**Unformatted text preview: **harris (tlh2479) – PostClass 11.6 – morales – (56815)
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001 10.0 points Determine whether the following series
∞ n=1 5n
(n + 2) 42n+1 is absolutely convergent, conditionally convergent, or divergent. 1. conditionally convergent correct
2. divergent
3. absolutely convergent
Explanation:
Since
∞ (−1)
n=1 n−1 1
e1/n
=−
4n
4 ∞ (−1)n
n=1 2. conditionally convergent
3. absolutely convergent correct
Explanation:
With
an = ∞ (−1)n
n=1 e1/n
,
n we have to decide if the series 1. divergent e1/n
n is absolutely convergent, conditionally convergent, or divergent.
First we check for absolute convergence.
Now, since e1/n ≥ 1 for all n ≥ 1,
1
e1/n
≥
> 0.
4n
4n 5n
(n + 2) 42n+1 But by the p-series test with p = 1, the series we see that
an+1
an 1 = 5n+1 (n + 2) 42n+1
((n + 1) + 2) 42(n+1)+15n = n+2
n+3 5
16 lim an+1
an = Thus
n→∞ 5
< 1,
16 so by the Ratio Test the given series is ∞ n=1 1
4n diverges, and so by the Comparison Test, the
series
∞
e1/n
4n
n=1 too diverges; in other words, the given series
is not absolutely convergent.
To check for conditional convergence, consider the series absolutely convergent . ∞ (−1)n f (n)
002 Determine whether the series
∞ (−1)n−1
n=1 n=1 10.0 points
e1/n
4n is absolutely convergent, conditionally convergent or divergent. where e1/x
.
4x
Then f (x) > 0 on (0, ∞). On the other hand,
f ( x) = 1+x
1 1/x e1/x
.
f (x) = − 3 e − 2 = −e1/x
4x
4x
4 x3
′ harris (tlh2479) – PostClass 11.6 – morales – (56815)
Thus f ′ (x) < 0 on (0, ∞), so f (n) > f (n + 1)
for all n. Finally, since 2 Consequently, by the Ratio test, the given
series is lim e1/x = 1 , absolutely convergent . x→∞ we see that f (n) → 0 as n → ∞. By the
Alternating Series Test, therefore, the series 004 10.0 points ∞ To apply the ratio test to the inﬁnite series
an , the value (−1)n f (n)
n=1
n is convergent.
Consequently, the given series is λ = lim n→∞ conditionally convergent . has to be determined.
Compute λ for the series
∞ keywords:
003 an+1
an n=1 10.0 points 6n
.
4n6 + 8 1. λ = 0
Determine whether the following series
∞ n=1 2. λ = (−7)n
n! is absolutely convergent, conditionally convergent, or divergent. 3
2 3. λ = 3
4 4. λ = 6 correct 1. conditionally convergent
5. λ = 2. divergent
3. absolutely convergent correct Explanation:
By algebra, Explanation:
The given series has the form
∞ an = an ,
n=1 (−7)n
.
n! But then
an+1
an = 7(n!)
7
=
,
(n + 1)!
n+1 lim an+1
6n+1
=
an
6n 4n6 + 8
4(n + 1)6 + 8 4n6 + 8
=
4(n + 1)6 + 8 4+ an+1
an = 0 < 1. . But 4 n+1
n 8
n6
6 8
+6
n Since in which case
n→∞ 1
2 n+1
n 6 = 1
1+
n 6 →1 . harris (tlh2479) – PostClass 11.6 – morales – (56815)
as n → ∞, we see that
lim n→∞ converges if the series an+1
an ∞ n→∞ k=1 8
n6 6 4+
= lim 3 6 n+1
4
n 8
+6
n = 6. Consequently, 1
4k converges. But this last series is a geometric
series with
1
< 1,
r=
4
hence convergent. Consequently, the given
series is λ=6. absolutely convergent .
005 10.0 points Determine whether the series
∞ k=1 006 sin2 (k )
(−1)k−1
4k 10.0 points Determine whether the following series is absolutely convergent, conditionally convergent or divergent ∞ n=1 1. divergent is absolutely convergent, conditionally convergent, or divergent. 2. conditionally convergent
3. absolutely convergent correct
Explanation:
To check for absolute convergence we have
to decide if the series
∞ k=1 sin2 (k )
4k is convergent. For this we can use the Comparison Test with
ak sin2 (k )
=
,
4k bk = 1
.
4k 1. absolutely convergent
2. divergent correct
3. conditionally convergent
Explanation:
Since the nth term of the series has the
form
(−1)n n n
,
an =
6
36
we use the Root Test, for then
|an |1/n = For then
0 ≤ a k ≤ bk , k=1 sin2 (k )
4k
1
61/n n
→∞
36 as n → ∞. The Root Test thus ensures that
the series is since 0 ≤ sin2 (k ) ≤ 1. Thus the series
∞ (−n)n
62 n+1 divergent . 007 10.0 points harris (tlh2479) – PostClass 11.6 – morales – (56815)
Consequently, for the given series, To apply the ratio test to the inﬁnite series
an , the value λ=1. n λ = lim n→∞ an+1
an 008 has to be determined, 1
sin
5n + 7 n=1 10.0 points Decide whether the series Compute λ for the series
∞ 4 ∞ 1
n n=1 . 1 n+1
4n
n n2 converges or diverges. 1. λ = 1
7 1. diverges 1
12 2. converges correct 2. λ = Explanation:
The given series has the form 3. λ = 1 correct ∞ an , 1
4. λ =
5 |an |1/n = Explanation:
By algebra, sin 1
n+1
1
n . 1
n+1
1
n+1 · 1
n sin 1
n · n
.
n+1 so
sin lim n→∞ 1
n+1
1
n+1 while
lim n→∞ 1
n sin 1
n lim |an |1/n = n→∞ n→∞ , n
n+1 n+1
n n = lim n→∞ 1+ 1
n n = e < 3. = 1,
converges .
=1 009 5n + 7
5n + 5 + 7 10.0 points To apply the root test to an inﬁnite series
an the value of 5n + 7
5(n + 1) + 7 n
= lim
n→∞ n + 1 e
< 1,
4 Consequently, the Root Test ensures that the
given series also. Thus
n→∞ n since
lim sin(x)
lim
= 1,
x→0
x λ = lim 1 n+1
4
n in which case
sin = But n2 But then 5. λ = 0 sin an n=1 1 n+1
=n
4
n n ρ = lim |an |1/n
n→∞ . has to be determined. harris (tlh2479) – PostClass 11.6 – morales – (56815)
Compute the value of ρ for the series
∞ n=1 6n + 5
n n 7
2 Explanation:
The given series has the form . ∞ n=1 |an |1/n = 4 3. ρ = 21 lim |an |1/n = n→1 2
7 5. ρ = . n−1
n n , in which case 10
7 4. ρ = n2 But then 7
correct
2 2. ρ = n−2
n a n = 4n an , 35
2 1. ρ = 5 4
< 1,
e2 since
lim Explanation:
After division, n→∞ 6n + 5
5
= 6 1+
n
6n n−2
n = lim n→∞ , n 1− 2
n n = 1
1
< 2.
2
e
2 Consequently, the Root Test ensures that the
given series so
1/n (an ) = 1/n 5
6 1+
6n 7
.
2 converges . But
lim 6 5
1+
6n 1/n n→∞ 1/n =1 011 Determine whether the series as n → ∞. Consequently,
ρ=
010 ∞ 7
.
2 n=0 10.0 points 4 n n=1 n−2
n converges or diverges. is absolutely convergent, conditionally convergent, or divergent. n2 2. conditionally convergent
3. divergent
Explanation:
We use the Ratio Test with 1. diverges
2. converges correct (−2)n
(2n)! 1. absolutely convergent correct Decide whether the series
∞ 10.0 points an = (−2)n
.
(2n)! harris (tlh2479) – PostClass 11.6 – morales – (56815)
For then
an+1
an for all x > 1. Thus f (x) is positive and
decreasing for all x > 1. But the improper
integral (−2)n+1 (2n)!
(2n + 2)! (−2)n = ∞ (−2)n+1 −2
(2n)!
.
=
=
n
(2n + 2)! (−2)
(2n + 1)(2n + 2)
Thus
lim n→∞ an+1
an
2
= 0 < 1.
(2n + 1)(2n + 2) = lim n→∞ 6 ∞ f (x) dx =
2 2 1
dx
x ln(x) does not converge. By the Integral Test,
therefore, the given series is not absolutely
convergent. It could still converge conditionally, however. To use the Alternating Series
Test to show that the series
∞ (−1)n f (n)
n=2 Consequently, the
series is absolutely convergent . is convergent, we have to check that
(i) f (n) > f (n + 1) for n ≥ 2, 012 10.0 points Determine whether the series
∞ (−1)n
n=2 6
n ln(n) converges conditionally, converges absolutely,
or diverges. 3. series converges absolutely
Explanation:
The given series can be rewritten as
∞ n n=2 (−1)n f (n) ,
n=2 where
f ( x) = 1
= (x ln(x))−1 > 0
x ln(x) for all x > 1. On the other hand, by the Chain
and Product Rules,
f ′ ( x) = − Since f is decreasing for all x ≥ 2, however,
we see that
n≥2 =⇒ 1
(ln(x) + 1) < 0
(x ln(x))2 f (n) > f (n + 1) . On the other hand,
x→∞ 2. series converges conditionally correct 6
(−1)
=6
n ln(n) x→∞ lim 1. series diverges ∞ (ii) lim f (x) = 0 . 1
= 0.
x ln(x) Consequently, by the Alternating series Test,
the given series is
conditionally convergent . ...

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