The sequence
is called summable if the sequence
converges, where
In this case,
is denoted by
and is called the sum of the sequence
The sequence
is summable if and only if
Therefore, if
is summable then
Grouping the series as
we see that the sum of the terms in each ( ) is
;
thus
is unbounded. Therefore the series
is not summable.
The infinite series
is called the geometric series. If
then each individual term does not converge to
so the series diverges. If
,
then it follows from the two equations
that
or
Since
,
it follows that
A nonnegative series is convergent if and only if the set of partial sums is bounded.
1. Suppose that
Then if
converges, so does
Proof. If
then
Since converges, is bounded. Therefore is bounded; consequently is convergent.
2. If , and , then converges if and only if converges.
Proof. Suppose that
converges. Since
,
there exists some
such that
Since the series converges, it follows that the series converges and this implies the convergence of the whole series which has only finitely many additional terms.
The converse follows at once since
Let
for all
,
and suppose that
(a) If then converges.
(b) If then the terms of do not approach so diverges.
Proof. (a) Suppose that
.
Fix
with
.
The assumption
implies that there exists some
such that
i.e.,
Thus
Since converges, it follows that converges. Therefore the whole series converges.
(b) Suppose that
.
Fix
with
.
Then there exists some
such that
which means
This shows that the individual terms of
do not approach
,
so
is not summable.
Let
for all
,
and suppose that
(a) If then converges.
(b) If then the terms of do not approach so diverges.
Proof. (a) Suppose that
.
Fix
with
.
The assumption
implies that there is some
such that
i.e.,
Thus
Since the geometric series
converges, it follows that
converges. Therefore the whole series
converges.
(b) Suppose that
.
Fix
with
.
Then there exists some
such that
which means
This shows that the individual terms of do not approach , so is not summable.
Let
be any positive number. The ratio
converges to
as
approaches infinity. Hence it follows from the ratio test that the series
converges for any positive number
.
Consequently we have
If the limit exists, it is denoted by , and called an improper integral.
Find if .
Show that does not exist.
Suppose that for and that exists. Show that if for all , then exists.
The improper integral is defined as . If both improper integrals and exist, then the improper integral exists and equals .
Note that does not exist although the limit exists.
Prove that if exists then exists and equals
There is another kind of improper integral in which the interval is bounded,
but the function is unbounded. If
is an unbounded function such that for any
with
is bounded on
and the limit
exists, then the improper integral
is said to exist and is equal to
Similarly, if
is an unbounded function such that for any
with
is bounded on
and the limit
exists, then the improper integral
is said to exist and is equal to
Suppose that
is positive and decreasing on
,
and that
for all
.
Then
converges if and only if the improper integral
Proof. The convergence of
is equivalent to the convergence of the series
Since
is decreasing we have
Therefore, if
exists, then the series
converges and so it follows from the comparison test that
converges, and so
converges.
If converges, the second half of the inequality shows that the series converges and so the improper integral exists.
Establish convergence or divergence by a comparison test
1. 2. 3. 4. 5.
6. 7. 8. 9. 10.
Decide for or against convergence based on the integral test
11. 12. 13. 14.
15. 16. 17. 18.
Decide convergence and name your test
19. 20. 21. 22. 23.
24. 25. (test all ) 26. (test all )
27. 28. (test all )
29. (a) Show that
(b) Express as a Riemann sum and show that approaches
30. Compute the sum of the telescoping series
31. Compute the sum
by considering the Taylor expansion of