Solutions-Assignment 1

MA541 : Real Analysis
Tutorial and Practice Problems - 1
Hints and Solutions
1. Suppose that S is a nonempty subset of real numbers that is bounded (i.e. bounded above as well
as below). Prove that inf S ≤ sup S. What can you say if inf S = sup S?
Solution: Second Part: inf S = sup S iﬀ S is a singleton.
2. Is the set {x ∈ R | x2 < x} bounded above (below)? If so, ﬁnd its supremum (inﬁmum).
Solution: Hint. {x ∈ R | x2 < x} = (0, 1).
3. Suppose n is a positive integer and a ∈ R, a ≥ 0. Show that there is a unique b ≥ 0 in R such that
bn = a.
Solution: Hint. Consider the set S = {x ∈ R : x ≥ 0, xn ≤ a}. Show that S is nonempty and
bounded above. The supremum b of S satisﬁes bn = a (show that bn < a and bn > a are not
possible). Finally show that that if c ≥ 0 and cn = a, then c = b.
4. Show that between any two distinct real numbers there is a rational number as well as an irrational
number.
Solution: (Refer to any text book.)
5. Prove Well-Ordering Principle of natural numbers: Every nonempty subset of positive integers has
a minimum.
Solution: Let S be a nonempty subset of positive integers. Then S is bounded below by 0, and
so has an inﬁmum, say c.
Claim: c is the least element of S. Since c + 1 is not a lower bound of S, there is m ∈ S such
that m < c + 1. Now, for any positive integer n < m we have n ≤ m − 1 < c and therefore,
n∈
/ S. Consequently, m is the least element of S. (Note that this also means that c = m.)
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6. Suppose that a real number a has the property that
1
n
≥ a for all n ∈ N. Show that a ≤ 0.
Solution: Hint. Follows from Archimedean Principle.
7. Let S and T be nonempty sets of real numbers such that every real number is in S or T and if s ∈ S
and t ∈ T , then s < t. Prove that there is a unique real number β such that every real number less
than β is in S and every real number greater than β is in T . (A decomposition of the reals into two
sets with these properties is a Dedekind cut. This is known as Dedekind’s theorem.)
Solution: Done in the Tutorial class.
8. Suppose that a sequence (xn ) converges to x and xn ≥ 0 for all n. Show that x ≥ 0.
Solution: Hint. Show that if x < 0, then xn < 0 for inﬁnitely many values of n.
9. A subset S of R is closed if every convergent sequence in S has its limit in S. Which of the following
sets are closed: [a, b), [a, b], Q, R \ Q? Justify your answer.
Solution: [a, b] is closed: If xn → x and a ≤ xn ≤ b for each n, then a ≤ x ≤ b.
[a, b) is not closed: If xn = b − b−a
/ [a, b).
n , then xn ∈ [a, b) and xn → b ∈
Q and R \ Q are not closed: Use density property.
10. Let (xn ), (yn ), (zn ) be sequences such that xn ≤ yn ≤ zn for all n ∈ N. If both (xn ) and (zn )
converges to the same limit c, then show that (yn ) converges to c. [This result is known as the
Sandwich Theorem.]
Solution: Refer to any text book.
)
(
xn+1 converges to L. Show that (xn ) (i) converges to 0, if
11. Let (xn ) be such that xn ̸= 0 and xn L < 1, and (ii) diverges, if L > 1. Show that (xn ) can converge or diverge, if L = 1.
Solution: Done in the Tutorial class.
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12. Let a, b ∈ R, and let (xn ) be deﬁned as follows: x1 = a, x2 = b, and xn = 12 (xn−2 + xn−1 ) for n ≥ 3.
Prove that (xn ) is convergent and ﬁnd its limit.
Solution: For n ≥ 3, we have xn = 12 (xn−2 +xn−1 ) and therefore xn −xn−1 = − 12 (xn−1 +xn−2 ),
i.e., |xn − xn−1 | = 12 |xn−1 + xn−2 |. Therefore, by the Result given below, (xn ) is a Cauchy
sequence and so convergent. Moreover,
xn − x1 = (xn − xn−1 ) + (xn−1 − xn−2 ) + · · · + (x2 − x1 )
]
[( )
( )n−3
1
1 n−2
+ −
+ · · · + 1 (b − a)
=
−
2
2
[
( )n−1 ]
2
1
=
(b − a).
1− −
3
2
Taking limits we get lim xn − a = 23 (b − a), i.e., lim xn = a + 32 (b − a)
13. Let (xn ) be a sequence and yn = n1 (x1 + x2 + · · · + xn ). Show that if (xn ) is convergent, then (yn )
is convergent. Is the converse true?
Solution: Suppose xn → a. Let ϵ > 0 be given. There exists N ∈ N such that |xn − a| < ϵ/2
for all n ≥ N . Now, for n ≥ N
1
|yn − a| = (x1 + · · · + xn ) − a
n
[
]
≤ |x1 − a| + |x2 a| + · · · + |xn − a|
=
N
n
1 ∑
1∑
|xi − a| +
|xi − a|
n
n
=
(n − N ) ϵ
K
+
.
n
n
2
i=1
i=N +1
∑
where K = N
i=1 |xi − a|. Choose positive integer M ≥ N such that K/M < ϵ/2. Then for
n ≥ M we have |yn − a| < ϵ.
The converse is not true. Take for example, xn = (−1)n .
14. Prove or disprove: If (xn ) converges to x, and (xn ) has a subsequence (xnk ) such that (−1)k xnk ≥ 0
for all k, then x = 0.
(
)
Solution: Suppose, if possible, x > 0. There exists m ∈ N such that xn ∈ x2 , 3x
2 for all n ≥ m
(Note: we took ϵ = x/2.) In particular, xn > 0 for all n ≥ m. Now, nm , nm+1 ≥ m and therefore
xnm , xnm+1 are positive. This implies that (−1)m xnm ≥ 0, (−1)m+1 xnm+1 ≥ 0 simultaneously
cannot hold, since one of them is negative. Thus, x ≤ 0. Similarly, it can be shown that x < 0
is not possible.
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15. Examine whether the sequences (xn ) converge, where xn is given below. Find also the limits, if
exist. Here, a, b are real numbers.
1
n
(iv)
sin3 n
(
)n
1 + n1
(vii)
nn
(i)
1
(iii)
an
n!
(an + bn ) n , where a, b > 0
(vi)
1
n+1
3·5·7···(2n−1)
2·5·8···(3n−1)
(ix)
1
n2
(ii)
an
(v)
(viii)
Solution: (i) − n1 ≤
1
sin3 n
n
(
+
1
n+2
+ ··· +
1
n+n
)
[a] + [2a] + · · · + [na]
≤ n1 . Now use Sandwich Theorem.
(ii) xn = an . If a = 0 or 1, then clearly (xn ) converges. If a = −1, then (xn ) = ((−1)n ) diverges.
Case: |a| > 1. Let |a| = 1 + h. Then
n(n − 1) 2
h + · · · + hn > nh.
2
Thus, (xn ) is not bounded, and so not convergent.
1
Case: 0 < a < 1. Then 1/a > 1. Put 1/a = 1 + h, i.e., a = 1+h
. Thus,
|an | = (1 + h)n = 1 + nh +
0 < x n = an =
1
1
1
=
≤
.
n
n
(1 + h)
1 + nh + · · · + h
1 + nh
Use Sandwich Theorem to conclude that xn → 0. Case: −1 < a < 0. Then |an | → 0 and
therefore an → 0.
|a|
(iii) xxn+1
= n+1 → 0 < 1. Thus xn → 0.
n
(
)n
(iv) Show that 2 < xn = 1 + n1 < 3 and (xn ) is monotonically increasing. Thus the sequence
converges. (lim xn is denoted by e.)
1
1
1
(v) Assume a ≤ b. Then b ≤ xn = (an + bn ) n ≤ (bn + bn ) n = 2 n b. Use Sandwich Theorem to
conclude xn → b. Thus, xn → max{a, b}.
1
1
n
1
+ n+2
+ · · · + n+n
≤ n+1
< 1, the sequence (xn ) is bounded. Moreover,
(vi) Since 0 < xn = n+1
1
1
1
xn+1 − xn = 2n+1 + 2n+2 − n+1 > 0, i.e., (xn ) is monotonically increasing. Hence (xn ) converges.
In fact,
[
]
∫ 1
1
1
1
1
dx
xn =
+
+ ··· +
→
= ln 2.
n 1 + 1/n 1 + 2/n
1 + n/n
0 1+x
1
1
(vii) xn = n n . We have xn > 1 for n ≥ 2. Put hn = xn − 1. Then n n = 1 + hn , i.e.,
n = (1 + hn )n = 1 + nhn +
n(n − 1) 2
n(n − 1) 2
hn + · · · >
hn .
2
2
√
2
2
i.e., 0 ≤ hn ≤ √n−1
. Using Sandwich Theorem, we
Therefore, for n ≥ 2 we have 0 ≤ h2n ≤ n−1
get hn → 0, i.e. xn → 1.
2n+1
(viii) Note that xxn+1
= 3n+1 → 2/3 < 1. Hence xn → 0.
n
(
)
(ix) xn = n12 [a] + [2a] + · · · + [na] . Note that for any real b, b − 1 < [b] ≤ b. Thus,
)
)
1(
1(
(a
−
1)
+
(2a
−
1)
+
·
·
·
+
(na
−
1)
<
x
≤
a
+
2a
+
·
·
·
+
na
.
n
n2
n2
Simplify this to get
1 a n+1
a n+1
− + ·
< xn ≤ ·
.
n 2
n
2
n
Use Sandwich Theorem and get xn → a2 .
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Result. Suppose (xn ) be a real sequence such that
|xn+1 − xn | ≤ c|xn − xn−1 | for all n.
(0.1)
If c < 1, then (xn ) is a Cauchy sequence and therefore convergent.
Proof. If x2 = x1 , then xn = x1 for all n, and the result follows.
Suppose x2 ̸= x1 . We have for all n
|xn+1 − xn | ≤ c|xn − xn−1 | ≤ · · · ≤ cn−1 |x2 − x1 |.
Moreover,
|xn+k − xn | ≤ |xn+k − xn+k−1 | + · · · + |xn−1 − xn |
≤ (ck−1 + cn−2 + · · · + c + 1)|xn−1 − xn |
1 − ck
=
|xn1 − xn |
1−c
1
≤
cn−1 |x2 − x1 |.
1−c
c(1 − c)
ϵ for all
|x2 − x1 |
n ≥ N . Then for any n ≥ N, k ≥ 1 we have |xn+k − xn | < ϵ. Thus (xn ) is a Cauchy sequence.
Let ϵ > 0 be given. Since 0 ≤ c < 1, we have cn → 0. Choose N ∈ N such that cn <
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