Answer Key for Quiz 4 (section B)
1)
∞
X
n=1
√n4−1
n4+ 1 <
∞
X
n=1
√n4
n4=
∞
X
n=1
n2
n4=
∞
X
n=1
1
n2
In the first step, we made the numerator greater and the denominator smaller, thus making the
fraction greater.
This last series we know to converge; it’s a p-series with p= 2. Since our original series is less than
this convergent series, our original series converges also.
2)
∞
X
n=1
99
100 =99
100 +99
100 +99
100 +... =∞
The terms in the series do not approach 0 as n→ ∞, so the series diverges.
3)
∞
X
n=1
100n(n+ 13)
n!The factorial and exponential are clues to use the Ratio Test.
L= lim
n→∞
¯
¯
¯
¯
¯
¯
100n+1((n+1)+13)
(n+1)!
100n(n+13)
n!
¯
¯
¯
¯
¯
¯
= lim
n→∞ ¯
¯
¯
¯
100n+1
100n·
n+ 14
n+ 13 ·
n!
(n+ 1)!¯
¯
¯
¯
= lim
n→∞ ¯
¯
¯
¯
100
1·
n+ 14
n+ 13 ·1
n+ 1¯
¯
¯
¯
= lim
n→∞ ¯
¯
¯
¯
100n+ 1400
n2+ 14n+ 13¯
¯
¯
¯
= 0
The value of this limit (0) can be obtained by using L’Hopital’s Rule or by dividing each term in
the fraction by nand then letting n→ ∞ or by simply looking at the ratio of the dominant terms
(100nover n2) as n→ ∞.
Since the ratio is less than 1, we know that the series converges.
4)
∞
X
n=1
n2
en3
Use the Integral Test with the integral Z∞
1
x2e−x3dx. Let w=−x3. Then dw =−3x2dx.
Z∞
1
x2e−x3dx =Zx=∞
x=1
ew−dw
3=−1
3e−x3¯
¯
¯
∞
1=−1
3¡e−∞ −e−1¢=−1
3(0 −1
e) = 1
3e
Since the improper integral converges, we know that our series must converge also (although not
to the same value as the integral).
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