1. Department of Statistics University of Rajshahi, Bangladesh
Lecture on
Sequence
Presented by
Dr. Md. Golam Hossain
Professor
Department of Statistics
University of Rajshahi, Bangladesh

2. Today’s learning objectives
Definition of sequence
Limit of the sequence
Convergent/divergent sequence
Bounded sequence
Least upper bound and greatest lower bound of sequence
Monotone sequence
Subsequence of a sequence
Nested interval and its property
Cauchy sequence and its property

3. What is sequence?
A sequence is a function whose domain is the set of natural numbers N and range is the subset of R.
Let f : N → S R be a function then f(1), f(2), f(3), is called a sequence of elements in S. Put an = f(n) for n ∈ N. Then a1, a2, a3, is a sequence of elements/terms in S. We usually denote this sequence by {an}.
Example: (1) {1,2,3,…n,…..}
(2) {1, 1/2, 1/3, 1/4,……..}

4. An infinite sequence of real numbers (in blue)
An infinite sequence of real numbers (in blue). This sequence is neither increasing, nor decreasing, nor convergent, nor Cauchy. It is, however, bounded.

5. Arithmetic Sequence Geometric Sequence
An Arithmetic Sequence is made by adding some value each time.
Example:
1, 4, 7, 10, 13, 16, 19, 22, 25, ...
This sequence has a difference of 3 between each number. The pattern is continued by adding 3 to the last number each time.
The nth term, xn = 3n-2
Geometric Sequence
A Geometric Sequence is made by multiplying by some value each time.
Example:
2, 4, 8, 16, 32, 64, 128, 256, ...
This sequence has a factor of 2 between each number. The pattern is continued by multiplying by 2 each time.
The nth term, xn = 2n

6. Special Sequences Triangular Numbers Sequence
This Triangular Number Sequence is generated from a pattern of dots which form a triangle. By adding another row of dots and counting all the dots we can find the next number of the sequence:
The nth term, xn = n(n+1)/2

7. Square Numbers Sequence Fibonacci Numbers Sequence
1, 4, 9, 16, 25, 36, 49, 64, 81, ...
The next number is made by squaring where it is in the pattern. The second number is 2 squared (22 or 2×2) . The seventh number is 7 squared (72 or 7×7) etc. The nth term, xn = n2
Cube Numbers Sequence
1, 8, 27, 64, 125, 216, 343, 512, 729, ...
The next number is made by cubing where it is in the pattern. The second number is 2 cubed (23 or 2×2×2). The seventh number is 7 cubed (73 or 7×7×7) etc. The nth term, xn = n3
Fibonacci Numbers Sequence
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
The Fibonacci Sequence is found by adding the two numbers before it together. The 2 is found by adding the two numbers before it (1+1). The 21 is found by adding the two numbers before it (8+13). The next number in the sequence above would be 55 (21+34). The nth term, xn = xn-1 + xn-2

8. Finite and Infinite sequences
Constant Sequence
A sequence {an} is said to be constant if an= aϵR, where a is a constant.
Example: {1, 1, 1, …} is a constant sequence.
Finite and Infinite sequences
The sequence is called finite or infinite according as there are or are not a finite number of terms.
Example: (i) The set of number {2, 7, 12, 17, ….,32} is a finite sequence and the nth term is given by an= f(n) =2+5(n-1) = 5n-3; n = 1, 2, …,7.
(ii) The set of numbers 1, 1/3, 1/5, 1/7, … is an infinite sequence with nth term, an = 1/2n-1; n= 1,2, 3, ….

9. Subsequence
A sequence {bk} is called a subsequence of the sequence {an} if there are natural numbers n1<n2<n3<… such that bk= ank for k= 1, 2, 3, …….
Example: (i) The sequence of prime numbers {2, 3, 5, 7, 11,…..} is a subsequence of the sequence of natural numbers {1, 2, 3, 4, ……} which is strictly increasing sequence of positive integers.
(ii) The sequence {2, 3, 7, 9, 5, 4,…} is not a subsequence of the sequence of natural number {1, 2, 3, 4, ……} which is not a strictly increasing sequence.

10. Definition: Let {an} be a sequence of real numbers
Definition: Let {an} be a sequence of real numbers. Let E be the set of numbers x (in the extended real number system) such that ank → x for some subsequence {ank} of {an}. This set E contains all subsequential limits, plus possibly the numbers +∞, −∞.
Put a∗ = supE, a∗ = inf E.
The numbers a∗ and a∗ are respectively called upper and lower limits of {an}; we use the notations a∗ = lim n→∞ sup
an and a∗ = lim n→∞ inf an.
Examples:
(a) Let {an} be a sequence containing all rationals. Then every real number is a subsequential limit, and lim n→∞ supan = +∞ and lim n→∞ inf an = −∞.
(b) Let an = (−1)n/[1 + (1/n)]. Then lim n→∞ sup an = 1 and lim n→∞ inf an = −1.

11. Limit of a Sequence
The limit of the sequence {an} is A, and is written as lim n→∞an = A if for any positive number (small) ɛ>0, there exists a natural number N, depending on ɛ such that |an – A| < ɛ n>N
i.e., A- ɛ< an <A+ ɛ n>N.
Example: If a sequence {4, 7/2, 10/3,.....} the nth term of this sequence, an = 3 + 1/n,
the limit of this sequence, lim n→∞an = 3.

12. Bounded and Unbounded Sequences
A sequence is said to be bounded if there exists a positive number M such that |an|≤M, n.
A sequence which is not bounded is called unbounded sequence.
Example: (i) {1/n} is a bounded sequence,
because 0 ≤an ≤1.
(ii) {3n-1} is an unbounded sequence,
because an 1, nϵN, and there is no
real number u such that an ≤u.

13. Bounded Above
A sequence {an} is said to be bounded above if there exists a real number M such that an≤M,  n = 1, 2, 3, …
Example: {-1, -2, -3, …} is a bdd above sequence as
an ≤(-1),  n N, and -1 is the upper bound.
Bounded Below
A sequence {an} is said to be bounded below if there exists a real number l such that an≥l  n = 1, 2, 3,….
Example: {1, 2, 3, 4, …, n, …} is a bdd below sequence an  1,  n N, and 1 is the lower bound.

14. Least Upper Bound and Greatest Lower Bound
A number M is called the least upper bound (l. u. b.) of the sequence {an} if an ≤M, n = 1, 2, 3, ……, while at least one term M-ɛ <M, where given ɛ>0.
A number m is called the greatest lower bound (g. l. b.) of the sequence {an} if an ≥m, n = 1, 2, 3, …,while at least one term m + ɛ>m, where given ɛ>0.

15. Convergent sequence Divergent Sequence
A sequence {an} is said to be convergent if limn→an exists. For example, an ={1, 1/2, 1/3, 1/4, …} = {1/n} is a convergent sequence, since limn→an = limn→ {1/n}=0.
Divergent Sequence
A sequence {an} is called divergent if it is not convergent, i.e., if limn→an does not exist.
Example: {an}= {3n} is a divergent sequences and diverges to +.
A sequence {an} is said to be properly divergent if limn→an = + or -. For example, {n} and {-nn} are properly divergent sequences.

16. Oscillate finitely sequences
A sequence {an} is said to be oscillatory if it is neither a convergent sequence nor a divergent one.
There are two types of oscillatory sequences:
A sequence {an} is said to be oscillate finitely if it is bounded and is not convergent.
Example: {(-1)n}, {1 + (-1)n} are oscillate finitely sequences.
A Sequence {an} is said to be oscillate infinitely if it is unbounded and properly divergent.
Example: {(-1)n n} is an oscillate infinitely sequence.

17. Null Sequence
If the sequence {an} is convergent to zero, i.e., limn→an =0, then the sequence {an} is called a null sequence.
Example: lim→an = limn→1/n = 0, then {1/n} is null sequence.

18. Theorem 1: If limn→∞an =l exists then it is unique
Proof. Let {an} be a convergent sequence. If possible let
lim n→∞an =l1 and limn→∞an =l2. Let us take l1 ≠l2. Now l1 ≠l2
 there is a p>0 such that | l1 -l2|>p (1)
Now let ɛ>p/2>0 there is a N1 such that | an –l1|<p/2,  n> N1 and also there is a N2 such that | an –l1|<p/2,  n> N2.
Let N = max {N1, N2}. Now if n>N then | an –l1|<p/2 and
| an –l2|<p/ (2)
Now | l1 –l2|= | l1-an+ an-l2| ≤ | l1- an |+ | an-l2 | = | an-l1 |+ | an-l2 | <p/2 +p/2=p using (2).
Therefore, | l1-l2 |<p, which contradicts (1).
Hence l1=l2. Thus, every convergent sequence has a unique limit.

19. Theorem 2: If limn→∞an = A and limn→∞bn = B then
(a) Limn→∞(an+ bn) = A+B
(b) Lim n→∞(an- bn) = A-B
(c) Lim n→∞kan = kA for any real number k
(d) Lim n→∞anbn = AB
(e) Lim n→∞ (1/ bn) = 1/B, bn ≠0, B ≠0
(f) Lim n→∞ (an/ bn) = A/B, bn ≠0, B ≠0
Proof. (a) Given any ɛ>0 there is a natural number N1 such that |an-A|< ɛ/2, n>N1, and there is a natural number N2 such that |bn-B|< ɛ/2, n>N2. Now, if n> max (N1, N2) then
|(an+ bn) – (A+B)| = |(an-A+ bn– B)| ≤|(an-A|+|bn– B)| < ɛ/2 + ɛ/2 = ɛ
lim n→∞(an+ bn) = A +B
(b) Given any ɛ>0 there is a natural number N1 such that |an-A|< ɛ/2, n>N1, and there is a natural number N2 such that |bn-B|< ɛ/2, n>N2. Now, if n> max (N1, N2) then
|(an- bn) – (A+B)| = |(an-A+ B- bn)|≤|an-A|+| B- bn| ≤|an-A|+| bn- B| < ɛ/2 + ɛ/2 = ɛ
lim n→∞(an-bn) = A –B

20. (c) If k=0, the result is obvious since |kan-kA|=0 for every natural number n. Suppose k≠0 and let ɛ>0 be given. Then there is a natural number N such that |an-A|< ɛ/|k|, n>N
Hence, if n>N, |kan-kA|=|k(an-A)|= |k|.|(an-A)|< |k| ɛ/|k|= ɛ
lim n→∞kan = kA for any real no. k.
(d) Let the sequence {an} is bounded and |an|≤M for every natural number n, where M is some positive real number. Now let ɛ>0 be given. There is a natural number N1 such that |an-A|< ɛ/2(1+|B|), n>N1 and there is a natural number N2 such that |bn-B|< ɛ/2M n>N2. .
Now, if n> max (N1, N2) then we have
|anbn-A|=|anbn-anB+ anB-AB|≤|anbn-anB|+ |anB-AB|=|an| |bn-B| + |B| |an-A|< M ɛ/M+ |B| ɛ/2(1+|B|)< ɛ
 lim n→∞anbn =AB

21. (e) We are assuming here that B and each bn is on zero
(e) We are assuming here that B and each bn is on zero. Let ɛ> 0 be given. There is a natural number N1 such that |bn-B|<|B|/2,  n>N1. Hence if n>N1 then |B|/2> |bn- B|= |B-bn|≥|B|-|bn|
|bn|>|B|/2. Also, there is a natural number N2 such that
|bn- B|< ɛ/2 |B|2, n >N2
Therefore, if n>max(N1, N2), we have
|1/ bn-1/B|=|(B-bn)/bnB| = |bn- B|/|bn|| B|<|B|2 ɛ/2 / | B|/2. | B|= ɛ
lim n→∞(1/bn) =1/B, bn≠0, B ≠0
(f) We are assuming here that B and each bn is nonzero.
ɛ> 0 be given. There is a natural number N1 such that |an-A|< |bn|/2ɛ, n>N1, and there is a natural number N2 such that |bn-B|< |bn|. |B|/|A|. ɛ/2, n>N2.
Now, if n> max (N1, N2) then we have
|an/ bn– A/B|= |(anB- bnA)/bnB|= |(anB- AB+AB- bnA)/bnB| ≤|(an -A||B|/ |bn||B| + |A||B-bn|/ |bn||B|= |(an -A|/ |bn| + |A|| bn-B|/|bn|| B|< | bn| ɛ/2 /|bn| + |A|/| bn-B| .|bn|| B|/| A| . ɛ/2 = ɛ
lim n→∞(an/bn) =A/B, bn≠0, B ≠0.

22. Theorem 3: Let {an} be a bounded sequence, and let
limn→∞supan = L and limn→∞infan = l. Then every ε > 0 corresponds a positive integer N such that for n ≥ N
l − ε < an < L + ε.
Proof. Let E be set of all subsequential limits of {an}. Thus, l and L are members of E. Let ε> 0 be given. If there were infinitely many n for which an ≥ L + ε, then there would have a subsequence {ank} of {an} such that ank → x and x ≥ L + ε.
But then x ∈ E and x ≥ L + ε. This contradicts the fact that L = supE. Hence there is a positive integer N1 such that n ≥ N1 implies that an < L + ε (1)
Similarly, we can find a positive integer N2 such that n ≥ N2 implies that an > l − ε (2)
Put N = max{N1,N2}. Then from (1) and (2) we have for n ≥ N
l − ε< an < L + ε

23. Theorem 4: If an ≤ bn for n ≥ N, where N fixed, then
lim n→∞ supan ≤ lim n→∞ supbn,
lim n→∞ infan ≤ lim n→∞ infbn.
Theorem 5: If an → a and bn → b, and if an > bn for n ≥ N, where N is a fixed positive integer, then a ≥ b.
Proof. Suppose a < b then b − a > 0. Put ε = b − a. Then since
an − bn → a − b, there exists a positive integer N′ such that
|(an − bn) − (a − b)| < ε whwnever n ≥ N′.
Hence we have for n ≥ N′
an − bn < ε + (a − b) = 0.
This contradicts the hypothesis of the theorem. Hence the theorem is proved.

24. Theorem 6: If {an} is a bounded sequence and bn → 0 then anbn → 0.
Proof. Since {an} is bounded, there exists a positive number K such that |an| ≤ K for all n ∈ N (1)
Now let ε > 0 be given. Since bn → 0, there exists a positive integer N such that n ≥ N implies that |bn| < ε/K (2)
From (1) and (2) we have whenever n ≥ N
|anbn| = |an||bn| < K ε /K= ε.
Hence anbn → 0.

25. Theorem 7: If an+1 > Kan when n ≥ N, where N is a fixed positive integer, and an > 0, K a constant > 1 then an→ ∞.
Proof. For any positive integer p, we have
aN+1/aN < K
aN+2/aN+1< K
aN+3/aN+2< K
. . .
aN+p/aN+p−1< K.
Multiplying the above inequalities, we have for p ≥ 1
aN+p < Kp aN (1)
If we put N + p = n in (1) we have
an < Kn aN/KN when n > N (2)
Since K > 1, Kn → ∞ by theorem 12. Hence from (2) we have
an → ∞.

26. Theorem 9: Let an > 0 and an+1/ an → l.
Theorem 8: If an+1 < Kan when n ≥ N, where N is a fixed positive integer, and an > 0, K a constant such that
0 < K < 1, then an → 0.
Proof. The proof is similar to that of theorem 7.
Theorem 9: Let an > 0 and an+1/ an → l.
(i) If l < 1 then an → 0.
(ii) If l > 1 then an → ∞.
Proof. (i) Let l < 1. Choose a number K such that l < K < 1 and put  = K − l. Then  > 0. Since an+1/an → l, there exists a positive integer N such that n ≥ N implies that
|an+1/an − l| < 
Hence an+1/an< l +  = K whenever n ≥ N.
Hence by Theorem 8
an → 0.

27. (ii) Let l > 1. Choose a number K such that l > K > 1 and put  = l − K. Then  > 0. Since an+1/an → l, there exists a positive integer N such that n ≥ N implies that
|an+1/an − l| < .
Hence an+1/an> l −  = K whenever n ≥ N.
Hence by Theorem 7
an → ∞.

28. Theorem 10: (Squeeze Theorem): If an ≤ bn≤ cn n ϵN and
limn→∞an = limn→∞cn = A then limn→∞bn =A.
Proof: Given that an ≤ bn≤ cn n
 an-A ≤ bn-A≤ cn-A n. Now given that limn→∞an = lim n→∞cn=A. There exists a N1 such that for given ɛ>0 and |an-A|< ɛ, n>N1 and there is a N2 such that |cn-A|< ɛ, n>N2.
Let N = max {N1, N2}. Now if n>N then |an-A|< ɛ, and |cn-A|< ɛ. Here bn-A≤ cn-A  (bn-A) ≤ |cn-A|< ɛ n>N …….(1)
Again an-A ≤ bn-A
 -(bn-A)≤ -( an-A) ≤ |an-A| |< ɛ n>N….(2).
Combining (1) and (2) we have |bn-A|< ɛ n>N
 lim n→∞bn =A.

29. Theorem 11: Every convergent sequence is bounded
Proof. Let {an} be a convergent sequence and limn→∞an = A. Now for given any ɛ>0 there is a natural number N such that |an-A| |< ɛ n>N
 |an| -|A| ≤ |an-A|< ɛ
 |an| ≤ |A| + ɛ for every n = N+1, N+2, …….
Let M = max {|a1| , |a2|, …, |aN|, |A| + ɛ} then |an| ≤ M for every natural number n. Therefore, {an} is bounded.

30. Theorem 12: Let a be a real number.
(i) If −1 < a < 1 then an → 0.
(ii) If a = 1 then an → 1.
(iii) If a = −1 then {an} oscillates between -1 and 1.
(iv) If a > 1 then an → +∞.
(v) If a < −1 then {an} oscillates infinitely.
Proof. (i) If 0 < a < 1, put p = 1 − a. Then 0 < p < 1 and hence
(1 − p)(1 + p) =1 − p2<1
=⇒ (1 − p) <1/1+p
=⇒ (1 − p)n <(1/1+p)n
=⇒ an < 1/(1+np+nC2p2+···+pn)
=⇒ an <1/np.
Thus given ε > 0 if we choose a positive integer N such that N > 1/pε, then |an| < ε whenever n ≥ N. Hence an → 0.
If −1 < a < 0, put b=−a. Then 0 < b < 1 and an = (−1)nbn. Hence bn → 0. Since {(−1)n} is bounded, by Theorem 6, an = (−1)nbn → 0.
If a = 0 then trivially an → 0. Thus we have proved that
an → 0 whenever − 1 < a < 1.

31. (ii) If a = 1 then an = 1 for all positive integer n. Hence an → 1.
(iii) If a > 1, put p = a − 1. Then p > 0 and a = 1 + p.
Hence an = (1 + p)n = 1 + np +nC2p2 + ・ ・ ・ + pn
=⇒ an > np > K for any positive number K
when n is sufficiently large. Hence an → +∞.
(iv) If a = −1, then an = 1 or −1 according as n is even or odd. Hence the sequence {an} oscillates between −1 and 1.
(v) If a < −1, put b = −a. Then b > 1 and a = −b. Hence bn → ∞. Since an = (−1)nbn, {an} oscillates between −∞ and +∞.

32. Monotone Sequence
A sequence is said to be monotone if it is either monotone decreasing or monotone increasing.
Example: (1) {1, 1/2, 1/3, …..}, {1, 2, 3, 4,……} etc. are monotone sequences.
The sequence {an} is called monotone increasing if an ≤ an+1 for every natural number n. For example {1, 2, 3,…..}, {1, 2, 2, 3, 3, 4, …..} etc. are monotone increasing sequences.
The sequence {an} is called monotone decreasing if an ≥ an+1 for every natural number n. For example {1, 1/2, 1/3,…..}, {1, 1/2, 1/2, 1/3, 1/3, 1/4, …..} etc. are monotone decreasing sequences.

33. Theorem 13: If {an} is monotone increasing and bounded above then {an} converges and lim n→∞(an)= L=sup nϵN an
Proof. By the least upper bound property of real numbers the set {an │nϵN} has a supremum, say L. Given any ɛ>0, there is a natural number N such that aN>L- ɛ. Also, since {an} is monotone increasing we have an>aN  n >N. Then
an≥ aN>L- ɛ n >N
an >L- ɛ n>N (1)
Also an≤L<L+ ɛ, n (2)
Combining (1) and (2) we get
L- ɛ <an<L+ ɛ, n>N
│an-L│< ɛ, n>N
Therefore, {an} converges and lim n→∞(an)= L=sup nϵN an

34. Theorem 14: If {an} is monotone decreasing and bounded below then {an} converges and lim n→∞(an)= l=infnϵN an
Proof. By the greatest lower bound property of real numbers the set {an│nϵN} has a infemum, say l. Given any ɛ>0, there is a natural number N such that aN<l+ ɛ. Also, since {an} is monotone decreasing, then an≤aN,
n >N. Then an≤ aN<l+ ɛ n >N
an <l+ ɛ, n>N (1)
Also an≥l>l- ɛ,n (2)
Combining (1) and (2) we get,
l- ɛ <an<l+ ɛ, n>N
│an-l│< ɛ n>N
Therefore, {an} converges and lim n→∞(an)= l=inf nϵN an

35. Theorem 15: A necessary and sufficient condition for a monotone sequence to be convergent is that it is bounded.
Proof. The condition is necessary: Every convergent sequence is bounded.
Let {an} be a convergent sequence and lim n→∞{an} = l. Now for given any ɛ>0 there is a natural number N such that │an-l│< ɛ, n >N
│an│-│l│≤│an-l│< ɛ
│an│≤ ɛ +│l│, n = N+1, N+2,……..
Let M = max{ │a1│,│a2│,…,│aN│,│l│+ ɛ }
Then │an│≤M for every natural number n ,
{an} is bounded.

36. The condition is sufficient: Bound monotone sequence is convergent.
Suppose that {an} be a bounded monotone increasing sequence. Let l be the least upper bound of the sequence, i.e., the least upper bound of the range of the sequence. We shall show that lim n→∞(an) = l.
Let ɛ>0 be given. Since l is the least upper bound of the sequence, there exists N such that aN> l- ɛ.
Also, {an} is monotone increasing
 an≥aN, n>N
Thus an ≥aN > l- ɛ, n>N (1)
Also, an≤l<l+ ɛ, n (2)
From (1) and (2) we have
l- ɛ< an<l+ ɛ n>N,
 |an-l|< ɛ n>N
Therefore lim n→∞(an) = l.
Thus the sequence {an} is convergent. Similarly when {an} is a bounded monotone decreasing sequence we also shown that {an} converges.

37. Theorem 16: If a sequence {an} converges to A, then every subsequence of {an} is also converge to A and the converges is also true.
Proof. Let the sequence {an} converges to A. Let us suppose that at least one of the subsequence {bn} = {ank} converges to B≠A.
Let B-A = t =2ɛ
Since lim n→∞(an) = A
A- ɛ<an< A+ɛ, n>N (1)
Also since lim n→∞(bk) = B
Thus, B- ɛ<bk< B+ɛ, k>K
A+ɛ<bk< B+ɛ, k>K (2)
If (2) holds, infinite bk, i.e., infinite ank lies outside of [A- ɛ, A+ɛ] which contradicts (1).
Thus, our hypothesis lim n→∞(bk) = B≠A is not true.
Therefore {bk}→A.

38. 2nd part:
Let us suppose that all subsequences of {an} converge to A (3)
Now let us take lim n→∞an = L ≠A.
Since we have shown above that if a sequence converges to a point then all its subsequences converge to that point. And since we have assumed that lim n→∞an = L ≠A then all subsequences of {an} will converge to L ≠A which contradicts (3). Hence our hypothesis lim n→∞an = L ≠A is false,
i.e., lim n→∞an = A.

39. Theorem 17: If {an} has one subsequence converging to A and a second subsequence converging to B and A ≠B then {an} diverges
Proof. Let us suppose that one subsequence {bk} of {an} is converge to A and a second subsequence {ck} of {an} converges to B, B≠A. Let B-A =t= 2ɛ.
Since limn→∞bk = A then
ɛ<bk< A+ɛ, k>K (1)
Also, since lim n→∞ck = B then
B-ɛ<ck< B+ɛ, k>K
 A+ɛ<ck< B+ɛ, k>K (2)
If (2) holds, infinite bk lies outside of [A- ɛ, A+ɛ] which contradicts (1). Also, we know that if lim n→∞ak = l exists then it is unique. But {bk} →A and {ck} →B and A ≠B then we may conclude that if one subsequence converging to A and another subsequence converging to B and A ≠B then {an} diverges.

40. Theorem 18: (a) If p > 0, then 1/np → 0.
(b) If p > 0, then (p)1/n → 1.
(c) (n)1/n → 1.
(d) If p > 0 and α is real, then nα /(1 + p)n → 0.
Proof. (a) Let  > 0 be given. Since (1/ )1/p is a real number, there exists a positive integer N such that N > (1/ )1/p. Hence if n ≥ N, then n > (1/ )1/p and this implies
1/np < .
Hence 1/np → 0.
(b) If p > 1, put xn = (p)1/n − 1. Then xn > 0, and, by the binomial theorem,
1 + nxn ≤ (1 + xn)n = p,
so that 0 < xn <p − 1/n
Since (p − 1)/n → 0, xn → 0. Hence (p)1/n → 1.
If p = 1, (b) is trivial, and if 0 < p < 1, put q = 1/p. Then q > 1 and
hence (q)1/n → 1. Hence by previous theorem
(p)1/n =1/(q)1/n → 1.

41. (c) Put xn = (n)1/n − 1. Then xn ≥ 0, and, by the binomial theorem,
n = (1 + xn)n ≥n(n + 1)/2 xn2
Hence 0 ≤ xn ≤ (2/n − 1)1/ (n ≥ 2).
Since (2/n − 1)1/2 → 0, xn → 0 and (c) follows.
(d) Let k be an integer such that k > α, k > 0. For n > 2k,
(1 + p)n > nCkpk = {n(n + 1) ・ ・ ・ (n − k + 1)}/ k!.pk >nkpk/2kk!
Hence
0 <nα/(1 + p)n <2kk!/pk .nα−k (n > 2k).
Since α − k < 0, nα−k → 0, by (a).
nα /(1 + p)n → 0.

42. Nested Interval Property
Let us consider a set of intervals [an, bn], n = 1, 2, 3, …, where each interval is contained preceding one and lim n→∞(bn-an) = 0. Such intervals are called nested intervals.
Nested Interval Property
Suppose I1 = [a1, b1], I2 = [a2, b2], I3 = [a3, b3],…,where I1 I2  I3 ,… and lim n→∞(bn-an) = 0. Then there is exactly one real number common to all the intervals i.e., In= a real number.
Proof. Given that I1 I2  I3 ,…, where [bn-an]. here an+1an and bn+1≤bn for n= 1, 2, 3, ….. Then {an} is a monotone increasing sequence and bounded above by b1, i.e., lim n→∞an = supnϵN an =x, say.
Now a1 ≤an ≤ bn≤ b1
bn  an  limn→∞(bn-an)  0. But given that limn→∞(bn-an) =0
 limn→∞bn = limn→∞an
 x=y
 an ≤x ≤ bn n and so In= x, a real number.

43. Theorem 19: (Bolzano-Weierstrass) Every bounded sequence has a convergent subsequence.
Proof. Let {an} be a bounded sequence. Then there is a M>0 such that |an|≤M  nϵ.
Hence an ϵ [-M, M]  nϵN.
Let us consider the intervals [-M, 0] and [0, M], then at least one of these two intervals must contain an for infinitely many natural numbers.
Let us define such interval I0. Next dividing I0 into two sub-intervals must contain an for infinitely many natural numbers. Let us say such an interval I1. Continuing in this manner, we obtain a sequence of intervals I0, I1, I2,….. with I0 I1 I2…..In....
Then the length of In is M/2n →0 as n→. By the nested interval property, there is exactly one point common to all these intervals, say A.
If we choose an1ϵI1, an2 ϵI2 with n2>n1, an3ϵI3 with n3>n2, an4ϵI4 with n4>n3 and so on. Then { an1, an2 an3, an4,……} is a subsequence of {an} and ank and A are both contained in Ik. Thus |ank - A|< M/2k, thus
limk→ ank = A.

44. Cauchy Sequence
A sequence {an} is called a Cauchy sequence if given any ϵ>0 there is a natural number N such that
|am – an |< ϵ  m, n >N |an+p – an |< ϵ,  m, n >N, p>0.
For example, {1/n} is a Cauchy sequence.
The plot of a Cauchy sequence (xn), shown in blue, as n versus xn. If the space containing the sequence is complete, the "ultimate destination" of this sequence, that is, the limit, exists.

45. Theorem 20: Every Cauchy sequence is bounded.
Proof. Let {an} be a Cauchy sequence of real numbers. Then for given >0 there is a natural number N such that
|am – an |< ϵ,  m, n >N.
Now if we choose n0>N, then |an |= |an – an0 +an0 |≤ |an – an0| +|an0 | |an |< ϵ + |an0 |  n >N.
Let M = max{ │a1│,│a2│,…,│aN│, ɛ +│an0│}, then
|an |≤M  n >N
 {an} is bounded.

46. Theorem 21: The necessary and sufficient condition for the convergence of real sequence is that the sequence must be Cauchy.
Proof. Necessary condition: Every convergent sequence is Cauchy. Let {an} be a convergent sequence and limn→ an = A. Then for given any ϵ>0 there is a natural number N such that
|an – A|< ϵ/2,  n >N.
For m, n >N we have
|am– an | = |(am –A) –(an –A)|≤ |am –A |+|an –A|< ϵ/2 + ϵ/2 = ϵ
 |am – an | < ϵ,  m, n>N .
 {an} is a Cauchy sequence.

47. Sufficient Condition: Every Cauchy sequence is convergent
Sufficient Condition: Every Cauchy sequence is convergent. Let {an} be a Cauchy sequence. Since every Cauchy sequence is bounded then {an} is also bounded. By Bolzano-Weierstrass theorem {an} has a convergent subsequence, say {ank}.
Let limn→ ank =a.
Now |an – a | = |an – ank + ank -a |≤|an – ank |+ |ank – a | (1)
Let ϵ>0 be given, since {an} is Cauchy there is a N1 such that
|an –ank|< ϵ/2,  n, nk >N (2)
and also there is a N2 such that
|ank –a|< ϵ/2  nk >N (3)
Let N = max{N1, N2}. Then for n, nk>N, (2) and (3) both are true. Then from (1) we have
|an – a |< ϵ/2+ ϵ/2 = ϵ,  n>N
|an | is convergent.

48. Theorem 22: A sequence of complex numbers is convergent if and only if it is a Cauchy sequence.
Proof. Let {an} be a sequence of complex numbers such that
Lim n→∞ an = a, and let ε > 0 be given. Then there exists a positive integer N such that n ≥ N implies that |an − a| < ε/2 . Thus if m ≥ N and n ≥ N, then |am − an| = |(am − a) + (a − an)| ≤ |am − a| + |an − a| ≤ε/2+ε/2= ε.
Hence {an} is a Cauchy sequence.
Conversely, let {an} be a Cauchy sequence. Then for ε = 1 there exists a positive integer N such that m ≥ N and n ≥ N implies that |am − an| < 1.
Now since the set {|a1|, |a2|, ・ ・ ・ , |aN|} is a finite set of real numbers, hence it has a maximum.
Let M = max{|a1|, |a2|, ・ ・ ・ , |aN|}.
Then if 1 ≤ n ≤ N
|an| ≤ M, (1)

49. and if n ≥ N
|an| = |(an − aN) + aN|≤ |an − aN| + |aN| < 1 +M (2)
Hence from (1) and (2), we have |an| < 1 + M for all n ∈ N. Thus {an} is bounded. Hence {an} has a convergent subsequence. Let {ani} be a convergent subsequence of {an} and ani → a. Let ε > 0 be given. Then there exists a positive integer N such that |am − an| <ε/2 when m ≥ N and n ≥ N, and there exists a positive integer N′ such that |ani − a| <ε/2
when ni ≥ N′. Let N′′ = max{N, N′}. Then m ≥ N′′, n ≥ N′′ and ni ≥ N′′ imply that |am − an| <ε/2 and |ani − a| <ε/2
Hence if n ≥ N′′ and ni ≥ N′′
|an − a| = |(an − ani) + (ani − a)|≤ |an − ani | + |ani − a| < ε/2 + ε/2 = ε
Thus an → a.

50. Theorem 23:
(Cauchy I): If an → a, then
(a1 + a2 + ・ ・ ・ + an )/n → a
(b) (Cauchy II): If an > 0 and an → a, then
(a1 , a2 , ・ ・ ・ an )1/n→ a
(c) (Cauchy III): If an > 0 and an+1/an → a, then
(an )1/n→ a.
Proof. (a) Let bn = an − a. Then bn → 0. Let ε > 0 be given, then there exists a positive integer N such that |bn| <ε/2
whenever n ≥ N.
Now, if m > N, then
|(b1 + b2 + …+bm )/m| ≤ |b1 /m| + |b2/m| +…. +|bN /m|+|bN+1 /m|+…+ |bm /m| = |b1 /m| + |b2/m| +…. +|bN /m|+1/m(|bN+1 |+…+ |bm|)< |b1 /m| + |b2/m| +…. +|bN /m|+(m-N)/m. ε/2
<|b1 /m| + |b2/m| +…. +|bN /m|+ ε/2
(Since (m-N)/m <1).

51. Now, for any fixed r, 1 ≤ r ≤ N,
br/m → 0 as m → ∞.
Hence for ε > 0 there exists a positive integer N′ such that
|br|/m< ε/2N when m > N′.
Thus if m > max{N, N′}, then
| (b1 + b2 + ・ ・ ・ + bm)/m |< N. ε/2N+ ε/2= ε
Hence lim m→∞ (b1 + b2 + ・ ・ ・ + bm)/m = 0
=⇒ lim n→∞ (b1 + b2 + ・ ・ ・ + bn)/n= 0
=⇒ lim n→∞ {(a1 − a) + (a2 − a) + ・ ・ ・ + (an − a)}/n = 0
=⇒ lim n→∞ {(a1 + a2 + ・ ・ ・ + an)/n} − a = 0
=⇒ lim n→∞ (a1 + a2 + ・ ・ ・ + an)/n= a

52. (b) Since an > 0 and an → a, a ≥ 0.
If a > 0, then ln an → ln a. Put bn = ln an, then by Cauchy I,
lim n→∞ (b1 + b2 + ・ ・ ・ + bn)/n= ln a
=⇒ lim n→∞ (ln a1 + ln a2 + ・ ・ ・ + ln an)/ n = ln a
=⇒ lim n→∞ (a1a2 ・ ・ ・ an)1/n = ln a
=⇒ lim n→∞ e (ln a1a2···an )1/n= eln a
=⇒ lim n→∞ (a1a2 ・ ・ ・ an)1/n = a.
If a = 0, then an → 0. Hence by Cauchy I,
(a1 + a2 + ・ ・ ・ + an)/n → 0.
Thus given ε > 0 there exists a positive integer N such that n ≥ N implies that
|(a1 + a2 + ・ ・ ・ + an)/n|< ε.
Since G.M. ≤ A.M.,
| (a1a2 ・ ・ ・ an)1/n |≤ |(a1 + a2 + ・ ・ ・ + an)/n |< ε when n ≥ N.
Hence (a1a2 ・ ・ ・ an )1/n→ 0.

53. (c) Since an > 0 and an+1/an → a, a ≥ 0.
If a > 0, then
ln an+1 / an → ln a.
Put bn =an+1 / an , , then by Cauchy I,
lim n→∞ (b1 + b2 + ・ ・ ・ + bn)/n= ln a
=⇒ lim n→∞ (ln a2 /a1+ ln a3/a2+ ・ ・ ・ + ln an+1/ an)/n
= ln a
=⇒ limn→∞(ln an+1 − ln a1)/n = ln a
=⇒ lim n→∞ (ln an − ln a1)/ (n − 1)= ln a
=⇒ lim n→∞ (ln an /n − 1) − limn→∞ (ln a1/n − 1) = ln a
=⇒ lim n→∞ ln an /n − 1 = ln a
=⇒ limn→∞ (ln an/n). limn→∞n/n − 1= ln a
=⇒ lim n→∞ ln (an)1/n = ln a
=⇒ lim n→∞ (an)1/n = a.
If a = 0, then for any positive number K there exists a positive integer N such that | an+1 /an|<1/K when n ≥ N.

54. Since an > 0,
0 < aN+1/aN< 1/K (1)
0 < aN+2/ aN+1< 1/K (2)
0 < aN+3/aN+2< 1/K (3)
・ ・ ・ ・ ・ ・ ・ ・ ・ (4)
0 < an/ an−1< 1/K (n > N) (5)
Multiplying above inequalities vertically we have
0 <an/aN< (1/K)n−N when n > N.
Hence for n > N
0 < an1/n ≤1/K(KN aN )1/n
Taking limit n → ∞ we have
0 ≤ lim n→∞ an1/n ≤1/K
Taking limit K → ∞ we have
Lim n→∞ an1/n = 0
The proof of the theorem is thus complete.

55. Thank You