SOLUTIONS TO EXAM III, MATH 10560
1. Determine which one of the following series is convergent.
Note: Comparison Tests may help.
P∞ cos2 n
(1)
2n
Pn=1
∞
ln n
(2)
n
Pn=1
∞
21/n
(3)
n
Pn=1
∞
1
(4)
n=1 n(cos2 n+1)
P∞
π n
(5)
n=1
2
Solution:
(1) Notice that 0 ≤ cos2 n ≤ 1 for all n. Therefore
0≤
cos2 n
1
≤ n .
n
2
2
P
P∞
1
Now ∞
n=1 2n is a geometric series with r = 1/2 so it converges. Thus
n=1
converges by the comparison test.
(2) Notice ln n > 1 for n > e. Therefore for n > e
cos2 n
2n
1
ln n
≤
.
n
n
P
P
Since ∞
p-series with p = 1, ∞
n=3 1/n is a P
n=3 1/n diverges and so by the com∞
ln n
parison test so does n=1 n .
(3) Notice 21/n > 1 for all n. Therefore for all n,
0≤
1
21/n
≤
.
n
n
P
P∞
Since ∞
n=1 1/n is a p-series with p = 1,
n=1 1/n diverges and so by the comP∞ 21/n
parison test so does n=1 n .
(4) Notice that cos2 n ≤ 1 for all n. Therefore for all n
0≤
1
1
≤
.
2
2n
n(cos n + 1)
P
P
Since ∞
with p = 1, ∞
n=1 1/(2n) is a p-series
n=1 1/(2n) diverges and so by the
P∞
comparison test so does n=1 n(cos12 n+1 .
P
n
(5) Notice limn→∞ 3(π/2)n = ∞, so by the divergence test ∞
n=1 3(π/2) diverges.
P∞ cos2 n
n=1
2n converges.
0≤
2. Which of the following statement is TRUE?
∞
P
(−1)n nn/2
(1)
converges absolutely.
nn
(2)
n=1
∞
P
n=1
(−1)n nn/2
nn
converges conditionally.
1
2
SOLUTIONS TO EXAM III
(3)
∞
P
(4)
n=1
∞
P
(5)
n=1
∞
P
n=1
(−1)n 2n
3n
converges conditionally.
(−1)n nn/2
nn
(−1)n 2n
3n
diverges.
diverges by divergence test.
Solution:
(1) Notice nn/2 /nn = 1/nn/2 . Using the root test:
!
(−1)n nn/2 1/n
1
L = lim = lim 1/2 = 0 .
n
n→∞ n→∞ n
n
∞
P
Therefore
n=1
(−1)n nn/2
nn
converges absolutely.
(2) See (1).
∞
P
(−1)n 2n
(3)
is a geometric series with r = −2/3 which satisfies |r| < 1 therefore
3n
n=1
∞
P
n=1
(−1)n 2n
3n
converges absolutely.
(4) See (4).
(5) See (3).
(1) is true.
3.
Determine if the two series diverge, converge conditionally, or converge absolutely.
∞
∞ n+1 n
X
X
(−1)n (n + 1)!
2
(I)
(II)
.
n2 · en
2n + 1
n=1
n=1
Solution:
(I) Notice (n + 1)! grows much faster than n2 en so limn→∞
n
limn→∞ (−1)n2(n+1)!
en
(n+1)!
n2 en
does not exist. Therefore by the divergence test,
diverges.
(II) We use the root test
n+1 n 1/n
2
2 · 2n
L = lim
=
lim
=2.
n→∞
n→∞ 2n + 1
2n + 1
n+1 n
P
2
Since L > 1, ∞
diverges.
n=1 2n +1
P
P∞ 2n+1 n
(−1)n (n+1)!
Both ∞
and
diverge.
2
n
n=1
n=1 2n +1
n ·e
= ∞ and so
P∞ (−1)n (n+1)!
n=1
4.
Determine the convergence or divergence of the following series:
∞
∞
∞
X
X
X
(−1)n
2n2
en
(I)
(II)
(III)
4
n
ln(n)
n +1
e +1
n=3
Solution:
n=2
n=1
n2 ·en
SOLUTIONS TO EXAM III
3
n
0
(I) Observe (−1)
ln(n) is alternating for n ≥ 3. Now let f (n) = 1/ ln(n) then f (n) =
1
1
−1/(n ln(n)2 ), meaning for n ≥ 3, ln(n)
is decreasing. Furthermore limn→∞ ln(n)
= 0.
P∞ (−1)n
Thus by the Alternating Series Test, n=3 ln(n) converges.
P
2
(II) Notice 0 < 2n2 /(n4 + 1) < 2n2 /n4 = 2/n2 . Now ∞
is p-series with p = 2,
n2
P∞ 2
P∞ 2nn=1
2
n=1 n2 converges. Then by the comparison test
n=2 n4 +1 converges.
P∞
n
en
(III) Notice limn→∞ ene+1 = 1 6= 0. Therefore by the Divergence Test,
n=1 en +1
diverges.
(I) and (II) converge. (III) diverges.
5.
Find the radius of convergence R for the power series
∞
d X (3n + 2)(x + 1)n
.
dx
5n (n + 1)
n=1
Solution: Let us try the ratio test
(3(n + 1) + 2)(x + 1)n+1 · 5n (n + 1) R = lim n+1
n→∞ 5
((n + 1) + 1) · (3n + 2)(x + 1)n (3n + 5)(x + 1)n+1 · 5n (n + 1) = lim n+1
n→∞ 5
(n + 2) · (3n + 2)(x + 1)n (3n + 5)(n + 1)
= lim
|x + 1|
n→∞ 5(n + 2)(3n + 2)
1
= |x + 1| .
5
We need R < 1 which is equivalent to 15 |x + 1| < 1 or |x + 1| < 5. Therefore, the radius
of convergence is
R = 5.
6.
Determine a power series representation for
Z
e2x dx
Solution: Recall that ex =
∞
X
xn
n=0
Z
e
n!
2x
. Now
dx =
Z X
∞
(2x)n
n=0
n!
dx
∞ Z
X
2n xn
=
dx
n!
=
n=0
∞
X
n=0
Finally,
2n
xn+1
+C
n! · (n + 1)
4
SOLUTIONS TO EXAM III
Z
e2x dx =
∞
X
2n
n=0
xn+1
+ C.
(n + 1)!
7.
Find the following limit using power series
cos(x10 ) − 1
x→0
x20
lim
Solution: We can write cos(x) as a series around x = 0 as follows
∞
X
(−1)n x2n
cos(x) =
=1−
(2n)!
n=0
x2 x4 x6
+
−
+ ....
2!
4!
6!
Now
lim
x→0
cos(x10 )
−1
x20
1−
= lim
x20
2!
+
x40
4!
x→0
x60
6!
+ ... − 1
x20
x→0
= lim
−
−
x20
2!
+
x40
4!
−
x60
6!
+ ...
x20
1
x20 x40
= lim − +
−
+ ...
x→0
2!
4!
6!
1
=− .
2
To sum up,
cos(x10 ) − 1
1
=− .
20
x→0
x
2
lim
8.
Which of the following series sums to 1?
Hint: some well known power series may help.
Solution: Notice that the power series of sin(x) is
∞
X
(−1)n 2n+1
x
(2n + 1)!
n=0
and the radius of convergence is R = ∞. Thus we have
∞
X
(−1)n π 2n+1
π
= sin( ) = 1
2n+1
2
(2n + 1)!
2
n=0
∞
X
(−1)n π 2n+1
22n+1 (2n + 1)!
n=0
9.
What is the fourth Taylor polynomial, T4 (x), for cos(2x) with center a = π?
SOLUTIONS TO EXAM III
5
Solution:
cos(2π) = 1
cos(2x)0 |x=π = −2 sin(2pi) = 0
cos(2x)(2) |x=π = −4 cos(2pi) = −4
cos(2x)(3) |x=π = 8 sin(2pi) = 0
cos(2x)(4) |x=π = 16 cos(2pi) = 16
Hence the Taylor polynomial at x = π is
2
1 − 2(x − π)2 + (x − π)4
3
1 − 2(x − π)2 + 23 (x − π)4
10.
The graph of the parametric curve shown above is the graph of which of the following
parametric equations?
Solution: Since the graph passes the points (−2, 0), (2, 0), (0, −3), (0, 3), only (a) and
(d) are possible. Observe that the curve is clockwise, so (d) is the correct one (because,
for example, the x component of the direction field of this curve at (0,3) should be
positive).
x(t) = 2 sin(t), y(t) = 3 cos(t)
11. For each of the following two series, determine whether the series converges or
diverges. Correct answers with no reasoning indicated will be worth only two points
each. If you wish to use the fact that a function is decreasing make sure you verify that
that is the case.
∞
∞
X
X
(−1)n 21/n
(−1)n ln(n)
(a)
(b)
2n
n2
n=2
n=2
6
SOLUTIONS TO EXAM III
Solution: The series in (a) is an alternating series, and note that
21/n
=0
n→∞ 2n
lim
since the numerator approaches one, while the denominator approaches infinity. Hence
series (a) is convergent by the alternating series test.
However we can say more! By looking at the absolute value of the terms, we obtain the
series
∞
X
21/n
n=2
2n
and notice the numerator is decreasing and approaches 1 from above. Furthermore, it is
always less than 2 (we start with the square root, and take higher order roots). Hence
we have the comparison
X 21/n X 1
≤
2n
2n
with a geometric series, so in fact we have absolute convergence.
Next we consider the series in (b). Again we take the limit of the positive part
ln(n)
=0
n→∞ n2
lim
by a simple application of L’Hôpital’s Rule, or by simply recalling the hierarchy of
asymptotic dominance as n → ∞:
log n < np < en < n! < nn
Hence
series (b) is convergent by the alternating series test.
12. Find the radius of convergence and interval of convergence of the following power
series:
∞
X
(−1)n (x + 3)n
√
4n n
n=1
Solution: Since the summand doesn’t involve raising a variable base to a variable
exponent, it is preferrable to use the ratio test over the root test:
√ √
an+1 (x + 3)n+1
4n n n
lim = lim n+1 √
·
= lim √
|x + 3|
n
an
4
n + 1 (x + 3)
4 n+1
√
n
and since lim √n+1
= 1, we have
an+1 1
= |x + 3|
lim an 4
By the root test, we need this to be less than 1 for convergence, i.e |x + 3| < 4. Hence
the radius of convergence is R = 4.
SOLUTIONS TO EXAM III
7
Our series is centered at a = −3, so the candidate for the interval of convergence is
going to be (−3 − 4, −3 + 4) = (−7, 1). The last step is to test the end points. We have
X (−1)n (−4)n X (−1)n (−1)n 4n X 1
√
√
√ = Div (p series with p = 1/2)
x = −7 =⇒
=
=
4n n
4n n
n
and
X (−1)n (4)n X (−1)n
√
√
x = 1 =⇒
=
= Conv (by alternating series test)
4n n
n
Hence
the interval of convergence is (−7, 1].
13. Consider the function f (x) = ex . a) Find the third Taylor polynomial T3 (x) of f (x)
centered at a = 0.
P
n
Solution: Recall the series for ex centered at a = 0 is simply n≥0 xn! , and to find the
third Taylor polynomial T3 , we take the partial sum up to n = 3 (so the degree of the
variable is 3):
T3 (x) = 1 + x +
x2 x3
+
2
6
b) Find the maximum of the function |f (4) (x)| on the interval [−1, 1]. Recall that
Taylor’s inequality says that if |f (n+1) (x)| ≤ M for all values of x in the interval −1 ≤
x ≤ 1, then
M
|Rn (x)| = |ex − Tn (x)| ≤
|x|n+1
(n + 1)!
for all values of x in the interval −1 ≤ x ≤ 1.
Solution: In general, n-th derivative of ex is ex . We need to maximize f (4) (x) = ex on
[−1, 1]. Since ex is increasing, the maximum is going to be at the right endpoint of the
interval, with the value
(4)
fmax = f (4) (1) = e1 = e
c) Use the above inequality and your results from part (b) to bound the remainder
|R3 (x)|. That is find a number K so that
|R3 (x)| = |ex − T3 (x)| ≤ K
for all values of x in the interval −1 ≤ x ≤ 1.
Solution: By Taylor’s inequality, we have (using M = e which we found in part (b)):
M
e
|R3 (x)| ≤
|x|3+1 = x4
(3 + 1)!
4!
so
e
e
|R3 (x)| ≤ x4 ≤
for x ∈ [−1, 1]
4!
4!
d) Using Taylor’s inequality, find the smallest value of n so that
e
|ex − Tn (x)| ≤
100
8
SOLUTIONS TO EXAM III
for all values of x in the interval −1 ≤ x ≤ 1.
Solution: We are trying to find the smallest n such that
e
|Rn (x)| <
100
for every x ∈ [−1, 1], and by following similar steps to part (c), all we need to do is find
the smallest n satisfying the inequality
e
e
<
(n + 1)!
100
In other words, we want 100 < (n + 1)!. Recall the factorial sequence
0! = 1,
1! = 1,
2! = 2,
3! = 6,
4! = 24,
5! = 120, . . .
so the smallest n which makes (n+1)! larger than 100 is n = 4, so then (4+1)! = 5! = 120
which is larger than 100. Hence
n=4