Math 231 Worksheet #14 Solutions
1. What is the difference between a series absolutely converging, conditionally converging
and diverging?
Solution:
P
P
A series
an is called absolutely convergent if the series of absolute values
|an |
is convergent.
P
A series an is called conditionally convergent if it is convergent but not absolutely
convergent.
P
P
A series
an is called divergent if limn→∞ sn doesn’t exist, where sn = nk=1 ak =
a1 + a2 + · · · + an (the sn are called the n-th partial sums of the series)
2. Determine if the series absolutely converges, conditionally converges or diverges.
∞
X
1
1 1 1
(−1)n−1 ·
(a) 1 − + − + − . . . =
2 3 4
n
n=1
Solution:
Alternating harmonic series. We know that the standard harmonic series doesn’t
converge, so the series is not absolutely convergent.
Alternating series test: bn ≥ bn+1 and limn→∞ bn = 0
Both of these conditions are satisfied, so the series is convergent. Since it’s not
absolutely convergent, it is conditionally convergent
∞ X
1
1
1
1
1
+ −... =
1− n
(b) 1 − + 1 − + 1 − + 1 −
2
4
8
16
2
n=1
Solution:
limn→∞ 1 −
(c)
∞
X
n=1
(−1)n
1
2n
= 1, so the series is divergent.
n+1
4n − 1
Solution:
n
1
1
n+1
limn→∞ 4n−1
= limn→∞ 4n−1
+ 4n−1
= limn→∞ 4−1 1 + 4n−1
=
n
Since the limit of the terms is not 0 the series is divergent.
∞
X
n+1
(d)
(−1)n 2
4n − 1
n=1
1
4
Solution:
n+1
> n+1
> 4nn2 = 14 n1 . Since the Harmonic series diverges, the series is not
4n2 −1
4n2
absolutely convergent by the comparison test.
Alternating series test:
limn→∞
n+1
4n2 −1
= limn→∞
f 0 (x) =
1
1+ n
1
4n− n
= 0. If f (x) =
x+1
4x2 −1
then
1 · (4x2 − 1) − (x + 1) · 8x
8x2 + x + 1
=
−
(4x2 − 1)2
(4x2 − 1)2
and so f 0 (x) < 0 for x ≥ 1 and the absolute value of the terms are decreasing. So
the series is convergent by the alternating series test. Since it is not absolutely
convergent, the series is conditionally convergent.
∞
X
n+1
(e)
(−1)n 3
4n − 1
n=1
Solution:
Limit comparison test: an =
1
1
n2
and bn =
n+1
4n3 −1
3
−1
n2
= limn→∞ n+1
= limn→∞ n4n3 −n
2 = 4
3
4n −1
P 1
converges, p-series with p = 2 > 1, so by the limit comparison
The series
n2
test the series is absolutely convergent.
limn→∞
an
bn
3. We will show that
ln 2 =
∞
X
(−1)n
n=0
n+1
=1−
1 1 1
+ − + −···
2 3 4
Using the alternating series estimation, how many terms of the series do we need to
add in order to find ln 2 to an error less than .05?
Solution:
|Rn | = |s − sn | ≤ bn+1
1
< 0.05
bn+1 =
n+1
n > 19
n = 20
4. We will show that
∞
π X (−1)n
1 1 1
=
= 1 − + − + −···
4
2n + 1
3 5 7
n=0
Using the alternating series estimation, how many terms of the series do we need to
add in order to find π4 to an error less than .005?
Solution:
|Rn | = |s − sn | ≤ bn+1
1
bn+1 =
< 0.005
2n + 1
n > 99.5
n = 100
5. Determine for which real numbers p ≥ 0 the series
∞
X
(−1)n
n=1
np
converges absolutely, conditionally converges or diverges.
Solution:
P
P 1
converges only if p > 1. Thus, the series ∞
We know that
n=1
np
converges when p > 1 but absolutely diverges when p ≤ 1.
By the alternating series test, the series converges for all p > 0.
Thus the series is conditionally converges for 0 < p ≤ 1.
The series diverges for p = 0.
(−1)n
np
absolutely