Math 414
Professor Lieberman
May 6, 2003
SOLUTIONS TO PRACTICE FINAL EXAM
1. Define a new sequence (bn ) by bn = an+1 − an . Then limn→∞ (bn + bn−1 ) = 0. Given ε > 0,
there is a natural number N such that, if n ≥ N , then |bn + bn−1 | < ε/2. We now prove by
induction (on k) that
ε
|bN +k | ≤ k + |bN |.
2
For k = 1, we have
ε
|bN +1 | = |bn − (bN + bN +1 )| ≤ |bN | + |bN + bN +1 | ≤ + |bN |.
2
If the inequality is true for k = m, then
|bN +(m+1) | = |bN +m − (bN +m + bN +(m+1) | ≤ |bN +m | + |bN +m + bN +(m+1) |
ε
ε
ε
≤ m + |bN | + ≤ (m + 1) + |bN |.
2
2
2
Now take M to be a natural number greater than N such that |bN |/M < ε/2. If n ≥ M ,
then
bn n − N ε |bN |
≤
+
< ε.
n
n 2
n
It follows that lim bn /n = 0.
n→∞
2. For n = 1, we have
1
X
Fk = F1 = 1 = F1+2 − 1.
k=1
If the statement is true for n = m, then
m+1
X
k=1
Fk =
m
X
Fk + Fm+1 = Fm+2 − 1 + Fm+1 = Fm+3 − 1.
k=1
3. In both parts, we will use the interval [0, 1].
(a) Use

1

 2 if x = 0,
f (x) = x if 0 < x < 1,

1
if x = 1.
2
(b) Use


1
f (x) = x


0
if x = 0,
if 0 < x < 1,
if x = 1.
1
2
4. By algebra,
a2 [f (x + bh) − f (x)] b2 [f (x + ah) − f (x)]
a2 f (x + bh) − b2 f (x + ah) + (b2 − a2 )f (x)
=
−
,
(a2 b − b2 a)h
(a2 b − b2 a)h
(a2 b − b2 a)h
so
a2 f (x + bh) − b2 f (x + ah) + (b2 − a2 )f (x)
lim
h→0
(a2 b − b2 a)h
f (x + bh) − f (x)
b2
f (x + ah) − f (x)
a2
lim
−
lim
= 2
2
a − ba h→0
bh
ab − b h→0
ah
a2
b2
−
f 0 (x) = f 0 (x).
=
a2 − ba ab − b2
5. The substitution u = ex (so x = ln u and dx = du/u) gives
Z ∞
Z ∞
sin u
x
sin(e ) dx =
du,
u
0
1
which converges. (More formally, we would say that
Z ∞
Z b
Z
x
x
sin(e ) dx = lim
sin(e ) dx = lim
b→∞ 0
0
b→∞ 1
ln b
sin u
du,
u
which exists because the improper integral
Z ∞
sin u
du
u
1
converges.)
6. Because the P
sequence (bk ) is bounded, there is a number M such that |bk | ≤ M for all
k. Because
|ak | converges, given ε > 0, there is a natural number N such that, if
n ≥ m ≥ N , then
n
X
ε
|ak | ≤
.
M
k=m
Now, if n ≥ m ≥ N , then
n
n
n
X
X
X
|bk ak | ≤
M |ak | = M
|ak | < ε,
k=m
which means that
P
k=m
k=m
ak bk converges absolutely.
7. This is an alternating series, so set ak = 1/(k ln k). Then ak > 0, (ak ) is a decreasing
sequence, and ak → 0 as k → ∞. Therefore the series converges by the alternating series
test.
8. FALSE. Take
f (x) =
(
1
1
x
if x = 0,
if x > 0,
and define xn = 1/n. Then lim f (xn ) = ∞, but lim f (x) = 0.
n→∞
x→∞
3
9. First, because the equation is a cubic, it has at least one root (by the intermediate value
theorem). If we define f (x) = x3 + ax + b, then the mean value theorem tells us that if it
has more than one root, then the equation f 0 (x) = 0 also has at least one root. However,
f 0 (x) = 3x2 + a, which is never zero because a > 0. Therefore the equation has exactly one
solution.
10. Take xn = nπ and tn = xn + 1/n. Then |xn − tn | → 0 as n → ∞ but
1 .
|f (xn ) − f (tn )| = tn sin
n Because lim sin h/h = 1, it follows that there is a natural number N such that sin(1/n) ≥
h→0
1/(2n) if n ≥ N . Hence, if n ≥ N , then
1 1
π
)
≥ .
n 2n
2
Therefore f is not uniformly continuous because a theorem in the book says that, if f is
uniformly continuous on D, then for any two sequences (xn ) and (tn ) of points in D, with
lim |xn − tn | = 0, we have lim f (xn ) − f (tn ) = 0.
|f (xn ) − f (tn )| ≥ (nπ +
n→∞
n→∞