Math 2260 Exam #3 Practice Problem Solutions
Math 2260 Exam #3 Practice Problem Solutions
1. Does the following series converge or diverge? Explain your answer. 2n 3n + n3 .
n=0
Answer: Since 3n + n3 > 3n for all n 1, it follows that
2n
2n
2n
3n + n3 < 3n = 3 .
Therefore,
2n
2n
1
3n + n3 <
3
n=0
n=0
=
1-
2 3
= 3.
Hence, the given series converges.
2. Does the following series converge or diverge? Explain your answer.
n 3n .
n=1
Answer: Use the Ratio Test:
lim
n
n+1 3n+1
n 3n
n + 1 3n
= lim
n
3n+1
?
n
1 n+1 = lim ?
n 3 n
=
1 .
3
Since
1 3
<
1,
the
Ratio
Test
implies
that
this
series
converges.
3. Does the following series converge or diverge? Explain your answer.
1
2n sin
.
n
n=1
Answer: Notice that the terms of this series are not going to zero:
1
1
lim 2n sin
= lim 2x sin
n
n
x
x
=
lim
x
sin
1
1 x
2x
=
lim
x
cos
1 x
?
-2
(2x)2
-2 x2
-2 cos
= lim
x
x2
1 x
4x2 ?
-2
1
= lim 4 cos
x
x
=4
where I went from the second to the third lines using L'H^opital's Rule. Since the limit of the terms is equal to 4, not zero, the series must diverge.
1
4. Does the following series converge or diverge? If it converges, find the sum. If it diverges, explain why. 2n + 3n 4n .
n=1
Answer: Re-writing slightly, the given series is equal to
2n 3n 4n + 4n
n=1
2n 3n
=
4n +
4n .
n=1
n=1
Since both of these series are convergent geometric series, I know the original series converges, so it remains only to determine the sum. Notice that
2n 2 4 8
2
24
2 2 n-1
2/4
1/2
4n = 4 + 16 + 64 + . . . = 4
1+ + +... 4 16
=
44
=
= = 1.
1 - 2/4 1/2
n=1
n=1
Similarly,
3n 3 9 27
3
39
3 3 n-1
3/4
3/4
4n = 4 + 16 + 64 + . . . = 4
1+ + +... 4 16
=
44
=
= = 3.
1 - 3/4 1/4
n=1
n=1
Therefore,
2n + 3n 2n 3n
4n = 4n + 4n = 1 + 3 = 4.
n=1
n=1
n=1
5. Find the interval of convergence of the power series
(2x - 5)n n23n .
n=1
Answer: We use the Ratio Test on the series of absolute values to first determine the radius of convergence:
lim
n
(2x-5)n+1 (n+1)2 3n+1
(2x-5)n n2 3n
|2x - 5|n+1
n3n
|2x - 5| n2
|2x - 5|
=
lim
n
(n + 1)23n+1
?
|2x - 5|n
=
lim
n
3
? (n + 1)2 =
. 3
Therefore,
the
given
series
converges
absolutely
when
|2x-5| 3
<
1,
meaning
when
|2x - 5|
<
3.
Now we check the endpoints. When 2x - 5 = 3, the series becomes
3n
1
n23n =
n2 ,
n=1
n=1
which converges.
Likewise, when 2x - 5 = -3, then series becomes
(-3)n (-1)n3n (-1)n
n23n =
n23n =
n2 ,
n=1
n=1
n=1
which also converges.
Therefore, the series converges for all x so that
-3 2x - 5 3,
which is the interval [1, 4].
2
6. Use the first two non-zero terms of an appropriate Taylor series to approximate
1
sin(x2) dx.
0
Estimate the error of your approximation (i.e. the difference between your answer and the actual value of the integral).
Answer: First, recall that the Taylor series centered at x = 0 for sin(x) is
x3 x5 x7 sin(x) = x - + - + . . . .
3! 5! 7! Therefore, the Taylor series centered at x = 0 for sin(x2) is
sin(x2) = (x2) - (x2)3 + (x2)5 - (x2)7 + . . . = x2 - x6 + x10 - x14 + . . . .
3!
5!
7!
3! 5! 7!
Hence,
1
sin(x2) dx =
1
x2 - x6 + x10 - . . .
dx =
x3 x7 -
+
x11
1 11 -... = - +
1
-. . . .
0
0
3! 5!
3 7 ? 3! 11 ? 5!
0 3 42 11 ? 5!
Therefore, we can approximate this number by
1 1 13 - =,
3 42 42
and we know the error is no bigger than
1
1
1
=
=,
11 ? 5! 11 ? 120 1320
so
in
particular
the
estimate
13 42
is
accurate
to
3
decimal
places.
(It
would
be
totally
fine
in
an
exam
situation
to
leave
your
answer
as
1 11?5!
.)
7. Find the radius of convergence of the Taylor series
(-1)n n2
1
+
n
(x
-
2)n.
n=2
Answer: Use the Ratio Test on the series of absolute values:
lim
n
(-1)n+1 1+(1+n) (n+1)2
(x
-
2)n+1
(-1)n n2
1+n
(x
-
2)n
2
+
n
|x
-
2|n+1
n2
= lim
n
(n + 1)2
? 1 + n |x - 2|n
2 + n n2
=
lim
n
|x
-
2|
?
1
+
n
?
(n
+
1)2
= |x - 2|.
Therefore, the series converges absolutely when |x - 2| < 1, so the radius of convergence is equal to 1.
3
8. Write the second-degree Taylor polynomial for f (x) = x centered at c = 25. Use this polynomial to
approximate 26 and estimate the error of this approximation.
Answer: I can write down the Taylor series centered at x = 25 for the function f (x) = x by first
computing the derivatives of f :
f (x) = 1 x-1/2 = 1
2
2x
f (x) = 1 ? -1 x-3/2 = -1
22
4x3/2
f (x) = -1 ? -3 x-5/2 = 3
42
8x5/2
Therefore,
f (25) = 25 = 5
f (25) = 1
1 =
2 25 10
-1
-1 -1
f
(25) =
4 ? 253/2
=
=
4 ? 125
500
3
3
3
f (25) =
=
=
8 ? 255/2 8 ? 3125 25, 000
Hence, the Taylor series centered at x = 25 for x is
(x - 25) (x - 25)2 3(x - 25)3
x=5+
-
+
-...
10
1000
150, 000
Therefore,
(26 - 25) (26 - 25)2
11
26 5 +
-
=5+ -
= 5.1 - 0.001 = 5.099,
10
1000
10 1000
with an error no more than
3(26 - 25)3
3
1
=
=
= 0.00002.
150, 000 150, 000 50, 000
(In fact, 26 5.0990195.)
9. Does the series
nn
n2
n=1
converge or diverge. Be sure to give a complete explanation.
Answer: Since limn n n = 1 as we discussed in class, a limit comparison to the series
natural:
lim
n
n n
n2 1 n2
nn
n2
= lim
n
n2
?
1
= lim n n = 1.
n
1 n2
is a
Therefore, since the series
1 n2
converges,
the Limit
Comparison Test
implies
that the
given
series
converges as well.
4
10. Does the following series converge or diverge? Explain your answer. n!(n + 1)! . (3n)!
n=1
Answer: Use the Ratio Test:
(n+1)!(n+2)!
lim
n
(3(n+1))! n!(n+1)!
(n + 1)!(n + 2)! (3n)!
(n + 1)(n + 2)
= lim
?
= lim
= 0,
n (3n + 3)! n!(n + 1)! n (3n + 3)(3n + 2)(3n + 1)
(3n)!
since the numerator is a polynomial of degree 2 but the denominator is a polynomial of degree 3. Therefore, since 0 < 1 the Ratio Test implies that the series converges.
11. Does the sequence
n2
arctan n2 + 1 n=1
converge or diverge? If it converges, find the limit; if it diverges, explain why.
Answer: First, notice that
n2
lim
n
n2
+1
=
1.
Therefore, the term inside the arctangent is going to 1, so
n2
lim arctan
n
n2 + 1
= arctan(1) = . 4
12. Does the series
1
n2 - n
n=2
converge or diverge? Explain your answer.
Answer:
For
large
n,
the
n2
should
dominate
the
n,
so
let's
do
a
limit
comparison
to
the
convergent
series
1 n2
.
lim
n
1 n2- n
1 n2
1
n2
=
lim
n
n2
-
n
?
1
n2
1
=
lim
n
n2
-n
=
lim
n
1-
1 n3/2
= 1.
Therefore, since
1 n2
converges,
so
does
n2
1 -n
.
13. Does the series converge or diverge? Explain your answer.
n! n5
n=1
Answer: Use the Ratio Test:
(n+1)!
lim
n
(n+1)5 n!
n5
(n + 1)! n5
=
lim
n
(n
+ 1)5
?
n!
n5
=
lim (n
n
+
1)
(n
+
1)5
=
since
the
expression
n5 (n+1)5
is
going
to
1
but
(n + 1)
is
going
to
.
Therefore, the Ratio Test implies that the series diverges.
5
14. Does the series converge or diverge? Explain your answer.
3n n3
n=1
Answer: Use the Ratio Test:
lim
n
3n+1 (n+1)3
3n n3
3n+1 n3
=
lim
n
(n
+
1)3
?
3n
(n + 1)3
= lim 3 ?
n
n3
= 3.
Since 3 > 1, the Ratio Test implies that the series diverges.
15. Does the series
2n + 3
(n2 + 3n + 6)2
n=0
converge or diverge? Explain your answer.
Answer: For n very large, the denominator will be dominated by the term n4, so do a limit comparison
to the convergent series
n n4
:
lim
n
2n+3 (n2 +3n+6)2
n n4
2n + 3
n4
=
lim
n
(n2
+
3n
+
6)2
?
n
2n + 3
n4
= lim
n
n
? (n2 + 3n + 6)2
= 2 ? 1 = 2.
Therefore, since the limit is finite and the series
n n4
=
1 n3
converges,
the
Limit
Comparison
Test
implies that the given series converges as well.
16. For which values of x does the series
(x - 4)n 5n
n=0
converge? What is the sum of the series when it converges?
Answer: First, use the Ratio Test on the series of absolute values:
lim
n
(x-4)n+1 5n+1
(x-4)n 5n
|x - 4|n+1
5n
|x - 4|
= lim
n
5n+1
? |x - 4|n =
, 5
so
the
given
series
converges
absolutely
whenever
|x-4| 5
<
1,
meaning
when
|x - 4|
<
5
(from
this
we
see that the radius of convergence of the series is 5).
Now check the endpoints. When x - 4 = 5, the series becomes
5n
5n = 1,
n=0
n=0
which diverges. Similarly, when x - 4 = -5, the series becomes
(-5)n 5n
=
(-1)n5n 5n
=
(-1)n,
n=0
n=0
n=0
which also diverges.
6
Therefore, the series converges for
-5 < x - 4 < 5,
which is to say, on the interval (-1, 9).
When the series does converge, it is just the geometric series
x-4 n
1
1
5
5
n=0
=
1
-
x-4 5
=
5-(x-4) 5
=
, 9-x
so
when
it
converges
the
series
converges
to
the
function
f (x)
=
5 9-x
.
17. Does the series
(-1)n(n2 + n3)
n4 + 1
n=1
converge absolutely, converge conditionally, or diverge? Explain your answer.
Answer: The series of absolute values
(-1)n(n2 + n3)
n2 + n3
n4 + 1
=
n4 + 1
n=1
n=1
should behave similarly to
n3 n4
,
so
do
a
limit
comparison
to
this
series:
n2 +n3
lim
n
n4 n3
n2 + n3 n4
= lim
n
n4
? n3
n2 + n3
= lim
n
n3
= 1.
n4
Therefore, since
n3 n4
=
values diverges as well.
1 n
diverges,
the
Limit
Comparison
Test
says
that
the
series
of
absolute
However, the given series satisfies the hypotheses of the Alternating Series Test, so it converges. Therefore, since the series converges but the series of absolute values diverges, we conclude that the series converges conditionally.
18. Does the series converge or diverge? Explain your answer.
(2n)! 2n(n!)2
n=1
Answer: Use the Ratio Test:
lim
n
(2(n+1))! 2n+1 ((n+1)!)2
(2n)!
(2n + 2)!
=
lim
n
2n+1((n
+
1)!)2
2n(n!)1 ?
(2n)!
2n (n!)2
(2n + 2)(2n + 1)
(n!)2
= lim
n
2
? ((n + 1)n!)2
(2n + 2)(2n + 1) 1
= lim
n
2
? (n + 1)2 .
Since (2n + 2)(2n + 1) = 4n2 + 6n + 2 and since (n + 1)2 = n2 + 2n + 1, the above limit is equal to
4n2 + 6n + 2 4
lim
n
2(n2
+ 2n + 1)
=
2
=
2.
Since 2 > 1, the Ratio Test implies that the given series diverges.
7
19. Does the series
(-1)n
n ln(n)
n=2
converge absolutely, converge conditionally, or diverge? Explain your answer.
Answer: The series of absolute values
(-1)n
1
=
n ln(n)
n ln(n)
n=2
n=2
diverges (see HW #12, Problem 3 for a proof of this). However, the series satisfies the hypotheses of the Alternating Series Test and hence converges, so we see that it converges conditionally.
20. For what values of p does the series
1
(n2 + 1)p
n=0
converge? Explain your answer.
Answer: For large n, the +1 will barely contribute, so do a limit comparison with the series
1 (n2 )p
:
lim
n
1 (n2 +1)p
1 (n2 )p
1
(n2)p
=
lim
n
(n2
+
1)p
?
1
= lim
n
n2 n2 + 1
p
= 1p = 1,
since
limn
n2 n2 +1
=
1.
Therefore, the given series and the series
1 (n2 )p
=
1 n2p
will
either
both
converge
or
both
diverge.
Since
1 n2p
converges
for
2p > 1
and
diverges
otherwise,
we
see
that
the
given
series
converges
when
2p > 1, which is to say when
1 p> .
2
21. What is the interval of convergence of the following power series? Explain your answer.
3n(x - 2)n
n2
.
n=1
Answer: Start by applying the Ratio Test to the series of absolute values:
lim
n
3n+1 (x-2)n+1 (n+1)2
3n (x-2)n n2
3n+1|x - 2|n+1
n2
n2
= lim
n
(n + 1)2
?
3n|x - 2|n
=
lim 3|x - 2| ?
n
(n + 1)2
= 3|x - 2|.
Therefore, the ratio test implies that the given series converges absolutely when 3|x - 2| < 1, meaning
when
|x
-
2|
<
1 3
.
Now,
check
the
endpoints.
When
x-2
=
1 3
,
the
series
becomes
3n
1 3
n2
n
=
1 n2 ,
n=1
n=1
which converges.
8
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