MAT 193 First Midterm
SHOW YOUR WORK!!
1. True or false?
(In my solutions I gave reasons for my answers to these true-false questions but you
didn’t need to do that.)
(a) If {an }∞
n=1 is an increasing sequence and every term an is less than 100,000 then
∞
{an }n=1 converges to a limit as n → ∞.
Solution: True. A bounded monotone sequence must have a limit. See p. 420
in the text.
(b) If {an }∞
n=1 is an increasing sequence and every term an is less than 10 then limn→∞ an =
10.
Solution: False. For example the sequence {1 − 1/n} tends to 1 as n → ∞.
(c) If
∞
X
|an | converges then lim an = 0.
n→∞
n=0
Solution: True. If the series converges then the terms must tend to 0 as n →
∞. See p. 426 in the text.
(d) If lim an = 0 then
n→∞
∞
X
an converges.
n=0
Solution: False. The harmonic series 1+1/2+1/3+1/4+1/5+· · · is a famous
counterexample. See p. 426 in the text.
(e) If
∞
X
n=0
|an | converges then
∞
X
an converges.
n=0
Solution: True. Absolutely convergent series are convergent. See p. 444 in the
text.
∞
X
1
converges. Show your
+ n2 + 1
n=1
work, briefly justify any inequalities that you use, and draw a diagram that explains
your use of the integral test.
2. Use comparison and integral tests to show that
n3
Solution: Let n > 0. Then n3 + n2 + 1 > n3 . Hence 1/(n3 + n2 + 1) < 1/n3 .
Therefore
∞
∞
X
X
1
1
<
3
2
n + n + 1 n=1 n3
n=1
By the integral test we have
Z ∞
∞
X
1
1
dx
≤1+
=1+ .
3
3
n
x
2
1
n=1
y ‡ €€€€13€
x
¼®
1
2
3
4
Hence
∞
X
n=1
n3
1
3
<
2
+n +1
2
It follows that the series converges.
∞
X
1
converges to a limit L. Suppose you approximate L by adding
3
n
n=1
up k terms of the series. How big does k need to be so that the error is less than 10−6 ?
3. The series
(−1)n+1
Solution: Solution #1. This is an alternating series with terms of decreasing size.
Therefore the alternating series test says that the absolute value of the error is no
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larger than the absolute value of the first term that is left out. So we need
1
< 10−6 .
3
(k + 1)
Since (1/100)3 = 10−6 it’s clear that k = 100 is large enough.
Solution:
Solution #2. It’s also possible but harder to use the integral test to estimate the
error of this alternating series.
∞
X
n=1
n+1
(−1)
k
∞
X
X
1
1
n+1 1
=
+
(−1)
(−1)n+1 3
3
3
n
n
n
n=1
n=k+1
so if we add up only the first k terms then the absolute value of the error is
¯
¯
¯
∞
∞ ¯
¯
¯ X
X
¯
¯
1
¯
¯
n+1 1 ¯
n+1
¯
(−1)
≤
(−1)
¯
¯
¯
¯
n3 ¯
n3 ¯
n=K+1
n=k+1
∞
X
1
n3
n=k+1
Z ∞
1
dt
≤
t3
k
1
= 2.
2k
=
p
Therefore we need 1/(2k 2 ) < 10−6 which means that k > 1/ (106 )/2 ≈ 707.1. So
k ≥ 708 would be enough.
This is a much worse estimate than we obtained with the alternating series test.
It’s worse because it doesn’t take account of all the cancellations that go on in an
alternating series.
4. Show how to use the “multiplication by r” trick to find the sum
S=
∞ µ ¶n
X
2
n=5
(Note that the sum starts at n = 5).
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3
.
Solution: Let S be the sum.
¡ ¢5
¡ ¢6
¡ ¢7
¡ ¢8
¡ ¢9
S = 23
+ 32
+ 32
+ 32
+ 32 + . . .
¡ 2 ¢6
¡ ¢7
¡ ¢8
¡ ¢9 ¡ ¢10
¡2¢
S =
+ 32
+ 32
+ 32 + 23
+ ...
3
3
Subtract the second row from the first. Almost everything on the right cancels and
we obtain
µ ¶5
1
2
S=
.
3
3
So
µ ¶5
2
S=3
.
3
5. (a) Find the radius of convergence of the series
µ
¶
∞
X
n
n
(−1)
2n (x − 3)n .
n
+
1
n=0
Solution: I’ll use the ratio test. (The root test would work just as well. Actually in this case the root test is a little easier).
¯
¯
¯ (−1)n+1 ¡ n+1 ¢ 2n+1 (x − 3)n+1 ¯
n + 1 n + 1 2n+1 |x − 3|n+1
¯
¯
¡n+2n ¢
=
lim
lim ¯
¯
n→∞ ¯
2n |x − 3|n
(−1)n n+1 2n (x − 3)n ¯ n→∞ n + 2 n
= 2|x − 3|
For absolute convergence we need
2|x − 3| < 1
hence
1
|x − 3| < .
2
Thus 1/2 is the radius of convergence.
(b) On what interval does the series in part (a) converge absolutely?
Solution: The series converges absolutely when |x−3| < 1/2 which means that
2.5 < x < 3.5.
6. Compute the first five terms of the Taylor series of the function f (x) = ln(x) expanded
about the point x = 3.
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Solution: If f (x) = ln(x) then f ′ (x) = 1/x = x−1 , f ′′ (x) = −1x−2 , f ′′′ (x) = 2x−3 ,
and the fourth derivative is f (4) (x) = −6x−4 . At x = 3 the values of f and its
first four derivatives are f (3) = ln(3), f ′ (3) = 1/3, f ′′ (3) = −1/9, f ′′′ (3) = 2/27,
f (4) (3) = −6/81 = −2/27. Therefore the first four terms of its Taylor series are
1
−1 1
2 1
−2 1
f (x) ≈ ln(3) + (x − 3) +
(x − 3)2 +
(x − 3)3 +
(x − 3)4
3
9 2
27 3!
27 4!
(If you included the fifth-derivative term (· · · )(x − 3)5 in your Taylor series that’s
OK.)
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