calculus lesson 9.4 comparison...
TRANSCRIPT
Calculus Lesson 9.4 Comparison Tests Theorem 9.4: Direct Comparison Test (p. 481) Similar to Direct Comparison Test from lesson 7.8. Let
€
an∑ be a series with nonnegative terms. a)
€
an∑ converges if there is a convergent series
€
bn∑ such that
€
bn ≥ an for all n > N. In simple terms, if the larger series converges, then the smaller must converge.
b)
€
an∑ diverges if there is a divergent series
€
bn∑ such that
€
bn∑ has nonnegative terms and
€
bn ≤ an for all n > N. In simple terms, if the smaller series diverges, then the larger diverges.
Example: Does
€
1n3 + 5nn=1
∞
∑ converge or diverge?
Solution: Start with
€
5n ≥ 0 (for
€
n ≥1), then
€
n3 + 5n ≥ n3, giving
€
1n3 + 5n
≤1n3
.
€
1n3n=1
∞
∑ is a p-series with p = 3 > 1, therefore it converges.
Since
€
1n3
≥1
n3 + 5n and
€
1n3n=1
∞
∑ converges, then by the Direct Comparison Test
€
1n3 + 5nn=1
∞
∑
also converges.
Example: Does
€
5n +12n −1n=1
∞
∑ converge or diverge?
Solution: Start with
€
1≥ 0, then
€
5n +1≥ 5n , which would give
€
5n +12n −1
≥5n
2n −1, which would then give
€
5n +12n −1
≥5n
2n.
€
5n
2nn=1
∞
∑ is a geometric series with
€
r =52
>1, therefore it diverges.
Since
€
5n
2n≤5n +12n −1
and
€
5n
2nn=1
∞
∑ diverges, then by the Direct Comparison Test
€
5n +12n −1n=1
∞
∑ also
diverges.
Theorem 9.5: Limit Comparison Test (p. 483) Suppose an > 0 and bn > 0 for all n > N.
(a) If
€
limn→∞
anbn
≥ 0 (except ∞) and
€
bn∑ converges, then
€
an∑ converges. This means if the limit is 0
or any number but ∞, and the comparison series converges, then the original series will converge.
(b) If
€
limn→∞
anbn
> 0 (including ∞) and
€
bn∑ diverges, then
€
an∑ diverges. This means if the limit
equals any number but 0 and the comparison series diverges, then the original series will diverge.
Example: Does
€
1n3 − 5nn=1
∞
∑ converge or diverge?
Solution: Compare
€
1n3 + 5n
to
€
1n3
.
€
1n3n=1
∞
∑ is a p-series with p = 3 > 1, therefore it converges.
€
limn→∞
1
n3−5n1n3
€
= limn→∞
n3
n3 − 5n (divide all terms by n3)
€
= limn→∞
11− 5
n2
€
=1
1− 5∞2
.
Since the 0 < limit < ∞ and
€
1n3n=1
∞
∑ converges, then by the Limit Comparison Test
€
1n3 − 5nn=1
∞
∑ also converges.
Example: Does
€
ln n( )( )3
n3n=1
∞
∑ converge or diverge?
Solution: Compare
€
ln n( )( )3
n3 to .
€
1n2n=1
∞
∑ is a p-series with p = 2 > 1, therefore it converges.
€
limn→∞
ln n( )( )3
n3
1n2
€
= limn→∞
ln n( )( )3
n=∞∞
, this is indeterminate, therefore use L'Hôpital's Rule,
€
limn→∞
3 ln n( )( )2
n=∞∞
, still indeterminate, therefore use L'Hôpital's Rule again,
€
limn→∞
6ln n( )n
=∞∞
, still indeterminate, therefore use L'Hôpital's Rule again,
€
limn→∞
6n1
=6∞
= 0 .
Since the limit = 0 and
€
1n2n=1
∞
∑ converges, then by the Limit Comparison Test
€
ln n( )( )3
n3n=1
∞
∑
also converges.
Example: Does
€
1n ln n( )n=2
∞
∑ converge or diverge?
Solution: Compare
€
1n ln n( )
to .
€
1nn=2
∞
∑ is the harmonic series, therefore it diverges.
€
limn→∞
1
n ln n( )1n
€
= limn→∞
nln n( )
=∞∞
, this is indeterminate, therefore use L'Hôpital's Rule,
€
limn→∞
1
2 n−1
2
1n
€
= limn→∞
n2 n
€
= limn→∞
n2
.
Since the limit = ∞ and
€
1nn=2
∞
∑ diverges, then by the Limit Comparison Test
€
1n ln n( )n=2
∞
∑
also diverges.