Math 2433 Homework #6
Solutions
Section 12.6.
Determine whether the given series is absolutely convergent, conditionally convergent or
divergent.
P∞ n2
2.
n=1 2n .
Solution: We have an =
n2
2n
so an+1 =
(n+1)2
2n+1
an+1 = lim
lim
n→∞ an n→∞
and thus
n+1 2
n
2
=
1
<1
2
so by Ratio Test, the series is absolutely convergent.
P
n
4.
(−1)n 2n4 .
n
2n+1
Solution: Here, we have We have an = (−1)n 2n4 so an+1 = (−1)n+1 (n+1)
4 and thus
an+1 2
= lim
lim
=2>1
n→∞ n+1 4
n→∞ an n
so by Ratio Test, the series is divergent.
5.
P (−1)n+1
√
4
n
.
1
n+2 √ 1
Solution: Here, we have We have an = (−1)n+1 √
and thus
4 n so an+1 = (−1)
4
n+1
r
an+1 n
= lim 4
=1
lim
n→∞
n→∞ an n+1
so by Ratio Test is inconclusive. Thus, to check
P 1 whether the series is absolutely
√
convergent, we look at the absolute value series
4 n which we know diverges (p-series
with p = 1/4. So the given series is not absolutely convergent. To see whether the
P (−1)n+1
√
series
is convergent, we use the Alternating series Test. Here, we have
4n
1
bn = n1/4
. Clearly limn→∞ bn = 0. Also, the function f (x) = x−1/4 has the derivative
f ′ (x) = − 41 x−5/4 , which is negative for x ≥ 1. Hence bn is decreasing for n ≥ 1. By
Alternating Series Test, the given series is convergent. Since it is convergent but not
absolutely convergent, it is conditionally convergent.
P
10. (−1)n √nn3 +2 .
Solution: As above, the Ratio Test turns out to be inconclusivr here (you have to check
1
2
Solutions
P n
√
this), so we have to look at the absolute value series, which is
. We can use the
P n n3 +2P 1
√
Limit Comparison Test here. The series for comparison is
which is a
=
n1/2
n3
divergent p-series. We compute
r
√ n
n3
n3 +2
=1>0
lim
= lim
n
3
√
n→∞
n→∞
n +2
3
n
so by the Limit Comparison Test, the absolute value series diverges. This means that
the given series is not absolutely
To check if it is conditionally convegent,
P convergent.
n
n
√
we look at the original series (−1) n3 +2 and see that it is an alternating series of the
P
form (−1)n bn where bn = √nn3 +2 . To check whether bn+1 ≤ bn , we look at the function
x
f (x) = (x3 +2)
1/2 . We have
3
(x3 + 2)1/2 − 23 x3 (x3 + 2)−1/2
(x3 + 2) − 23 x3
1 − x2
f (x) =
=
=
< 0,
(x3 + 2)
(x3 + 2)3/2
(x3 + 2)3/2
′
for x ≥ 2.
Hence bn is decreasing for n ≥ 2. Also, we can show that limn→∞ bn = 0. Therefore, by
the Alterating Series Test, the given series conveges. So the final conclusion is that the3
series is conditionally convergent.
P
18. nn!n .
Solution: We have
an+1 = lim
lim
n→∞ an n→∞
(n+1)!
n!
(n+1)n+1
nn
= lim = lim
n→∞
so by Ratio Test, the series is absolutely convergent.
n→∞
1
n+1 n
n
=
1
<1
e
31. For which of the following series is the Ratio Test inconclusive?
P 1
(i)
3.
P nn
(ii)
2n .
P (−3)n−1
√
(iii)
.
n
√
P n
.
(iv)
1+n2
Solution: The Ratio Test is inconclusive for (i) and (iv). I am giving you the final
answer but you have to check each case and decide.
Section 12.7
Determine whether the given series is convergent or divergent.
Math 2433 Homework #6
4.
3
P
(−1)n n2n+2 .
P
Solution: The series is an alternating series of the form (−1)n bn where bn = n2n+2 .
Clearly bn > 0 for n ≥ 1. We look at the corresponding function f (x) = x2x+2 . We have
2
f ′ (x) = (x2−x
2 +2)2 , which is negative for x ≥ 2. Thus, we also have that bn is decreasing.
Also, limn→∞ bn = 0 so bn satisfies the two properties required and the given series is
convergent by the Alternating Series Test.
13.
P 3n n2
n!
.
Solution: We use the Ratio Test here.
2
an+1 3 n+1
n+1
n
=
= lim 3
=0<1
lim
n→∞
n→∞ an n+1
n2
so by Ratio Test, the series converges.
21.
P (−2)2n
nn
.
Solution: Here an =
(−2)2n
nn
so an+1 =
(−2)2n+2
(n+1)n+1
so
an+1 4
= lim
lim n→∞ (n + 1)
n→∞
an
so by Ratio Test, the series converges.
26.
P n2 +1
5n
n+1 n
n
=0<1
.
Solution: Here again we can use the Ratio Test.
(n+1)2 +1
an+1 1
n2 +1
= lim
= <1
lim n→∞
n→∞
an
5
5
so by Ratio Test, the series converges.