Math 152 (honors sections)
Harold P. Boas
boas@tamu.edu
Reminder
Second examination is Wednesday, November 3.
The exam covers through section 10.4.
An old exam is posted at http://www.math.tamu.edu/˜boas/courses/
152-2002c/exam2.pdf
Math 152
October 29, 2004 — slide #2
Convergence tests so far
∞
• If an 6→ 0, then ∑ an diverges.
n=1
• A geometric series with |ratio| < 1 converges.
• Integral test for positive decreasing functions: the improper integral
∞
corresponding series
∑
f (n) either both converge or both diverge.
Z
∞
1
f (x) dx and the
n=1
• Inequality comparison test for positive terms: if 0 < a n < bn (at least for n large) and if
∞
∞
n=1
n=1
∑ bn converges, then ∑ an converges too.
∞
an
exists (finite limit), and if ∑ bn conn→∞ bn
n=1
• Limit comparison test for positive terms: if lim
∞
verges, then ∑ an converges too.
n=1
Math 152
October 29, 2004 — slide #3
Root test (not in book)
∞
Example:
n
∑ 2n converges
n=1
∞
This is not a geometric series, and it is bigger than the convergent geometric series
comparison Ã
test does
! not seem to help.
n
Since n =
2
n1/n
2
n
, and since lim
n1/n
n
= 1, we have n <
2
µ
1.1
2
1
, so the
2n
n=1
∑
¶n
when n is large, so we
¶
∞ µ
1.1 n
can use the comparison test after all: compare to the convergent geometric series ∑
.
2
n=1
n→∞
Math 152
October 29, 2004 — slide #4
Root test and ratio test
∞
> 1, then
Root test: If 0 < a n , and if lim a1/n
< 1, then ∑ an converges. Moreover, if lim a1/n
n
n
n→∞
∞
∑ an diverges (because then a n 6→ 0). If lim
n→∞
n=1
n=1
a1/n
n
n→∞
= 1, the test gives no information.
Ratio test: Exactly the same as the root test, except look at lim
n→∞
an+1
instead of lim a1/n
n .
n→∞
an
3n (n!)2
Example: ∑
.
(2n)!
n=1
Á
3
an+1
3n+1 (n + 1)!(n + 1)! 3n (n!)(n!)
3(n + 1)2
lim
= lim
= lim
= < 1,
n→∞ an
n→∞
n→∞
(2n + 2)!
(2n)!
(2n + 2)(2n + 1)
4
so the original series converges.
∞
Math 152
October 29, 2004 — slide #5
Series with some negative terms
Negative terms can only help with convergence:
∞
∞
n=1
n=1
if ∑ |an | converges, then so does ∑ an .
An absolutely convergent series converges.
∞
Example: ∑
n=1
(−1)n
n2
∞
converges because ∑
n=1
1
n2
converges.
∞
The two series sum to different values, however: ∑
n=1
∞
Example: ∑
yet
1
n
(−1)n
n2
=
−π 2
12 ,
∞
and ∑
n=1
1
n2
=
π2
6 .
diverges (harmonic series),
n=1
∞
(−1)n
∑ n converges
n=1
(in fact to the value ln( 21 )).
Math 152
October 29, 2004 — slide #6
Alternating series test
∞
If an ↓ 0, then ∑ (−1)n an converges.
n=1
Error estimate: When the alternating series test applies, the sum of the series is trapped between any two consecutive partial sums.
(−1)n
1
1
1
1
∑ n4 = −1 + 24 − 34 + 44 − 54 + · · ·
n=1
∞
(−1)n
1
1
1
1
1
1
1
−1 + 4 − 4 + 4 − 4 < ∑
<
−1
+
−
+
4
2
3
4
5
24 34 44
n=1 n
∞
Example:
(−1)n
< −0.9459
−0.94754 < ∑
4
n=1 n
∞
∞
Exact value: ∑
n=1
Math 152
(−1)n
n4
=
−7π 4
720
≈ −0.947033.
October 29, 2004 — slide #7
Homework
• Read section 10.4, pages 605–610.
• Do the Suggested Homework problems for section 10.4.
Monday we will review for the exam and look at an old exam.
Math 152
October 29, 2004 — slide #8