Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2015
Exam #3
Solutions to the sample
This is a take-home examination. The exam includes 5 questions and one bonus problem.
The total mark is 100 points, not including the bonus question. Please show all the work,
not only the answers.
1 [20 Points]. Determine the radius of the convergence R of the following series and discuss
whether or not they converge at x = R and x = −R :
P∞ xn
√
(a)
n=1 n
(b)
P∞
(c)
P∞
(d)
P∞
n=1
(−1)n−1 xn
n4
nn
n
n=1 (n!)2 x
√
n=1
n
n2 + 2n − n xn
Solution:
(a) Let cn =
√1 .
n
Then
√
n
lim
n→∞
cn = 1.
Hence R = 1. When x = 1 we have
n
X
1
√ ,
cn =
n
n=1
n=1
n
X
and hence the series diverges at x = 1. When x = −1 we have
n
X
n
X
1
√ (−1)n ,
cn (−1) =
n
n=1
n=1
n
and hence, according to the Leibnitz (alternating series) test, the series converges at
x = −1.
(b) Let cn =
(−1)n−1
.
n4
Then
lim
n→∞
p
n
|cn | = 1.
Hence R = 1. When x = −1 or x = 1, the series converges according to the power test.
1
(c) Let cn =
nn
.
(n!)2
Then
c h
e
1 n n + 1 i
n+1 ·
= lim
= 0.
lim = lim 1 +
2
n→∞ n + 1
n→∞ cn
n→∞
n
(n + 1)
Hence R = +∞.
√
n
(d) Let cn =
n2 + n − n . Then
√
n
lim
n→∞
cn = lim
n→∞
√
n2 + 2n − n = lim √
n→∞
n2
2n
= 1.
+ 2n + n
Hence R = 1. When x = 1 we have
n
X
ck =
n X
k=1
k=1
n
n
k X 2k k X
1 k
2k
√
>
=
1−
.
2k + 1
2k + 1
k 2 + 2k + k
k=1
k=1
n
1
Since limn→∞ 1 − 2n+1
= e−1/2 6= 0, the series diverges. Similarly, since cn doesn’t
converge to zero as n → ∞, the series diverges also when x = −1.
P
nn
2n
0
2 [20 Points]. Let f (x) = ∞
n=0 (n!)2 x , where we use the usual convention that 0 = 1.
Note that f (x) is well-defined for every x ∈ R, according to the result in Problem 1(c) above.
(a) Compute f 0 (0) or show that the derivative doesn’t exist.
(b) Prove that f is a convex function.
(c) Show that for any x > 0,
1
f (x) − 1 <
2ex
Z
x
f (x) dx <
0
1
f (x) − 1 .
x
Hint: Integrate part by part. To estimate the result of the integration you may use
without proof the following bounds:
1 n
1< 1+
< e,
∀ n ∈ N.
(1)
n
Solution:
(a) By Theorem 12 on p. 211 of the textbook,
∞
X
nn
f (x) =
2nx2n−1 .
2
(n!)
n=1
0
Thus f 0 (0) = 0.
2
(b) By Theorem 12 on p. 211 of the textbook,
∞
X
nn
f (x) =
(2n)(2n − 1)x2n−2 .
2
(n!)
n=1
00
f is convex since f 00 (x) > 0 for all x ∈ R.
(c) Theorem 11 on p. 211 of the textbook along with Theorem 8 on p. 208 imply that
Z
0
x
Z x
∞
∞
X
X
nn
nn
1
2n
x dx =
x2n+1 .
f (x) dx =
2
2
(n!) 0
(n!) 2n + 1
n=0
n=0
Thus
Z x
∞
∞
X
nn
1 X (n + 1)n+1 2n+2
(n + 1)2 nn
1
2n+1
f (x) dx =
x
=
x
·
(n!)2 2n + 1
x n=0 [(n + 1)!]2
(n + 1)n+1 (2n + 1)
0
n=0
∞
1 X (n + 1)n+1 2n+2 n + 1 n n
x
·
= (substitute k = n + 1)
=
x n=0 [(n + 1)!]2
2n + 1 n + 1
∞
1 X k k 2k
k k − 1 k−1
. =
x
·
.
x k=1 (k!)2
2k − 1
k
It follows then from (1) that, with n = k − 1,
k k − 1 k−1
n + 1 n n
n+1 1 −n
=
=
1+
,
2k − 1
k
2n + 1 n + 1
2n + 1
n
and hence
k k − 1 k−1
1
<
< 1,
2e
2k − 1
k
k ≥ 2.
Thus
Z
0
x
∞
1 X k k 2k
1
f (x) dx <
x
=
f
(x)
−
1
x k=1 (k!)2
x
and
Z
0
x
∞
1 X k k 2k
1
x
=
f
(x)
−
1
.
f (x) dx >
2ex k=1 (k!)2
2ex
3 [20 Points].
(a) Prove that
lim
n→∞
1 Z
n!
∞
n −x
x e dx = 1.
1
3
(b) Prove or disprove: If a sequence of real-valued functions fn converges to f on [a, b],
and
Z b
Z b
lim
fn (x)dx =
f (x)dx,
n→∞
a
a
then fn converges to f uniformly on [a, b].
Solution:
(a) Let
1
fn =
n!
∞
Z
1
x e dx = −
n!
n −x
1
Z
∞
xn (e−x )0 dx.
1
Integrating by part, we obtain
fn =
e−1
+ fn−1 .
n!
Thus
fn − f0 =
n
X
(fk − fk−1 ) =
k=1
n
X
e−1
k=1
n!
,
which implies
fn = f0 + e−1
n
n
n
X
X
X
1
1
1
= e−1 + e−1
= e−1
.
n!
n!
n!
k=1
k=0
k=1
Therefore,
lim fn = lim e
n→∞
−1
n→∞
n
X
1
= 1.
n!
k=0
To verify
last equality above, one can, for instance, use Taylor expansions for
P∞ the
xn
x
e = n=0 n! at x = 1.
(b) Counterexample:
x2
if x > 0
x + (1 − nx)2
gn (x) =
x2

 − 2
if x ≤ 0.
x + (1 + nx)2



2
We have
lim gn (x) = 0,
n→∞
x ∈ R.
Also,
since gn is a continuous function for all n ∈ N and gn (x) = −gn (−x), we get
R1
g (x) = 0 for all n ∈ N. But, for any n ∈ N,
−1 n
max |gn (x) − 0| = gn (1/n) = −gn (−1/n) = 1,
x
and hence fn does not converge to zero uniformly.
4
4 [20 points].
(a) Prove or disprove: Every point-wise converging sequence of functions fn : R → R
contains a uniformly converging subsequence.
(b) Construct sequences (fn )n∈N and (gn )n∈N of real-valued functions on R which converge
uniformly but such that the product fn gn doesn’t converge uniformly.
Solution:
(a) Counterexample: fn (x) =
x2
.
x2 +(1−nx)2
We have
x ∈ R.
lim fn (x) = 0,
n→∞
But, for any n ∈ N,
max |fn (x) − 0| = fn (1/n) = 1,
x
and hence fn does not converge to zero uniformly.
(b) Let fn (x) = gn (x) = ex + n−1 . Then both fn and gn converge uniformly on R to
h(x) = ex . However,
|fn gn (x) − e2x | =
2ex
1
+ 2
n
n
is unbounded on R for every fixed n ∈ N, and hence fn gn does not converge uniformly
to h2 (though fn gn of course converges to h2 point-wise).
5 [20 points]. Let (fn )n∈N be a sequence of real-valued continuous functions which converges
uniformly on a set E. Prove that
lim fn (xn ) = f (x)
n→∞
for every sequence of points (xn )n∈N such that limn→∞ xn = x in E.
Solution: By Theorem 1 on p. 203 of the textbook, f is continuous on E. Fix any ε > 0
and x ∈ E. Since f is continuous at x, there exists δ > 0 such that d(x, y) < δ implies
|f (x) − f (y)| < ε/2. If limn→∞ xn = x, there exists N such that n > N implies both
sup |fn (y) − f (y)| <
y∈E
ε
2
and
|xn − x| < δ.
Thus, for n > N we have
|f (x) − fn (xn )| ≤ |f (x) − f (xn )| + |f (xn ) − fn (xn )| ≤
≤
ε ε
+ = ε.
2 2
5
ε
+ sup |fn (y) − f (y)|
2 y∈E
Hence limn→∞ fn (xn ) = f (x).
6 [Bonus question]. Prove the upper bound in (1) in the statement of Problem 2.
Solution: The lower bound is trivial. We will next prove the upper bound. Define
1 n
an = 1 +
n
. Thus f (n) = log an . It suffices to show that an is increasing and
and f (x) = x log 1+x
x
limn→∞ log an = limx→∞ f (x) = 1. To this end, first observe that
lim f (x) = lim x log(1 + x−1 ) − log 1
x→∞
t→∞
=
=
log(1 + t) − log 1
substitute t = x−1 lim
t→∞
t
0 1
log(1 + t) t=0 =
= 1.
1 + t t=0
It remains to show that f (x) is increasing for x ≥ 1. To this end observe that f (1) = log 2
and
1
1
1+x
0
+x
−
= sx − 1 − log sx ,
f (x) = log
x
1+x x
where sx =
x
1+x
∈ (0, 1). Now, let g(s) = s − 1 − log s. Notice that g(1) = 0 and
g 0 (s) = 1 − s−1 < 0,
s ∈ (0, 1).
Thus
s ∈ (0, 1).
g(s) > g(1) = 0,
In particular, f 0 (x) = g(sx ) > 0, and hence f is increasing.
6