Improper integrals and series have a lot in common. The integral test
bridges the two notions. Notice that series do possess tools which
are not available for improper integrals (such as the ratio and root
tests) and the improper integrals possess other tools not available
for series (such as the techniques of integration). So depending on
the nature of the problem, you may switch from one to the other one
via the integral test.
The Integral Test. Consider a decreasing function . Hence for any , we have
which implies
Set , then we have
for . If we add these inequalities from n=1 to n=N, we get
If is the sequence of partial sums associated to the series , then we have
or equivalently
Since for any , then we know that
is bounded.
The Integral test
Remark. Note that it may happen that f(x) is not decreasing on
the entire interval but only on some subinterval
(where A > 1). The above conclusion is still valid.
Example. Establish convergence or divergence of
Answer. Set
Then we have
Clearly the function for . Hence f(x) is decreasing on . So the integral test implies that the improper integral
is convergent if and only if the series
is convergent. We recognize Bertrand's series. So we conclude that the improper integral
is convergent if and only if
are all divergent while the improper integrals
are all convergent.
Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.
Mohamed A. Khamsi