M1220-2
Quiz 9 - Homework Quiz
Spring 2005
Quiz Scores (out of 10): n = 121; mean = 7.8
25th percentile = 7 ; median (50th percentile) = 8 ; 75th percentile = 9.5
1. (6 pts) State whether the following series are absolutely convergent, conditionally convergent, or divergent. Justify your answer by listing the test used and showing the test
results.
i.
+∞
X
n(−3)n
n=1
4n−1
n+1
Note that limn→+∞ |an+1 /an | = limn→+∞ (n+1)3
4n
by the Ratio Test, this series converges absolutely.
ii.
4n−1
n3n
= limn→+∞
n+1 3
n 4
=
3
4
< 1. So,
+∞
X
n!
(−5)n
n=1
(n+1)! 5n
(n+1)! 5n
5n+1 n! = limn→+∞
n!
5n+1
P+∞ n!
Test, n=1 (−5)n diverges.
Note that limn→+∞ |an+1 /an | = limn→+∞
limn→+∞
n+1
5
= +∞. So, by the Ratio
iii.
+∞
X
n=1
(−1)n
=
n
3 + n2
Note that limn→+∞ |an+1 /an |
= limn→+∞
n+1
3+n2
3+(n+1)2 n
=1
P n
P
so the Ratio Test is inclusive. However, the series,
1/n
3+n2 is similar to the series,
which diverges. Since
h
i
n
limn→+∞ 3+n2 / n1 = 1,
P n
by the Limit Comparison Test, the series
3+n2 also diverges. In that case, the given
series cannot converge absolutely.
However, since the terms of the given series, ignoring their sign, are decreasing and since
P+∞
n
n n
limn→+∞ 3+n
2 = 0, by the Alternating Series Test, the series
n=1 (−1) 3+n2 converges
conditionally.
2. (2 pts) Write out the first 4 terms in the power series given below. Then find the
convergence set for the power series and the radius of convergence.
+∞
X
(−3)n xn
√
n+1
n=o
The first four terms are: 1 −
3x
√
2
+
2
9x
√
3
n+1
−
n+1
27x3
√
4
√
√
n+1
x
√n+1
limn→+∞ |an+1 /an | = limn→+∞ | 3 √n+2
3n xn | = limn→+∞ |3x n+2 | = |3x|. But |3x| <
1when −1/3 < x < 1/3 so the power series converges when −1/3 < x < 1/3. At the
P (−1)n
√
endpoints of the interval: when x = 1/3, the above series becomes
which converges
n+1
P 1
√
by the Alternating Series Test; when x = −1/3, the above series becomes
which
n+1
diverges as it is a p-series with p ≤ 1. So, the convergence set for this series is (−1/3, 1/3]
and the radius of convergence is R = 1/3.
3. (2 pts) Write out the first 4 terms in the power series given below. Then find the
convergence set for the power series and the radius of convergence.
+∞
X
(x − 2)n
nn
n=1
The first four terms are: (x − 2) + (x − 2)2 /22 + (x − 2)3 /33 + (x − 2)4 /44 .
n+1
n
(x−2)
n
x−2
limn→+∞ |an+1 /an | = limn→+∞ | (n+1)
n+1 (x−2)n | = limn→+∞ | n+1
1
n n
n n
n+1 ( n+1 ) . Since limn→+∞ ( n+1 )
1
n n
limn→+∞ ( n+1
( n+1
) = 0. So, the convergence set
|x − 2| limn→+∞
radius of convergence is R = +∞.
h
n
n+1
in
= 1/e and limn→+∞
|=
1
n+1
= 0,
for this series is (−∞, +∞) and the