Answer Key for Quiz 4 (section A)
1)
∞
X
n=1
n3+ 2n2
7n4>
∞
X
n=1
n3
7n4=
∞
X
n=1
1
7n=1
7
∞
X
n=1
1
n
This last series we know to diverge; it’s the harmonic series (a p-series with p= 1). Since our original
series is greater than this divergent series, our original series diverges also.
2)
∞
X
n=1
1
2=1
2+1
2+1
2+... =∞
The terms in the series do not approach 0 as n→ ∞, so the series diverges.
3)
∞
X
n=1
(2n)!
(n!)25nThe factorial and exponential are clues to use the Ratio Test.
L= lim
n→∞
¯
¯
¯
¯
¯
¯
(2(n+1))!
((n+1)!)25n+1
(2n)!
(n!)25n
¯
¯
¯
¯
¯
¯
= lim
n→∞
¯
¯
¯
¯
(2n+ 2)!
(2n)! ·5n
5n+1 ·(n)!(n)!
(n+ 1)!)(n+ 1)!) ¯
¯
¯
¯
= lim
n→∞
¯
¯
¯
¯
(2n+ 2)(2n+ 1)
1·1
5·1
(n+ 1)(n+ 1)¯
¯
¯
¯
= lim
n→∞
¯
¯
¯
¯
4n2+ 6n+ 2
5n2+ 10n+ 5 ¯
¯
¯
¯
=4
5
The value of this limit (4/5) can be obtained by using L’Hopital’s Rule or by dividing each term in
the fraction by n2and then letting n→ ∞ or by simply looking at the ratio of the dominant terms
(4n2over 5n2).
Since the ratio is less than 1, we know that the series converges.
4)
∞
X
n=1
(−1)n
n+ 2003 =−1
2004 +1
2005 −1
2006 +...
The terms alternate, decrease in magnitude, and approach zero, so the series passes the Alternating
Series Test and therefore converges.
1