Homework 11
Math 501
Due November 21, 2014
Exercise 1
Let (an ) ⊂ R be a sequence such that an ≥ 0 for all n and
1
2
2 (x
We note first that if x, y > 0 then xy ≤
easily from the fact that (x − y)2 ≥ 0.
P∞
n=1
an converges.
+ y 2 ). Indeed, this follows
It follows from the above remark that
√
an
1
1
≤
an + 2
n
2
n
for all n ≥ 1. Therefore since
∞
X
∞
X
1
2
n
n=1
an ,
n=1
both converge, it follows by the comparison test that
P∞ √
n=1
an /n converges.
Exercise 2
Consider the series
∞
X
1
.
n logp n
n=1
Note that
Z
e
∞
dx
=
x logp x
Z
1
∞
dx
xp
converges whenever p > 1 and diverges whenever p ≤ 1. It follows by the
integral test that the series converges and diverges respectively for the same
ranges of p.
Exercise 3
1
Suppose that (an ), (bn ) ⊂ R are sequences of positive real numbers and
L = lim sup
n→∞
Moreover, suppose that
P∞
n=1 bn
an
< ∞.
bn
converges. We claim that
P∞
n=1
an converges.
Indeed, let K be any real number greater than L. Note then that there
exists N such that
an
≤K
bn
P∞
for all n ≥ N . It follows that n=N an converges because
∞
X
an ≤ K
n=N
It follows that the whole sum
∞
X
bn < ∞.
n=N
P∞
n=1
an must then converge.
Exercise 4
P∞
Let (an ) ⊂ R be a sequence
of a positive real numbers. Prove that n=1 an
P∞
converges if and only if n=1 log(1 + an ) converges. [Hint: use L’Hôpital’s rule
and the limit comparison test.]
P∞
Suppose that
converges. Observe that by the limit comparison
n=1 an P
∞
test, in order to show that n=1 log(1 + an ) converges, it suffices to show that
lim sup
n→∞
log(1 + an )
< ∞.
an
(1)
P
Note that because
an converges, it follows that an → 0. Morevover, by
L’Hôpital’s rule, we have that
log(1 + x)
1
= lim
= 1.
x→0
x→0 1 + x
x
lim
It follows that (1) holds. The reverse implications follows similarly.
Exercise 5
Let cn = 1 + an where an ≥ 0 for all n or an ≤ 0 for all n.
P∞
P∞Suppose that n=1 an converges. Observe that by the previous exercise,
n=1 log(1 + an ) converges. That is, the sequence of partial sums
Sn =
n
X
log(1 + ak )
k=1
2
converges. By the continuity of the function f (x) = ex , it follows that the
sequence f (Sn ) converges as well. Moreover, note that
!
n
n
n
n
X
Y
Y
Y
f (Sn ) = exp
log(1 + ak ) =
exp (log(1 + ak )) =
1 + ak =
ck .
k=1
It follows that
k=1
Q∞
k=1 ck
k=1
k=1
converges. The reverse implication follows similarly.
Exercise 6
Let (an ), (bn ) ⊂ R be sequences of real numbers. Suppose that
converges and (bn ) is bounded by M . Observe that
∞
X
|an bn | ≤ M
P∞
n=1
n=1
|an | < ∞.
n=1
n=1
Therefore, since
∞
X
P∞
an bn converges absolutely, it converges.
Exercise 7
Let fn , f : [a, b] → R be functions. Suppose that fn → f pointwise and
Z
b
Z
fn dx →
a
b
f dx
a
as n → ∞. Prove or disprove: fn ⇒ f .
Consider fn , f : [0, 1] → R given by
fn (x) = xn ,
f (x) =
0
1
x ∈ [0, 1)
.
x=1
Note that fn → f pointwise. Moreover,
Z 1
Z 1
1
→0=
f dx.
fn dx =
n+1
0
0
However fn does not converge to f uniformly.
3
|an |