that , . But
so diverges to .
Example
For we have , hence . Therefore
and
so converges.
Related
Stated in
- Proposition 4.31 (Comparison Test for Integrals)§4.5 Improper Integrals