MA2108S Tutorial Solution
Zhang Liyang, Xiong Xi, Sun Bo
Tutorial on limit superior and inferior
(1). Given lim(an ) ≥ 0 and (bn ) is bounded, show that
lim an bn = lim an lim bn
n→∞
n→∞
n→∞
Proof. Let u := limn→∞ an bn , a := lim(an ), b := limn→∞ bn .
When a > 0, let x be a cluster point of (an bn ). Then there is a convergent
subsequence (ank bnk ) that converges to x. Since (bn ) is bounded, (bnk ) is
also bounded, by the Bolzano-Weierstrass Theorem, (bnk ) has a convergent
subsequence (bnkl ), say lim(bnkl ) = p. Since (an ) is convergent, (ankl ) also
converges to a, so x = lim(ank bnk ) = ap, since u = limn→∞ an bn , so ap ≤ u,
p ≤ ua , since b = limn→∞ bn , we have b ≤ ua , so ab ≤ u.
Let q be a cluster point of (bn ). Then there is a convergent subsequence
(bnk ), lim(bnk ) = q, since (an ) is convergent, (ank ) is convergent, (ank bnk ) is
convergent, then lim(ank bnk ) = aq, q ≤ limn→∞ bn , so aq ≤ ab, so ab is an
upper bound for S(an bn ), since u = limn→∞ an bn , then ab ≥ u, therefore, we
have limn→∞ an bn = limn→∞ an limn→∞ bn
When a = 0, since (bn ) is bounded, ∃ M > 0 such that bn ≤ M, ∀ n ∈ N. Let
² > 0 be given. Then since (an ) is convergent, there exists K ∈ N such that
if n ≥ K, then |an | < M² , that is, an < M² , so if n ≥ K, |an bn | < an |bn | < ²,
so lim(an bn ) = 0. Therefore, limn→∞ an bn = 0 = lim(an )limn→∞ bn .
(3). Let an > 0 for n ∈ N. If limn→∞ an+1 /an = r,show that {an }∞
n=1
converges if r < 1. What if r ≥ 1?
Proof. Case(i): r < 1. Let ² = 1−r
. Since lim an+1
= r and ² > 0,
2
an
an+1
∃ n0 ∈ N such that when n > n0 , an < r + ² = 1+r
< 1. Then
2
0 < an+n0 < an+n0 1+r
< · · · < ( 1+r
)n−1 an0 . Since lim( 1+r
)n+1 an0 = 0,
2
2
2
lim an+n0 = 0. Hence lim an = 0
1
Case(ii) r > 1: inconclusive. While the sequence (rn ) diverges if r > 1, the
sequence an = r/n when n is odd and 1/n if n is even has limn→∞ an+1 /an = r
and the sequence (an ) converges to 0.
Case(iii) r = 1: inconclusive.
The sequence (1, 1, · · · , 1) converges, while (1, 12 , 1, 32 , 13 , 23 , 1, 43 , 42 , 14 , 42 , 34 , 1, · · · , )diverges.
(4). Show that
Z
lim (
n→∞
π/6
0
1
sinn xdx)1/n = .
2
Proof. When 0 ≤ x ≤ π6 , we have 0 ≤ sin x ≤
Z
0≤(
π/6
0
1
2
Thus
1 π
sinn xdx)1/n ≤ ( )1/n
2 6
Since lim 21 ( π6 )1/n = 12 lim( π6 )1/n = 12 , therefore, we have
Z
lim (
n→∞
π/6
0
On the other hand, since
1 π
1
sinn xdx)1/n ≤ lim ( )1/n = .
n→∞ 2 6
2
1
lim sin x = ,
x→π/6
2
for 0 < ² < 12 , there exist 0 < δ < π6 such that 12 − ² < sin x < 12 .
R π/6
R π/6
Thus,( 0 sinn xdx)1/n ≥ ( π/6−δ sinn xdx)1/n . Therefore,
Z
limn→∞ (
π/6
0
1 1
1
sinn xdx)1/n ≥ limn→∞ δ n ( − ²) = − ².
2
2
Since ² can be arbitrary small, therefore,
Z
limn→∞ (
π/6
0
1
sinn xdx)1/n ≥ .
2
2
Thus,
Z
lim (
n→∞
π/6
0
1
sinn xdx)1/n = .
2
(6). Show that the set of clutser points of the sequence ((−1)n ) is {−1, 1}.
Proof. Clearly there are subsequences ((−1)2n ) = (1) and ((−1)2n−1 ) = (−1)
that converge to 1 and -1 respectively.
Now we show there is no other cluster point other than −1 and 1. If there
is a subsequence (ank ) that converges to l, l 6= 1 and l 6= −1. Let ² =
min{|1 − l|, |1 + l|}. Clearly ² > 0. But for all N ∈ N, there exists k > N ,
ank = 1 or ank = −1, |ank − l| ≥ min{|1 − l|, |1 + l|} = ², which is a
contradiction.
3.6.9
Let (xn ) and (yn ) be sequences of positive numbers such that lim(xn /yn ) =
+∞
(a) Show that if lim(yn ) = +∞, then lim(yn ) = +∞.
Proof. (a) Note that (xn ) and (yn ) are all positive sequences. Because
lim(xn /yn ) = +∞, there exist N ∈ N, such that for all n > N , xn /yn > 1,
i.e., xn > yn . For any M, as lim(yn ) = +∞, there exist N2 ∈ N, such that
when n > N2 , yn > M . Hence when n > max{N, N2 }, xn > yn > M . Hence
lim(xn ) = +∞.
(b) Show that if (yn ) is bounded, then lim(xn ) = 0.
Note that (xn ) and (yn ) are all positive sequences. As xn is bounded, xn ≤ M
for all n ∈ N. For all ² > 0, because lim(xn /yn ) = +∞, there exists N ∈ N,
such that for all n > N ,
Hence lim(yn ) = 0.
xn
yn
>
M
,
²
which implies |yn − 0| = yn <
3
²xn
M
≤ ².
3.7.1
Let
P
an be a given series and let
P
bn be the series in which the terms are
P
the same and in the same order as in
an except that the terms for which
P
an = 0 have been omitted. Show that
an converges to A if and only if
P
bn converges to A.
Proof. Since the terms for which an = 0 has been omitted, the sequence
P
P
P
of initial sums ( kn=1 bn ) is a subsequence of ( kn=1 an ). Hence if
an
P
converges to A, then
bn converges to A.
P
Conversely, if
bn converges to A. For all ² > 0, there exists N ∈ N,
P
such that when n > N , | nk=1 bk − A| < ². According to the definition of
P
PN2
(bn ), there exists N2 ∈ N, such that | N
k=1 bk | = |
k=1 ak |. When n > N2 ,
Pn
Pn0
P
| k=1 ak −A| = | k=1 bk −A| < ², for an n0 , n0 > N . Hence an converges
to A.
3.7.2
Show that the convergence of a series is not affected by changing a finite
number of its terms.(Of course, the value of the sum may be changed.)
P
Proof. Let (an ) denote the series. Let S = {n : an is changed}. As |S| is
finite, there exists N ∈ N, such that when n > N , an is not changed. For all
P
² > 0, as (an ) converges, there exists N2 ∈ N, such that when m > n > N2 ,
P
Pm
Pm
| m
a
|
<
².
When
m
>
n
>
max{N,
N
},
|
b
|
=
|
n
2
n
n+1
n+1
n+1 an | < ².
P
Hence (bn ) also converges.
3.7.3
By using partial fractions, show that
P
1
(a) ∞
n=0 (n+1)(n+2) = 1
4
1
1
1
= n+1
− n+2
,we have
(n+1)(n+2)
1
1
1
1
− 3 + · · · + n+1 − n+2 = n+1
. Hence
2
n+2
Solution: Since
sn = 1 − 12 +
(b)
P∞
1
n=0 (α+n)(α+n+1)
Solution:
sn =
1
(α+n)(α+n+1)
lim(sn ) = 1
= α1 , if α > 0
=
1
α+n
−
1
.
α+n+1
Hence
1
1
1
1
1
1
1
1
−
+
−
+· · ·+
−
= −
.
α α+1 α+1 α+2
α+n α+n+1
α α+n+1
Thus lim(sn ) =
1
.
α
3.7.4
If
P
xn and
P
yn are convergent, show that
P
xn + yn is convergent.
P
Proof. Let sn denotes the nth partial sum of
xn and tn denotes the nth
P
P
partial sum of
yn . Then sn + tn is the nth partial sum of (xn + yn )
Assume lim(sn ) = s and lim(tn ) = t. By limit theorems, lim(sn + tn ) = s + t,
P
thus
(xn + yn ) is convergent.
3.7.5
P
Can you give an example of a convergent series
xn and a divergent series
P
P
yn such that
(xn + yn ) is convergent? Explain.
P
P
P
Solution: No. If
xn converges, then
(−xn ) converges. By 3.7.4,
yn =
P
xn + yn − xn is convergent, resulting in a contradiction.
P
P
Alternative solution: Assume
xn and
xn + yn are both convergent.
Given ² > 0, by definition, there exist M1 ∈ (N ) such that |xn+1 +...+xm | < 2²
for all m > n ≥ M1 . There exist M2 ∈ (N ) such that |(xn+1 + yn+1 )... +
(xm + ym )| < 2² for all m > n ≥ M2 .
Let M = maxM1 , M2 , then for all m > n ≥ M ,we have
|yn+1 + ... + ym | = |(xn+1 + yn+1 + ... + (xm + ym ) − (xn+1 + ...xm )| ≤
5
|(xn+1 + yn+1 )... + (xm + ym )| + |xn+1 + ... + xm | < 2² + 2² = ². Therefor
P
yn is convergent.
P
P
Thus, we can not find a convergent
xn and a divergent
yn such that
P
(xn + yn ) is convergent.
3.7.6
P
(a) Show that the series ∞
n=1 cos n is divergent.
Since (cos n) does not converge to 0, the series diverges.
P
2
(b) Show that the series ∞
n=1 (cos n)/n is convergent.
P
P
n
1
2
Since | cos
| ≤ n12 and
is convergent, by the Comparison Test, ∞
n=1 |(cos n)/n |
n2
n2
P∞
P
2
is convergent. So n=1 (cos n)/n2 is absolutely convergent, hence ∞
n=1 (cos n)/n
converges.
3.7.8
If
P
an with an > 0 is convergent, then is
P
a2n always convergent? Either
prove it or give a counterexample.
P
P
Solution: Since
an , converges, there exists M > 0 such that
an < M ,
P
2
since an > 0, we have 0 < an < M , 0 < an < M an . Since
M an converges,
P 2
an converges by the comparison test.
P
Alternative solution: Let ² > 0 be given, since
an is convergent, there
√
√
exists M ( ²) ∈ N such that if m > n ≥ M ( ²), then 0 < an+1 + an+2 +
√
· · · + am < ², so we have
a2n+1 + a2n+2 + · · · + a2m ≤ (an+1 + an+2 + · · · + am )2 < ²
P 2
so
an converges.
Question(C)
Show that if we define the positive real numbers as above, then the set of
“real number” satisfies properties 2.1.5.
6
(1) We show that if [(an )] ∈ P , [(bn )] ∈ P , then [(an + bn )] ∈ P , [(an bn )] ∈ P .
Since [(an )] ∈ P , ∃ (a0n ) ∈ [(an )] with a0n > r, for all n for some positive
rational number r; similarly, ∃ (b0n ) ∈ [(bn )] with b0n > r2 for all n for some
positive rational number r2 .
Thus, a0n + b0n > r1 + r2 > 0, a0n b0n > r1 r2 > 0 for all n ∈ N. Therefore,
[(an + bn )] = [(a0n + b0n )] ∈ P , [(an bn )] = [(a0n b0n )] ∈ P .
(2) We show that for any [(an )], exactly one of the following statement is
true :
(i)[(an )] ∈ P (ii) [(an )] = [(0)] (iii) −[(an )] ∈ P .
Case(i): If [(an )] = 0, then for all r > 0 and for all (a0n ) ∈ [(an )], there
exists K ∈ N such that |a0m | < r ∀ m ≥ K, that is, am < r and −am < r for
all m ≥ K. Thus [(an )] ∈
/ P and −[(an )] ∈
/ P.
Case(ii): if [(an )] 6= 0, we show that exactly one of the following is true:
(i) [(an )] ∈ P (ii) −[(an )] ∈ P.
To see this, recall that (an ) 6∼ (0) implies there exists r ∈ Q+ such that for
all K ∈ N, there exists NK ≥ K such that |aNK | ≥ r. Thus, there exists a
subsequence (ank ) of (an ) with |ank | ≥ r for all k. Hence it will have either
a subsequence with all terms ≥ r or a subsequence with all terms ≤ −r.
However, as (an ) is a Cauchy sequence, at most one of them holds. In the
first case where all terms ≥ r > r/2, we have [(an )] ∈ P , and in the second
case, clearly −[(an )] ∈ P .
7