Math 2260 Exam #3 Practice Problem Solutions
1. Does the following series converge or diverge? Explain your answer.
∞
X
2n
.
3n + n3
n=0
Answer: Since 3n + n3 > 3n for all n ≥ 1, it follows that
n
2n
2n
2
<
=
.
3n + n3
3n
3
Therefore,
∞ n
X
1
2
2n
<
=
n
3
3 +n
3
1−
n=0
n=0
∞
X
2
3
= 3.
Hence, the given series converges.
2. Does the following series converge or diverge? Explain your answer.
∞
X
n
.
n
3
n=1
Answer: Use the Ratio Test:
n+1
3n+1
n→∞ nn
3
lim
Since
1
3
n + 1 3n
1 n+1
1
·
= lim ·
= .
n→∞ 3n+1
n→∞
n
3
n
3
= lim
< 1, the Ratio Test implies that this series converges.
3. Does the following series converge or diverge? Explain your answer.
∞
X
2n sin
n=1
1
.
n
Answer: Notice that the terms of this series are not going to zero:
1
1
lim 2n sin
= lim 2x sin
n→∞
x→∞
n
x
sin x1
= lim
1
x→∞
=
=
2x
1
−1
x · x2
lim
−2
x→∞
(2x)2
− cos x1
4x2
lim
·
x→∞
x2
−2
cos
= lim 2 cos
x→∞
1
x
=2
where I went from the second to the third lines using L’Hôpital’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.
∞
X
2n + 3 n
.
4n
n=1
Answer: Re-writing slightly, the given series is equal to
X
∞ n
∞
∞
X
2
3n
2n X 3n
+
=
+
.
4n
4n
4n n=1 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
X
n−1
∞
∞
X
2
2/4
4
8
2
2
4
1/2
2n
2 2
=
=
+
+
+
.
.
.
=
1
+
+
+
.
.
.
=
=
= 1.
n
4
4
16
64
4
4
16
4
4
1
−
2/4
1/2
n=1
n=1
Similarly,
X
n−1
∞
∞
X
3
9
27
3
3
9
3/4
3/4
3n
3 3
=
= +
+
+ ... =
1+ +
+ ... =
=
= 3.
n
4
4
16
64
4
4
16
4
4
1
−
3/4
1/4
n=1
n=1
Therefore,
∞
∞
∞
X
X
2n + 3n
2n X 3n
=
+
= 1 + 3 = 4.
4n
4n n=1 4n
n=1
n=1
5. Find the interval of convergence of the power series
∞
X
(2x − 5)n
.
n2 3 n
n=1
Answer: We use the Ratio Test on the series of absolute values to first determine the radius of
convergence:
(2x−5)n+1 (n+1)2 3n+1 n3n
|2x − 5|
n2
|2x − 5|
|2x − 5|n+1
= lim
lim ·
=
lim
·
=
.
n
2
n+1
n
2
(2x−5) n→∞
n→∞
n→∞ (n + 1) 3
|2x
−
5|
3
(n
+
1)
3
n2 3n 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
∞
∞
X
X
3n
1
=
,
2 3n
2
n
n
n=1
n=1
which converges.
Likewise, when 2x − 5 = −3, then series becomes
∞
∞
∞
X
X
X
(−3)n
(−1)n 3n
(−1)n
=
=
,
n2 3 n
n2 3 n
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
Z 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
sin(x) = x −
x5
x7
x3
+
−
+ ....
3!
5!
7!
Therefore, the Taylor series centered at x = 0 for sin(x2 ) is
sin(x2 ) = (x2 ) −
(x2 )3
(x2 )5
(x2 )7
x10
x14
x6
+
−
+ . . . = x2 −
+
−
+ ....
3!
5!
7!
3!
5!
7!
Hence,
Z
0
1
sin(x2 ) dx =
Z
0
1
3
1
x10
x
x7
x11
1 1
1
x6
+
− . . . dx =
−
+
− ... = − +
−. . . .
x2 −
3!
5!
3
7 · 3! 11 · 5!
3 42 11 · 5!
0
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
1
situation to leave your answer as 11·5!
.)
7. Find the radius of convergence of the Taylor series
√
∞
X
(−1)n 1 + n
(x − 2)n .
2
n
n=2
Answer: Use the Ratio Test on the series of absolute values:
(−1)n+1 √1+(1+n)
n+1 √
(x
−
2)
(n+1)2
2 + n |x − 2|n+1
n2
√
lim
=
lim
·
√
n
(−1) 1+n
n→∞
n→∞
(n + 1)2
1 + n |x − 2|n
(x − 2)n n2
√
2+n
n2
√
= lim |x − 2| ·
·
n→∞
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 :
1 −1/2
1
x
= √
2
2 x
1 −1 −3/2
−1
f 00 (x) = ·
x
= 3/2
2 2
4x
−1 −3 −5/2
3
000
f (x) =
·
x
= 5/2
4
2
8x
f 0 (x) =
Therefore,
f (25) =
√
25 = 5
1
1
f 0 (25) = √ =
10
2 25
−1
−1
−1
=
=
f 00 (25) =
4 · 125
500
4 · 253/2
3
3
3
f 000 (25) =
=
=
8 · 3125
25, 000
8 · 255/2
√
Hence, the Taylor series centered at x = 25 for x is
√
x=5+
(x − 25) (x − 25)2
3(x − 25)3
−
+
− ...
10
1000
150, 000
Therefore,
√
26 ≈ 5 +
(26 − 25) (26 − 25)2
1
1
−
=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
∞ √
n
X
n
2
n
n=1
converge or diverge. Be sure to give a complete explanation.
P 1
√
Answer: Since limn→∞ n n = 1 as we discussed in class, a limit comparison to the series
n2 is a
natural:
√
n n
√
n
√
n n2
n2
lim 1 = lim
·
= lim n n = 1.
2
n→∞
n→∞
n→∞
n
1
n2
P 1
Therefore, since the series
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.
∞
X
n!(n + 1)!
.
(3n)!
n=1
Answer: Use the Ratio Test:
(n+1)!(n+2)!
(3(n+1))!
lim
n!(n+1)!
n→∞
(3n)!
= lim
n→∞
(n + 1)!(n + 2)!
(3n)!
(n + 1)(n + 2)
·
= lim
= 0,
n→∞
(3n + 3)!
n!(n + 1)!
(3n + 3)(3n + 2)(3n + 1)
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
= 1.
n→∞ n2 + 1
Therefore, the term inside the arctangent is going to 1, so
n2
π
lim arctan
= arctan(1) = .
n→∞
n2 + 1
4
lim
12. Does the series
∞
X
1
√
2−
n
n
n=2
converge or diverge? Explain your answer.
2
Answer:
P 1For large n, the n should dominate the
series
n2 .
1√
n2 − n
1
n→∞
n2
lim
Therefore, since
P
1
n2
√
n, so let’s do a limit comparison to the convergent
1
n2
n2
1
√
√ = lim
·
=
lim
= 1.
1
2
n→∞ n2 −
n→∞
n→∞
n 1
n − n
1 − n3/2
= lim
converges, so does
P
13. Does the series
1√
.
n2 − n
∞
X
n!
n5
n=1
converge or diverge? Explain your answer.
Answer: Use the Ratio Test:
(n+1)!
(n+1)5
lim
n!
n→∞
n5
since the expression
n5
(n+1)5
n5
(n + 1)! n5
·
=
lim
(n
+
1)
=∞
n→∞ (n + 1)5
n→∞
n!
(n + 1)5
= lim
is going to 1 but (n + 1) is going to ∞.
Therefore, the Ratio Test implies that the series diverges.
5
14. Does the series
∞
X
3n
n3
n=1
converge or diverge? Explain your answer.
Answer: Use the Ratio Test:
3n+1
(n+1)3
lim
3n
n→∞
n3
3n+1
n3
(n + 1)3
·
=
lim
3
·
= 3.
n→∞ (n + 1)3 3n
n→∞
n3
= lim
Since 3 > 1, the Ratio Test implies that the series diverges.
15. Does the series
∞
X
n=0
(n2
2n + 3
+ 3n + 6)2
converge or diverge? Explain your answer.
Answer: For n very large,
denominator will be dominated by the term n4 , so do a limit comparison
P the
n
to the convergent series
n4 :
2n+3
(n2 +3n+6)2
lim
n
n→∞
n4
2n + 3
n4
2n + 3
n4
·
=
lim
·
= 2 · 1 = 2.
n→∞ (n2 + 3n + 6)2
n→∞
n
n
(n2 + 3n + 6)2
= lim
Therefore, since the limit is finite and the series
implies that the given series converges as well.
16. For which values of x does the series
P
n
n4
=
1
n3
converges, the Limit Comparison Test
∞
X
(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:
(x−4)n+1 5n+1 |x − 4|n+1
5n
|x − 4|
=
lim
lim ·
=
,
n+1
n
(x−4)n n→∞
n→∞
5
|x
−
4|
5
5n so the given series converges absolutely whenever |x−4|
< 1, meaning when |x − 4| < 5 (from this we
5
see that the radius of convergence of the series is 5).
Now check the endpoints. When x − 4 = 5, the series becomes
∞
∞
X
X
5n
=
1,
5n
n=0
n=0
which diverges.
Similarly, when x − 4 = −5, the series becomes
∞
∞
∞
X
X
X
(−5)n
(−1)n 5n
=
=
(−1)n ,
n
n
5
5
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 x − 4 n
1
1
5
=
= 5−(x−4) =
,
x−4
5
9−x
1− 5
n=0
5
so when it converges the series converges to the function f (x) =
17. Does the series
5
9−x .
∞
X
(−1)n (n2 + n3 )
n4 + 1
n=1
converge absolutely, converge conditionally, or diverge? Explain your answer.
Answer: The series of absolute values
∞ ∞
X
(−1)n (n2 + n3 ) X
n2 + n3
=
n4 + 1
n4 + 1
n=1
n=1
should behave similarly to
P n3
n4 ,
so do a limit comparison to this series:
n2 +n3
n4
n3
n→∞
n4
lim
n2 + n3 n4
n2 + n3
· 3 = lim
= 1.
4
n→∞
n→∞
n
n
n3
= lim
P n3
P1
Therefore, since
n4 =
n diverges, the Limit Comparison Test says that the series of absolute
values diverges as well.
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
∞
X
(2n)!
n (n!)2
2
n=1
converge or diverge? Explain your answer.
Answer: Use the Ratio Test:
(2(n+1))!
2n+1 ((n+1)!)2
(2n)!
n→∞
2n (n!)2
lim
(2n + 2)!
2n (n!)1
·
n→∞ 2n+1 ((n + 1)!)2
(2n)!
= lim
(2n + 2)(2n + 1)
n→∞
2
(2n + 2)(2n + 1)
= lim
n→∞
2
= lim
(n!)2
((n + 1)n!)2
1
·
.
(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
= = 2.
n→∞ 2(n2 + 2n + 1)
2
lim
Since 2 > 1, the Ratio Test implies that the given series diverges.
7
19. Does the series
∞
X
(−1)n
n ln(n)
n=2
converge absolutely, converge conditionally, or diverge? Explain your answer.
Answer: The series of absolute values
∞ ∞
X
(−1)n X
1
=
n ln(n) n ln(n)
n=2
n=2
diverges by the integral test:
∞
Z
2
1
dx = lim
b→∞
x ln(x)
so, using the substitution u = ln x, we see that du =
Z
ln(b)
lim
b→∞
ln(2)
1
x
Z
b
2
1
dx,
x ln(x)
dx and so the above integral becomes
du
ln(b)
= lim [ln(u)]ln(2)
b→∞
u
= lim (ln(ln(b)) − ln(ln(2)))
b→∞
=∞
Therefore the improper integral
P∞
1
n=2 n ln(n) also diverges.
R∞
2
1
x ln(x)
dx diverges, and so the integral test says that the series
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
∞
X
1
2 + 1)p
(n
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 +1)p
lim
1
n→∞
(n2 )p
since limn→∞
n2
n2 +1
1
(n2 )p
= lim
·
= lim
n→∞
n→∞ (n2 + 1)p
1
n2
n2 + 1
p
P
1
(n2 )p :
= 1p = 1,
= 1.
P 1
P 1
Therefore, the given series and the series
(n2 )p =
n2p will either both converge or both diverge.
P 1
Since
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.
∞
X
3n (x − 2)n
.
n2
n=1
8
Answer: Start by applying the Ratio Test to the series of absolute values:
n+1
3 (x−2)n+1 (n+1)2
3n+1 |x − 2|n+1
n2
n2
=
lim
·
=
lim
3|x
−
2|
·
= 3|x − 2|.
lim n
3 (x−2)n n→∞
n→∞
n→∞
(n + 1)2
3n |x − 2|n
(n + 1)2
n2 Therefore, the ratio test implies that the given series converges absolutely when 3|x − 2| < 1, meaning
when |x − 2| < 31 .
Now, check the endpoints. When x − 2 = 13 , the series becomes
∞
X
3n
n=1
1 n
3
n2
=
∞
X
1
,
n2
n=1
which converges.
Likewise, when x − 2 = − 13 , the series becomes
∞
X
3n − 31
n2
n=1
n
=
∞
X
(−1)n
,
n2
n=1
which also converges.
Therefore, the series converges for
1
1
− ≤x−2≤ ,
3
3
5 7
meaning the interval of convergence is 3 , 3 .
22. What are the first four nonzero terms of the Taylor series centered at x = 0 for the function f (x) = xe3x ?
Answer: Since
ex = 1 + x +
we see that
e3x = 1 + (3x) +
x3
x2
+
+ ...,
2!
3!
(3x)2
(3x)3
9x2
27x3
+
+ . . . = 1 + 3x +
+
+ ....
2!
3!
2!
3!
Therefore,
9x2
27x3
9x3
27x4
xe3x = x 1 + 3x +
+
+ . . . = x + 3x2 +
+
+ ....
2!
3!
2!
3!
23. Which of the following gives the value of
Explain your answer.
(a)
1
2
1
1
(b) −
2 120
(c)
R 1/2
0
cos(x2 ) dx correct to within 0.0001 (i.e. within 1/10, 000)?
1
1
−
24 2688
1
1
(d) −
2 320
Answer: Since
cos(x) = 1 −
1
1
(e) −
2 384
x2
x4
+
− ...,
2!
4!
the Taylor series for cos(x2 ) centered at x = 0 is
cos(x2 ) = 1 −
(x2 )2
(x2 )4
x4
x8
+
− ... = 1 −
+
− ....
2!
4!
2
24
9
(f)
1
1
−
24 1024
Therefore,
1/2
Z
1/2
x4
x8
+
− . . . dx
2
24
0
1/2
5
9
x
x
= x−
+
− ...
10 9 · 24
0
1
1
1
= −
− ....
+
2 320 9 · 24 · 29
cos(x2 ) dx =
0
Z
1−
Hence, choice (d) is clearly the best approximation.
24. What is the limit of the following sequence? Explain your answer.
∞
sin(arctan(ln(n)))
n=2
Answer: Since limn→∞ ln(n) = +∞ and since limx→π/2+ tan(x) = +∞, it follows that
lim arctan(ln(n)) =
n→∞
π
.
2
Therefore,
lim sin(arctan(ln(n))) = sin(π/2) = 1.
n→∞
25. Does the series
√
∞
X
n=2
n2
n
−n+1
converge or diverge? Explain your answer.
Answer:
For large n, we should expect that n2 will dominate the denominator, so do a limit comparison
P √n
to
n2 :
√
√
n
n
n2
n2
n2 −n+1
√
√
lim
=
lim
=
lim
·
= 1.
n
n→∞
n→∞ n2 − n + 1
n n→∞ n2 − n + 1
2
n
Therefore, since the series
26. Does the series
P √n
n2
=
P
1
n3/2
converges, so does the given series.
∞
X
(−1)n (n2 + n3 )
n4 + ln(n)
n=1
converge absolutely, converge conditionally, or diverge? Explain your answer.
Answer: First, notice that the series satisfies the Hypotheses of the Alternating Series Test, so it
definitely converges. To see whether it converges absolutely, consider the series of absolute values
∞ ∞
X
(−1)n (n2 + n3 ) X
n2 + n3
=
.
n4 + ln(n) n4 + ln(n)
n=1
n=1
10
Now, you should expect the n3 to dominate the numerator and the n4 to dominate the denominator,
P n3
so do a limit comparison to the series
n4 :
n2 +n3
n4 +ln(n)
lim
n3
n→∞
n4
n4
n2 + n3
·
n→∞ n4 + ln(n) n3
= lim
n4
n2 + n3
· 4
3
n→∞
n
n + ln(n)
1
+1
1
= lim n
·
n→∞
1
1 + ln(n)
4
= lim
n
=1·1
= 1.
P n3
P1
Therefore, since the series
n4 =
n diverges, the Limit Comparison Test implies that the series of
absolute values diverges, and hence the given series only converges conditionally.
27. For which values of p does the following series converge? Explain your answer.
∞
X
√
1 + np .
n=1
Answer: Notice that for any p, 1 + np > 1, meaning that
√
√
1 + np > 1 = 1.
Therefore,
∞
X
√
1 + np >
n=1
∞
X
1,
n=1
which obviously diverges, so the given series also diverges.
28. What is the interval of convergence of the following series? Explain your answer.
∞
X
(x − 1)n
.
n2
n=1
Answer: First, use the Ratio Test on the series of absolute values:
(x−1)n+1 (n+1)2 |x − 1|n+1
n2
n2
= lim
lim ·
=
lim
|x
−
1|
·
= |x − 1|.
n
(x−1) n→∞
n→∞ (n + 1)2
n→∞
|x − 1|n
(n + 1)2
n2 Thus, the Ratio Test says that the given series converges absolutely when |x − 1| < 1.
Now check the endpoints. When x − 1 = 1, the series becomes
∞
∞
X
X
1n
1
=
,
2
2
n
n
n=1
n=1
which converges.
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Likewise, when x − 1 = −1, the series becomes
∞
X
(−1)n
,
n2
n=1
which also converges.
Therefore, the given series converges for
−1 ≤ x − 1 ≤ 1,
so the interval of convergence is [0, 2].
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