Math 172.503
Exam 3 Solutions
April 27, 2010
1. D : A simple matter of checking the various formulas reveals that the correct one is
2. B : Because 0 ≤ cos2 (n) ≤ 1 for all n, we have 0 ≤
2 n
= 0, and so
limn→∞ cos3(n)
lim
n→∞
cos2 (n)
3
n
+
cos2 (n)
3
1
3
=
≤
1
3
(−1)n+1
.
n2 + 1
for all n. Therefore, we have
1
.
3
3. C : We break up the series,
n X
n
∞
∞
∞
∞
∞
X
X
X
1
1
4
1
22n − 1 X 22n
1
=
−
=
·
−
·
.
n+1
n+1
n+1
5
5
5
5
5
5
5
n=0
n=0
n=0
n=0
n=0
The two sums on the right are both convergent geometric series, the first with common ratio r = 45
and the second with common ratio r = 51 . Therefore, using the formula for the sum of a geometric
series, the total sum is
∞
1
1
X
22n − 1
3
5
5
=
−
= .
4
1
n+1
5
4
1− 5
1− 5
n=0
4. B : We want to find N so that for all n > N , we have |an − L| <
1
100 .
First we see that
4−n
= −1.
n→∞
n
L = lim
1
Therefore substituting into the inequality |an − L| < 100
,
4 − n
1
n + 1 < 100 ⇐⇒
we want n to satisfy
4
< 1 .
n 100
After cross-multiplying, we see that we need n > 400. Therefore, N = 400 and N = 1000 are both
suitable, but N = 100 is not.
5. A : The Alternating Series Estimation Theorem states that the error in summing up the first 10 terms
1
is bounded above by the absolute value of the 11-th term, which is 121
.
∞ X
1
1
6. The series
−
is a telescoping series. For part (a), we have the finite sum
n n+2
n=1
1
1 1
1 1
1 1
1
1
1
1
sk = 1 −
+
−
+
−
+
−
+ ··· +
−
+
−
.
3
2 4
3 5
4 6
k−1 k+1
k k+2
After canceling the telescoping terms, we are left with
sk = 1 +
1
1
1
−
−
.
2 k+1 k+2
For part (b), we can find the sum of the series by taking the limit,
∞ X
1
1
1
1
3
1
1
−
= lim sk = lim 1 + −
−
=1+ −0−0=
.
k→∞
k→∞
n n+2
2 k+1 k+2
2
2
n=1
3n
= ∞ : This limit can be found in several different ways. One is several quick applications
n→∞ n3
of l’Hôpital’s rule (since the form of this limit if ∞
∞ ):
7. (a) lim
(ln 3)3n
(ln 3)2 3n
3n
(ln 3)3 3n
=
lim
=
lim
=
lim
= ∞.
n→∞ n3
n→∞
n→∞
n→∞
3n2
6n
6
lim
Another way is simply to observe that the numerator (3n ) grows exponentially, while the denominator
(n3 ) is a polynomial. As n → ∞, the exponential swamps the polynomial, taking the value to ∞.
(b)
lim (ln(3n) − ln(n − 2)) = ln 3 : The familiar properties of logarithms yield,
n→∞
lim (ln(3n) − ln(n − 2)) = lim ln
n→∞
8. Recalling that
n→∞
3n
n−2
= ln(3).
∞
∞
X
X
1
1
=
=
x2n . Therefore, if we multiply by x, we
xn , it follows that
2
1−x
1
−
x
n=0
n=0
find
∞
∞
X
X
x
2n
=
x
·
x
=
x2n+1 .
1 − x2
n=0
n=0
∞
X
tan−1 (n)
converges . The justification is that since | tan−1 (n)| ≤
n
3
n=1
of n, we have the inequality of series,
9. (a) This series
π
2
for all values
∞
∞
X
tan−1 (n) X π/2
≤
.
3n
3n
n=1
n=1
The series on the right is a geometric series with r = 31 and therefore converges. So by the direct
comparison test , our series converges. It is also possible to use the limit comparison test or the ratio
test to show convergence here.
∞
X
1
cos
(b) The series
diverges . We calculate the limit of terms,
n
n=1
lim cos
n→∞
1
= cos(0) = 1 6= 0,
n
and so by the test for divergence , or n-th term test, the series diverges. It is also possible to use a
comparison test to show divergence here.
∞ √ 4
∞
X
X
n −1
1
(c) The series
converges
.
Apply
the
limit
comparison
test
with
the
series
:
5
n
n3
n=2
n=2
√
c = lim
n→∞
n4 −1
n5
1
n3
√
= lim
n→∞
n4 − 1 n3
·
= lim
n→∞
n5
1
√
n4 − 1
= lim
n→∞
n2
r
1−
1
= 1.
n4
P∞
Since 0 < 1 < ∞, the limit comparison test says that our series converges if and only if n=2 n13
converges. That series converges by the p-series test with p = 3 > 1. It is also possible to apply a
direct comparison test to show convergence.
10. (a) We use the ratio test,
ρ=
|x−2|n+1
(n+1)2n+1
lim |x−2|n
n→∞
n2n
n2n
n
|x − 2|n+1
|x − 2|
|x − 2|
·
= lim
·
=
.
n+1
n→∞ n + 1
n→∞ (n + 1)2
|x − 2|n
2
2
= lim
The open interval of convergence can be found by solving for ρ < 1, so we solve the inequality,
|x − 2|
<1
2
⇐⇒
|x − 2| < 2
⇐⇒
−2 < x − 2 < 2
⇐⇒
0<x<4.
(b) We check the endpoints. At x = 0, the series is
∞
∞
X
X
(−1)n
(−2)n
=
f (0) =
.
n · 2n
n
n=1
n=1
This sum is the alternating harmonic series, which we know converges . Or this can be shown easily
by the alternating series test. At x = 4, the series is
f (4) =
∞
∞
X
X
1
2n
=
.
n
n·2
n
n=1
n=1
This series is the harmonic series, which diverges .
(c) We differentiate term-by-term,
"∞
#
∞
∞
X
X
d X (x − 2)n
n(x − 2)n−1
(x − 2)n−1
f (x) =
=
=
.
n
n
dx n=1 n · 2
n·2
2n
n=1
n=1
0
(d) To calculate f 0 (3), we substitute into the series from (c):
f 0 (3) =
This sum is a geometric series with r =
1
2
∞
∞
X
X
1n−1
1
=
.
n
n
2
2
n=1
n=1
and first term 21 . Therefore,
f 0 (3) =
1
2
1−
1
2
= 1.