**Unformatted text preview: **Real Analysis Homework IV solutions
October 11th, 2019 Exercise 2.3.3
Suppose {xn } is a bounded sequence and {xnk } is a subsequence. Prove
lim inf xn ≤ lim inf xnk .
n→∞ k→∞ Proof. Let an = inf{xm : m ≥ n}, bn = inf{xnm : m ≥ n}. Then,
lim an = lim inf xn , lim bn = lim inf xnk For any n, consider the sets Sn = {xm : m ≥ n} and Tn = {xnm : m ≥ n}. Notice that
nm ≥ m (this was proven in class), so if m ≥ n, then nm ≥ n . It follows that if xnm ∈ Tn ,
then xnm ∈ Sn , so Tn ⊂ Sn . By Exercise 1.1.4, this means inf Sn ≤ inf Tn , or an ≤ bn . Since
n was arbitrary, we can take the limit (Prop 2.2.3) to conclude that
lim inf xn ≤ lim inf xnk .
n→∞ k→∞ Exercise 2.3.5
a. Let xn = (−1)n
,
n b. Let xn = (n−1)(−1)n
,
n (−1)n a. Since n ≤ find lim sup xn and lim inf xn . 1
n find lim sup xn and lim inf xn . and 1
n (−1)n → 0, Prop 2.2.10 implies n → 0. Hence, by Prop. 2.3.5, 0 = lim (−1)n
(−1)n
(−1)n
= lim sup
= lim inf
.
n
n
n 1 b. Let an = {sup{xk : k ≥ n}, then
lim sup xn = lim an .
We claim that for all n, an = 1 . To prove this, we first note that
1≥ (n − 1)(−1)n
n so 1 is an upper bound . On the other hand, if b < 1 , then by the Archimedean property
there exists an m such that
1
<1−b
m
Then, taking k = 2(m + n) , we see that k > n , so xk ∈ {xk : k ≥ n} , and furthermore,
(k − 1)(−1)k
k
1
=1−
k
1
≥1−
m
>b xk = so b is not an upper bound and an = 1. It follows that
1 = lim an = lim sup xn .
Similarly, let bn = inf{xk : k ≥ n}. We claim bn = −1 for all n. To see this, observe
that −1 is a lower bound of {xk : k ≥ n} for all n, and that for any b > −1 there is an
m such that
1
< b − (−1),
m
and hence for k = 2(m + n) + 1,
(k − 1)(−1)k
k
1
= −1 +
k
<b xk = so −1 = inf{xk : k ≥ n} = bn . Thus,
lim inf xn = lim bn = −1. 2 Exercise 2.4.1
Prove that n n2 −1
n2 o is a Cauchy sequence using directly the definition of Cauchy sequences. Proof. Let > 0 be arbitrary. Then, since n22 → 0 , there exists an M such that for all n ≥ M ,
2
< . Choose M , and let n, k ≥ M .
n2
Without loss of generality, assume n ≥ k. Then,
n2 − 1 k 2 − 1
n2 − k 2
−
=
n2
k2
n2 k 2
(n − k)(n + k)
≤
n2 k 2
n(2n)
≤ 2 2
nk
2
≤ 2
k
< Exercise 2.5.3
Decide the convergence or divergence of the following series.
P∞
3
a.
n=1 9n+1
P∞
1
b.
n=1 2n−1
c. P∞ d. P∞ e. P∞ n=1 (−1)n
n2 1
n=1 n(n+1) n=1 2 ne−n We make the following observation:
P
P
Lemma . If
xn diverges and α 6= 0, then
αxn diverges.
Proof. We prove
P the contrapositive.
Suppose
αxn converges. Then, by proposition 2.5.12(i),
X xn = X1
1X
αxn =
αxn
α
α converges.
3
12
3
a. Observe that 4 9n+1
= 9n+1
≥ n1 . Thus, by example 2.5.11 and the comparison test, 4 9n+1
P 3
diverges. By our lemma, this implies
diverges as well.
9n+1 3 b. We have that
diverges. 1
2n−1 ≥ 11
.
2n By the lemma and example 2.5.11, P11
2n diverges, so P 1
2n−1 P (−1)n P 1
P (−1)n
c. Observe that
converges by the p-test (Prop 2.5.17). Thus, n2 =
n2
n2
converges absolutely, and so it converges (Prop. 2.5.15).
P
1
d. Let SN = N
n=1 n(n+1) . Then,
SN =
= N
X
n=1
N
X
n=1 =1−
so lim SN = 1 , and P 1
n(n+1) 1
n(n + 1)
1
1
−
n n+1
1
N +1 converges to 1 . e. We have that
lim n→∞ ne−n 2 1/n = lim n1/n e−n
n→∞
1/n
−n
= lim n
lim e
n→∞ n→∞ =1·0
=0
By Proposition 2.3.5, if limn→∞ xn exists, then lim supn→∞ xn = limn→∞ xn . Thus,
lim sup ne −n2 1/n =0<1 n→∞ so the series P ne−n converges by the root test (Prop 2.6.1).
2 Exercise 2.5.4
a. Prove that if P∞ n=1 xn converges, then P∞ n=1 (x2n + x2n+1 ) also converges. b. Give an explicit example where the converse does not hold.
a. Proof. Let
Sn =
Tn = n
X
k=1
n
X
k=1 4 xk
(x2k + x2k+1 ) P
P∞
be the partial sums of ∞
x
and
We want to prove
n
n=1
n=1 (x2n + x2n+1 ), respectively.
P
that limn→∞ Tn exists. Observe limn→∞ Sn converges , since ∞
x
n=1 n converges. Moreover, the partial sums are related by
Tn = S2n+1 − x1 .
Using the rules of limit arithmetic,
lim Tn = lim S2n+1 − x1 = n→∞ so P∞ n=1 (x2n n→∞ ∞
X xn − x1 . n=1 + x2n+1 ) converges. b. Let xn = (−1)n . Then, ∞
∞
X
X
(x2n + x2n+1 ) =
0=0
n=1 but n=1
∞
X xn = n=1 ∞
X (−1)n n=1 cannot converge, since the terms do not go to zero (Prop 2.5.9). Exercise 2.5.6
Let P xn be a series a. If there is an N and a ρ < 1 such that
|xn+1 |
<ρ
|xn |
for all n ≥ N , then the series converges absolutely.
b. If there is an N such that for all n ≥ N ,
|xn+1 |
≥1
|xn |
then the series diverges.
(Although it is not explicitly given, we will assume xn 6= 0 for n ≥ N in order to ensure all the
fractions are well-defined).
P
a. Proof. By Prop 2.5.6, it is enough to prove that ∞
n=N +1 |xn | converges.
For n ≥ N ,
|xn | |xn−1 |
|xN +1 |
···
|xN |
|xn−1 | |xn−2 |
|xN |
≤ ρn−N |xN |
P
n−N
By Prop P
2.5.5, 2.5.6 and Prop 2.5.12(i), ∞
|xN | converges, so by the comparin=N +1 ρ
son test
|xn | converges.
|xn | = 5 b. Proof. Observe that for n ≥ N ,
|xn | |xn−1 |
|xN +1 |
···
|xN |
|xn−1 | |xn−2 |
|xN |
≥ |xN | |xn | = Thus, |x
Pn | ≥ |xN | > 0 for all n ≥ N , so xn does not converge to 0 . By Prop 2.5.9, this
shows
xn diverges. Exercise 2.5.8
Show that converges.
P
(−1)n
The partial sums of ∞
are
n=1
n
P∞ n=1 (−1)n
n Sn = n
X
(−1)k
k=1 k . We wish to prove that {Sn } converges. First, let us consider S2n . By grouping terms, we find
that
n
n
X
X
1
1
−1
S2n =
−
=
2k 2k − 1 k=1 2k(2k − 1)
k=1
which are exactly the partial sums of
∞
X
n=1 By comparison to the p-series 1
n=1 n2 , P∞ −1
.
2n(2n − 1) we see that this series converges, so lim S2n = n→∞ ∞
X
k=1 −1
2k(2k − 1) exists. Let us call this limit S.
Certainly, if Sn converges, it must converge to S. Let’s show lim Sn = S. Let > 0 be
arbitrary. Since S2m convergesto S, there
exists an M such that for all m ≥ M , |S2m − S| < . 2
Choose M , and let N ≥ max , 2M .
Consider n ≥ N . There are two cases: either n is even or n is odd.
If n is even, n = 2k. It follows that k ≥ M , so
|S2k − S| <
2 by the choice of M .
On the other hand, if n is odd, n = 2k − 1. Then,
|Sn − S| ≤ |Sn − Sn+1 | + |Sn+1 − S|
6 As above,
|S2k − S| <
On the other hand,
2 (−1)n |Sn − Sn+1 | = n 1
=
n
1
≤
N
≤
2 so |Sn − S| ≤ |Sn − Sn+1 | + |Sn+1 − S| <
P (−1)n
So, |Sn − S| < and
converges.
n Exercise 2.5.9
a. Prove that if P xn and P yn converge absolutely, then P xn yn is absolutely convergent. b. Find an example where the converse is not true.
c. Find P
an explicit
P example
P where all three series are convergent, are not just finite sums,
and ( xn ) ( yn ) 6= xn yn . That is, series are not multiplied term-by-term.
P
a. Since
yn converges, yn → 0 by Prop 2.5.9, so yn P
is bounded . Let B be such that
|yn | ≤ B for all n . Then, |xn yn | ≤ B|xnP
|. Since
B|xn | converges (Prop 2.5.12(i)
and
|xn yn | converges as well, which shows that
P the definition of absolute convergence),
xn yn is absolutely convergent.
b. Let
(
1 n is even
xn =
0 n is odd
(
0 n is even
yn =
1 n is odd
P
P
P
Then,
xn yn =
0 is absolutely convergent. However,
|xn |
Pxn yn = 0 for all n, so
and
|yn | do not converge since |xn |, |yn | 6→ 0 (Prop 2.5.9).
n
c. Let xn = yn = 12 . Then, by Prop 2.5.5,
X
X
X
X
|xn | =
|yn | =
xn =
yn = 1
P
P
so
xn ,
yn are absolutely convergent, but
X
X 1
1 X X
xn y n =
=
6=
xn
yn .
4n
3
7 Exercise 2.5.11
Prove that if an > 0 and bn > 0 for all n, and
0 ≤ lim
then P an and P an
≤∞
bn bn either both converge or both diverge. Proof. By limit arithmetic, if
then 0 ≤ lim an
≤∞
bn
−1
bn
an
0 ≤ lim
= lim
≤∞
an
bn
P
so it is enough to prove that if
bn converges, then an converges, since we can reverse the
roles of an and
to get the general result.
P bn and take the contrapositive
Suppose
bn converges, and let L = lim abnn . Then, there exists an M such that for n ≥ M ,
an
<L+1
bn
P
P∞
choose N . By Prop 2.5.6, ∞
n=N bn converges, so, since 0 < an <
P(L + 1)bn , n=N an converges
(Prop 2.5.12 and the comparison test). Finally, by Prop 2.5.6,
an converges. Exercise 2.5.16
Use the Cauchy condensation principle to decide the convergence of
P ln n
a.
n2
P 1
b.
n ln n
P 1
c.
n(ln n)2
P
1
d.
n ln n(ln ln n)2
P
P
P
P
To simplify notation, we will write
an ∼
bn to mean
an converges if and only if
bn
converges.
Also, where necessary, we will star the summation at n = 2 or n = 3 instead of n = 1 to
avoid undefined terms.
a. When n ≥ 3, ln n increases
slowly than n2 , so
P lnmore
P
n
ln n
Prop 2.5.6, we see that
∼ ∞
n=3 n2
n2 ln n
n2 is decreasing after n = 3. Using On the other hand, we can apply the Cauchy condensation test to the second series to
obtain
X ln n X n ln 2
∼
n2
22n
Since n ln 2 1/n
1
1 lim n = lim n1/n lim (ln 2)1/n lim = ,
n→∞
n→∞
n→∞
n→∞ 2
2
2
this last sequence converges, so the original sequence converges.
8 b. This sequence of terms is decreasing (here, we must begin with n = 2 to avoid undefined
terms). Applying Cauchy condensation gives that
X 1
1
∼
n ln n
n ln 2
P1
which diverges by comparison with
, so the original series diverges.
n
X c. Since the terms in the series are decreasing (n ≥ 2), we can apply Cauchy condensation
to get
X
X
1
1
∼
2
2
n(ln n)
n (ln 2)2
P 1
which converges by comparison with the p-series
, so the original series converges.
n2
d. (We assume n ≥ 3 so all the terms are defined). The series is decreasing, so one application
of condensation gives
X X
1
1
∼
n ln n(ln ln n
n ln 2(ln(n ln 2))2 By logarithm arithmetic, ln(n ln 2) = ln n + ln ln 2. Thus,
1
1
1
1
=
≤
n ln 2(ln(n ln 2))2
n(ln n + ln ln 2)2
ln 2 n(ln n)2
so, by comparison with the series in part (c), the original sequence converges. 9 ...

View
Full Document