Math 55 Third Long Exam
Sample Only (Answer Key)
1. Determine if the following sequence/series converges converges or diverges. (a)
en
ln(1 + ) n
∞
n=1
en en = lim = 1. n→∞ n→∞ 1 + en n→∞ en n Therefore Therefore the sequence sequence is convergent.
lim
Solution.
∞
ln(1 + en )
= lim lim
en )
ln(1 + (b) n
n=1
Solution. Since ∞
(c)
n=1
lim
ln(1 + en ) n
n→∞
= 1 = 0, the series diverges by Divergence Test.
( 1)n ln(n + 1)
−
1 . ln(n + 1) 1 i. lim bn = lim lim =0 n→∞ n→∞ ln(n + 1) 1 1 ii. bn+1 = = b n , so bn is decreasing < ln(n + 2) ln(n + 1) Therefore, the series converges by the Alternating Series Test. Let b n =
Solution.
{ }
∞
5n
−2 (d) n
n + 1
n=1
Solution.
lim an
n→∞
|
n
5n
n
5n
−2 Let = − +2 1 | = lim + 1 a n
n
1
n
n→∞
n
.
1 n
= lim lim
n→∞
5
2 n
n + 1
= 25 > 1. 1 .
Therefore the series diverges by Root Test. ∞
(e)
n=1
n5 + n4 n7 + n3 n + 1
−
1 n5 + n4 n5 b and = = . n n7 + n3 n + 1 n7 n2 an n5 + n4 n7 + n6 2 lim = lim 7 = 1 > 0 n = lim 7 n→∞ bn n→∞ n + n 3 n→∞ n + n3 n + 1 n + 1
Solution.
Let a n =
−
·
−
∞
1 is convergent ( -series with Since n=1
n2
Limit Comparison Test.
p
−
p = 2 >
∞
1), 1 ), the series
n=1
n5 + n4 also converges by n7 + n3 n + 1
−
∞
2 by looking at its partial sum. 2. Evaluate the sum − 1 2 . By partial fraction decomposition, Let = − 1 1 1 = − 1 − + 1 1 1 1 1 1 1 1 n=2
n th
n2
n
Solution.
s n
i=2
i2
n
sn
i
i=2
=
1
∞
n=2
1 2
2 n2
+
−3
= 1+
Hence,
i
− 1
+
− 4
2
+
− 5
3
− 6
4
+ . . . +
1
n
− n1 − n −1 1
= lim sn = lim n→∞
n→∞
1 1+ 2
1
1
− n − n − 1
1
− 2 − n
+
1
n
− 1 −
3 = . 2 ∞
( 3. Detrmine the radius of convergence and interval of convergence of the power series n=0
Solution.
lim
Let a n =
− 2)
n
3n (n + 1)
.
−
−
+ 1 + 2 n n
· − |x − 2| < 1, i.e., |x − 2| < 3. The series converges when n→∞
x
(x 2)n . By Ratio Test, 3n (n + 1)
(x 2)n+1 3 n (n + 1) an+1 = lim n+1 = lim n→∞ 3 n→∞ (n + 2) (x 2)n an
1 n + 1
x
− 2 = |x − 2| 3
3
3 Hence, R = 3 and the series converges when ∞
• If x
n
= −1 the series becomes ,
n=0
Let bn =
−3 < x − 2 < 3, i.e., on the interval ( −1, 5). (−1) 3 = (−1) . (−3) =
3n (n + 1)
1 . n + 1
∞
n=0
n
n
3n (n + 1)
∞
n=0
n
n + 1
1 =0 n→∞ n→∞ n + 1 1 1 ii. bn+1 = < = b n so bn is decreasing n + 2 n + 1 Hence the series converges by Alternating Series Test. i. lim bn = lim
{ }
∞
• If x
∞
3n 1 = . n n + 1) n + 1 3 ( n=0 n=0
= 5, the series becomes
1 1 and b n = . n + 1 n an n Then lim = lim = 1 > 0 . n→∞ bn n→∞ n + 1 ∞ ∞ 1 1 Since diverges, also diverges by Limit Comparison Test. n n + 1 n=1 n=1 Let an =
Hence, the interval of convergence is [ 1, 5).
−
√
4. Find a power series representation for ln 1 + 2 x using the power series reprentation for
√
1 1 dx = ln(1 + 2x) + c = ln 1 + 2 x + c. 1 + 2x 2 1 Now, we obtain a power series for : 1 + 2x Solution. First,
notice that
ˆ
1 . (1 + 2x)
∞
∞
1 1 = = ( 2x)n = ( 2)n xn 1 + 2x 1 ( 2x) n n
−
−−
−
=0
=0
Therefore,
ˆ √ ln 1 + 2x =
1 dx = 1 + 2x
∞
∞
n
xn+1
ˆ (−2) = (−2) + 1 √ (−2) = ln1 = 0. Hence, ln 1 + 2 = . n
n
x dx
n=0
n=0
∞
Letting x = 0, we get C
n
x
n=0
n
xn+1
n + 1
∞
5. Find the Maclaurin series for f (x) = e
2x
∞
Since e
=
n=0
n
(2ln2) and use this to find the sum . n=1
x
+ C
n!
xn , we have n! ∞
∞
x n n!
n
(2 ) = 2 =
2x
e
n=0
n=0
n!
xn
Letting x = ln 2, we obtain ∞
2ln2
e
n
2 (ln2) ! (2ln2)
n
=
n=0
n
∞
ln 22
e
=
n=0 ∞
n
(2ln2) Hence,
n!
2
= e ln 2 = 4.
n!
n=0
n
6. Find the third degree Taylor polynomial for f (x) = ln(1 + x) centered at 0 and use it to approximate ln(1.5).
1
−
f (x) = (1 + x) f (x) = (1 + x)−2 f (x) = 2(1 + x)−3
−
3
f (i) i
(0) ( )= 1 ! 1
Therfore, T 3 x
i=0
Hence, ln(1.5)
≈ T 3
2
=
xi = 0 +
2
⇒ ⇒ ⇒
f (0) = ln1 = 0 f (0) = 1 f (0) = 1 f (0) = 2
3
2x x − + . 1! 2! 3! x
1 2 1 1 1 10 + = + = . − − 2 22 · 2 23 · 3! 2 8 24 24
−
