Section 11.4 Lecture Slides

1. Chapter 11: Sequences, Series, and Power Series
Section 11.4: Comparison Tests
Alea Wittig
SUNY Albany

2. Outline
The Direct Comparison Test
The Limit Comparison Test

3. ▶ Consider the series
∞
X
n=1
1
2n + 1
▶ This series looks very much like a convergent geometric series
we have examined:
∞
X
n=1
1
2n
= 1
▶ But
P 1
2n+1 isn’t quite a geometric series, since it can’t be
written in the form
P
arn−1
▶ Nevertheless, we can use what we know about the series
P 1
2n
to determine whether the series
P 1
2n+1 converges as well.

4. ▶ Let sn denote the nth partial sum of
P 1
2n+1:
sn =
n
X
i=1
1
2i + 1
▶ Observe that
0 <
1
2i + 1
<
1
2i
for all integers i
▶ So adding up the first n terms we have
0 <
n
X
i=1
1
2i + 1
| {z }
sn
<
n
X
i=1
1
2i
≤ 1
since
n
X
i=1
1
2i
<
∞
X
n=1
1
2n
= 1
therefore 0 < sn < 1 for all n, ie, sn is bounded

5. ▶ Since 1
2n+1+1
is positive for any n,
n
X
i=1
1
2i + 1
<
n
X
i=1
1
2i + 1
+
1
2n+1 + 1
sn < sn+1, ie, sn is increasing
▶ So by the bounded and monotonic convergence theorem,
sn =
Pn
i=1
1
2i +1
is a bounded, increasing sequence and thus it
is convergent.
▶ Thus by definition of a convergent series,
∞
X
n=1
1
2n + 1
is convergent

6. The Direct Comparison Test

7. The Direct Comparison Test
Theorem: Suppose that
∞
X
n=1
an and
∞
X
n=1
bn
are series with positive terms.
(i) If
P
bn is convergent and an ≤ bn for all n, then
P
an is also
convergent.
(ii) If
P
bn is divergent and an ≥ bn for all n, then
P
an is also
divergent.

8. On the left is an illustration of Direct Comparison (i).
On the right is an illustration of Direct Comparison (ii).
Here tn =
Pn
i=1 bi , the partial sum of bn.
And sn =
Pn
i=1 ai , the partial sum of an.

9. ▶ In general, to use direct comparison on the series
P
an, we
need a known series
P
bn to compare it to.
▶
P
bn will most often be:
▶ A geometric series: X
arn−1
▶ converges if |r| < 1 and diverges if |r| ≥ 1.
▶ A p-series:
X 1
np
▶ converges if p > 1 and diverges if p ≤ 1.

10. Example 1
1/3
Determine whether the series
∞
X
n=1
5
2n2 + 4n + 3
converges or diverges.

11. Example 1
2/3
▶ To use direct comparison we first check that we indeeed have
positive terms ✓
▶ Next we must find a series
P
bn to compare with.
∞
X
n=1
an =
∞
X
n=1
5
2n2 + 4n + 3
looks like a p-series so we find a p-series to compare to.
▶ As a general strategy for picking bn, take the numerator of an
over the highest power term in the denominator of an:
∞
X
n=1
bn =
∞
X
n=1
5
2n2

12. Example 1
3/3
▶ For all n we have
5
2n2 + 4n + 3
<
5
2n2
ie, an < bn.
▶ So if the series
P
bn converges, then direct comparison will
tell us that
P
an converges as well.
∞
X
n=1
5
2n2
=
5
2
∞
X
n=1
1
n2
converges p = 2 > 1
=⇒
∞
X
n=1
5
2n2 + 4n + 3
converges by direct comparison

13. Example 2
1/2
Test the series
∞
X
k=1
ln k
k
for convergence or divergence.

14. Example 2
2/2
▶ Here we have a series that doesn’t look quite like a p-series or
a geometric series.
▶ Nevertheless, we know that ln k > 1 for k > e, and the
harmonic series
∞
X
k=1
1
k
= 1 +
1
2
+
∞
X
k=3
1
k
is divergent
▶ So since
ln k
k
>
1
k
for k ≥ 3
by direct comparison,
∞
X
k=3
ln k
k
is divergent
and thus
∞
X
k=1
ln k
k
=
ln 2
2
+
∞
X
k=3
ln k
k
is divergent

15. The Limit Comparison Test

16. ▶ To determine whether the series
∞
X
n=1
1
2n − 1
converges or diverges, then we might think to use direct
comparison test since the terms 1
2n−1 are close to 1
2n .
▶ But this won’t work since, for all n,
1
2n − 1
>
1
2n
and
X 1
2n
converges.
▶ For this example we can use the following test

17. The Limit Comparison Test
Theorem: Suppose that
X
an and
X
bn
are series with positive terms.
If lim
n→∞
an
bn
= c where c > 0 is a finite number
then either both series converge or both diverge.

18. Example 3
1/2
Use the Limit Comparison Test to test
∞
X
n=1
1
2n − 1
for convergence or divergence.

19. Example 3
2/2
▶ We compare
1
2n − 1
with
1
2n
, which we know converges.
lim
n→∞
an
bn
= lim
n→∞
1
2n
1
2n − 1
= lim
n→∞
2n − 1
2n
= lim
n→∞
1 −
1
2n
= 1
▶ So the Limit Comparison Test tells us that
∞
X
n=1
1
2n − 1
converges.

20. Example 4
1/3
Determine whether the sequence
∞
X
n=1
2n2 + 3n
√
5 + n5
converges or diverges.

21. Example 4
2/3
▶ The series has positive terms ✓
▶ Using our general strategy for choosing
P
bn, we compare to
the following series:
2n2 + 3n
√
n5
∞
X
n=1
2n2 + 3n
√
n5
=
∞
X
n=1
2n2
√
n5
+
3n
√
n5
=
∞
X
n=1
2
n1/2
+
3
n3/2
>
∞
X
n=1
2
n1/2
← diverges
▶ We can’t use direct comparison because
2n2 + 3n
√
n5
>
2n2 + 3n
√
5 + n5

22. Example 4
3/3
▶ But taking the limit of the ratio of the terms of the series:
lim
n→∞
2n2 + 3n
√
5 + n5
2n2 + 3n
√
n5
= lim
n→∞
2n2 + 3n
√
5 + n5
√
n5
2n2 + 3n
= lim
n→∞
√
n5
√
5 + n5
=
s
lim
n→∞
n5
5 + n5
=
s
lim
n→∞
1
5
n5 + 1
= 1 0
▶ Thus by Limit Comparison, the series converges.