Math 414
Professor Lieberman
May 12, 2003
SOLUTIONS TO FINAL EXAM
1. First we check the case n = 1:
F1 = 1
and
1
√
5
"
√ !n
1+ 5
−
2
For n = 2, we have

√ !2
1  1+ 5
√
−
2
5
"
√ #
√ !n #
√
1− 5
1 1+ 5 1− 5
= 1.
−
=√
2
2
2
5

"
!
!#
√ !2
√
√
1− 5 
1
1+ 5
1− 5
=√
+1 −
+1
2
2
2
5
"
√
√ #
1 1+ 5 1− 5
=√
−
=1
2
2
5
= F2 .
Now we suppose that
1
Fm = √
5
"
√ !m
1+ 5
−
2
√ !m #
1− 5
2
for some natural number m ≥ 2. Then
Fm+1 = Fm + Fm−1


"
√ !m−1
√ !m−1
√ !m
√ !m #
1+ 5
1− 5
1
1+ 5
1− 5
1

−
−
+√ 
=√
2
2
2
2
5
5

!
!
√ !m−1
√
√ !m−1
√
1+ 5
1+ 5
1− 5
1− 5
1
+1 −
+1 
=√ 
2
2
2
2
5


√ !m−1
√ !2
√ !m−1
√ !2
1+ 5
1− 5
1− 5 
1  1+ 5
−
=√
2
2
2
2
5


√ !m+1
√ !m+1
1  1+ 5
1− 5
.
−
=√
2
2
5
2. There is no maximum or minimum. The supremum is 1 and the infimum is −1.
1
2
3. 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→∞
4. 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
h→0
sin(1/n) ≥ 1/(2n) if n ≥ N . Hence, if n ≥ N , then
π
1 1
)
≥ .
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→∞
5. We write f (x) = x/(x + 1), g(x) = ln(x + 1) and h(x) = x. Then f (0) = g(0) =
h(0) = 0, so we have equality if x = 0. In addition,
f 0 (x) =
1
,
(x + 1)2
g 0 (x) =
1
,
x+1
h0 (x) = 1.
For x > 0, we have f 0 (x) < g 0 (x) < h0 (x), so f − g and g − h are strictly decreasing
functions on (0, ∞). It follows that
x
< ln(x + 1) < x
x+1
for x > 0, so the inequality is true and it’s an equality only if x = 0.
6. We first use the definition to see that an is the Riemann sum for the integral corresponding to the partition Pn = {x0 , . . . , xn } with xj = j/n. Therefore
L(Pn , f ) ≤ an ≤ U (Pn , f ).
Because the mesh of Pn (the mesh is just the length of the longest subinterval in the
partition) is 1/n, it follows that
Z 1
lim L(Pn , f ) = lim U (Pn , f ) =
f (x) dx,
n→∞
n→∞
0
and the squeeze theorem says that
Z 1
f (x) dx = lim an .
0
n→∞
3
7. Method 1 (The root test)
s
k
1
1
1
=
≤
k
(ln k)
ln k
ln 3
for k ≥ 3 and 1/ ln 3 < 1, so the series CONVERGES.
Method 2 (The comparison test) For k ≥ 3,
1
1
≤
,
k
(ln k)
(ln 3)k
and
∞
X
k=2
1
(ln 3)k
is a convergent geometric series, so the original series CONVERGES.
8. (a) When n = 1, we have
n
X
ak bk = a1 b1 = s1 b1 = s1 b2 − s1 (b2 − b1 )
k=1
= s1 b 2 −
1
X
sk (bk+1 − bk ).
k=1
If the equation is true for n = m and m ∈ N, then
m+1
X
ak b k =
k=1
m
X
ak bk + am+1 bm+1
k=1
= sm bm+1 −
m
X
sk (bk+1 − bk ) + am+1 bm+1
k=1
m
X
= sm+1 bm+1 −
sk (bk+1 − bk )
k=1
= sm+2 bm+1 − sm+2 (bm+2 − bm+1 ) −
m
X
sk (bk+1 − bk )
k=1
= sm+1 bm+2 −
m+1
X
sk (bk+1 − bk ).
k=1
(b) From (a), we have
n
X
ak bk = sn bn+1 −
k=1
n
X
sk (bk+1 − bk ) ≤ M bn+1 −
m
X
M (bk+1 − bk )
k=1
k=1
because bn+1 ≥ 0 and (bk+1 − bk ) ≤ 0. Next,
M bn+1 −
m
X
k=1
M (bk+1 − bk ) = M
bn+1 −
n
X
k=1
!
(bk+1 − bk )
= M b1 ,
4
so
n
X
ak b k ≤ M b 1 .
k=1
The other inequality is proved similarly.
9. (a) It is not enough to show that the partial sums are bounded! Instead, start with
a given ε > 0 and write L for the bound on (sn ), so |sn | ≤ L for all n ∈ N. Then
there is a natural number N such that bN ≤ ε/L. If m ≥ n ≥ N , then
m
X
ak bk ≤ bm L ≤ bN L < ε.
k=n
P
Therefore the series
ak bk converges.
(b) Set b = lim bn . Suppose first that (bn ) decreases to b, and set cn = bn − b. Then
n→∞
P
(cn ) is a decreasing sequence
which
converges
to
zero,
and
the
partial
sums
of
ak
P
are bounded, so the series
ak ck converges by part (a). Next, we have
X
X
X
ak b k =
ak c k +
ak b,
and we have just shown that the first sum converges. The second sum converges
because it’s a constant b times a convergent series. When (bn ) increases to b, a
similar argument applies (with dn = b − bn defining the sequence which decreases to
zero).
10. In both parts, we will use the interval
(a) Use

1

2
f (x) = x

1
[0, 1].
2
if x = 0,
if 0 < x < 1,
if x = 1.


1
f (x) = x

0
if x = 0,
if 0 < x < 1,
if x = 1.
(b) Use
11. Here, we need a δ−ε type proof. Given ε > 0, there is a δ > 0 such that |f 0 (x)−A| < ε
if 0 < |x − c| < δ and a < x < b. If |h| < min{δ, c − a, b − c}, then there is a number
x between c and c + h such that
f (c + h) − f (c)
= f 0 (x),
h
so
f (c + h) − f (c)
< ε.
−
A
h
Since ε > 0 is arbitrary, it follows that
f (c + h) − f (c)
lim
= A,
h→0
h
which means that f 0 (c) exists and equals A.
5
√
12. The substitution u = x (so x = u2 and dx = 2u du) gives
√
Z 1
Z 1
exp( x)
√
dx =
eu 2 du = 2(e1 − e0 ) = 2e − 2.
x
0
0
To be technically correct, you need to recognize
that the original integral is an im√
√
proper integral (because the integrand exp( x )/ x ) is unbounded near x = 0) and
take some appropriate limits. Nobody noticed this fact, so no points were deducted
for missing it.