A sequence of real numbers or a sequence in R is a mapping f : N R.
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1 A sequence of real numbers or a sequence in R is a mapping f : N R.
2 A sequence of real numbers or a sequence in R is a mapping f : N R. Notation: We write x n for f(n), n N and so the notation for a sequence is (x n ).
3 A sequence of real numbers or a sequence in R is a mapping f : N R. Notation: We write x n for f(n), n N and so the notation for a sequence is (x n ). Examples: 1. Constant sequence: (a, a, a,...), where a R
4 A sequence of real numbers or a sequence in R is a mapping f : N R. Notation: We write x n for f(n), n N and so the notation for a sequence is (x n ). Examples: 1. Constant sequence: (a, a, a,...), where a R 2. Sequence defined by listing: (1, 4, 8, 11, 52,...)
5 A sequence of real numbers or a sequence in R is a mapping f : N R. Notation: We write x n for f(n), n N and so the notation for a sequence is (x n ). Examples: 1. Constant sequence: (a, a, a,...), where a R 2. Sequence defined by listing: (1, 4, 8, 11, 52,...) 3. Sequence defined by rule: (x n ), where x n = 3n 2 for all n N
6 A sequence of real numbers or a sequence in R is a mapping f : N R. Notation: We write x n for f(n), n N and so the notation for a sequence is (x n ). Examples: 1. Constant sequence: (a, a, a,...), where a R 2. Sequence defined by listing: (1, 4, 8, 11, 52,...) 3. Sequence defined by rule: (x n ), where x n = 3n 2 for all n N 4. Sequence defined recursively: (x n ), where x 1 = 4 and x n+1 = 2x n 5 for all n N
7 Convergence: What does it mean?
8 Convergence: What does it mean? Think of the examples: (2, 2, 2,...) ( 1 n ) (( 1) n 1 n ) (1, 2, 1, 2,...) (( 1) n (1 1 n )) (n 2 1)
9 Convergence: What does it mean? Think of the examples: (2, 2, 2,...) ( 1 n ) (( 1) n 1 n ) (1, 2, 1, 2,...) (( 1) n (1 1 n )) (n 2 1) Definition: The sequence (x n ) is convergent if there exists l R such that for every ε > 0, there exists n 0 N such that x n l < ε for all n n 0.
10 Convergence: What does it mean? Think of the examples: (2, 2, 2,...) ( 1 n ) (( 1) n 1 n ) (1, 2, 1, 2,...) (( 1) n (1 1 n )) (n 2 1) Definition: The sequence (x n ) is convergent if there exists l R such that for every ε > 0, there exists n 0 N such that x n l < ε for all n n 0. We say: l is a limit of (x n ): lim n x n = l
11 A sequence which is not convergent is called divergent.
12 A sequence which is not convergent is called divergent. Result: The limit of a convergent sequence is unique.
13 A sequence which is not convergent is called divergent. Result: The limit of a convergent sequence is unique. Ex. Examine whether the sequences listed earlier are convergent. Also, find their limits if they are convergent.
14 A sequence which is not convergent is called divergent. Result: The limit of a convergent sequence is unique. Ex. Examine whether the sequences listed earlier are convergent. Also, find their limits if they are convergent. Ex. Same question for the following sequences. (i) ( n+1 2n+3 ) (ii) (n ) (iii) (n3 + 1) (iv) (α n ), where α < 1
15 A sequence which is not convergent is called divergent. Result: The limit of a convergent sequence is unique. Ex. Examine whether the sequences listed earlier are convergent. Also, find their limits if they are convergent. Ex. Same question for the following sequences. (i) ( n+1 2n+3 ) (ii) (n ) (iii) (n3 + 1) (iv) (α n ), where α < 1 Definition: The sequence (x n ) is bounded if there exists M > 0 such that x n M for all n N. Otherwise (x n ) is called unbounded (not bounded).
16 A sequence which is not convergent is called divergent. Result: The limit of a convergent sequence is unique. Ex. Examine whether the sequences listed earlier are convergent. Also, find their limits if they are convergent. Ex. Same question for the following sequences. (i) ( n+1 2n+3 ) (ii) (n ) (iii) (n3 + 1) (iv) (α n ), where α < 1 Definition: The sequence (x n ) is bounded if there exists M > 0 such that x n M for all n N. Otherwise (x n ) is called unbounded (not bounded). Examples: (i) ( 3n+2 2n+5 ) (ii) (1, 2, 1, 3, 1, 4,...)
17 Result: Every convergent sequence is bounded. So, Not bounded implies Not convergent.
18 Result: Every convergent sequence is bounded. So, Not bounded implies Not convergent. Limit rules for convergent sequences
19 Result: Every convergent sequence is bounded. So, Not bounded implies Not convergent. Limit rules for convergent sequences Let x n x and y n y.
20 Result: Every convergent sequence is bounded. So, Not bounded implies Not convergent. Limit rules for convergent sequences Let x n x and y n y. Then (i) x n + y n x + y (ii) kx n kx for all k R (iii) x n x (iv) x n y n xy (v) xn y n x y if y 0
21 Result: Every convergent sequence is bounded. So, Not bounded implies Not convergent. Limit rules for convergent sequences Let x n x and y n y. Then (i) x n + y n x + y (ii) kx n kx for all k R (iii) x n x (iv) x n y n xy (v) xn y n x y if y 0 Ex. Similar results for divergent sequences?
22 Result: Every convergent sequence is bounded. So, Not bounded implies Not convergent. Limit rules for convergent sequences Let x n x and y n y. Then (i) x n + y n x + y (ii) kx n kx for all k R (iii) x n x (iv) x n y n xy (v) xn y n x y if y 0 Ex. Similar results for divergent sequences? Ex. If x n x and x 0, then show that there exists n 0 N such that x n 0 for all n n 0.
23 Ex. Examine the convergence and find the limits (if possible) of the following sequences. (i) ( 2n2 3n 3n 2 +5n+3 ) (ii) ( n + 1 n) (iii) ( 4n 2 + n 2n)
24 Ex. Examine the convergence and find the limits (if possible) of the following sequences. (i) ( 2n2 3n 3n 2 +5n+3 ) (ii) ( n + 1 n) (iii) ( 4n 2 + n 2n) Ex. Show that both the following sequences are convergent with limit 1. (i) (α 1 n ), where α > 0 (ii) (n 1 n)
25 Ex. Examine the convergence and find the limits (if possible) of the following sequences. (i) ( 2n2 3n 3n 2 +5n+3 ) (ii) ( n + 1 n) (iii) ( 4n 2 + n 2n) Ex. Show that both the following sequences are convergent with limit 1. (i) (α 1 n ), where α > 0 (ii) (n 1 n) Sandwich theorem: Let (x n ), (y n ), (z n ) be sequences such that x n y n z n for all n N.
26 Ex. Examine the convergence and find the limits (if possible) of the following sequences. (i) ( 2n2 3n 3n 2 +5n+3 ) (ii) ( n + 1 n) (iii) ( 4n 2 + n 2n) Ex. Show that both the following sequences are convergent with limit 1. (i) (α 1 n ), where α > 0 (ii) (n 1 n) Sandwich theorem: Let (x n ), (y n ), (z n ) be sequences such that x n y n z n for all n N. If both (x n ) and (z n ) converge to the same limit l, then (y n ) also converges to l.
27 Ex. Examine the convergence and find the limits (if possible) of the following sequences. (i) ( 2n2 3n 3n 2 +5n+3 ) (ii) ( n + 1 n) (iii) ( 4n 2 + n 2n) Ex. Show that both the following sequences are convergent with limit 1. (i) (α 1 n ), where α > 0 (ii) (n 1 n) Sandwich theorem: Let (x n ), (y n ), (z n ) be sequences such that x n y n z n for all n N. If both (x n ) and (z n ) converge to the same limit l, then (y n ) also converges to l. Examples: (i) ( 1 n sin2 n) ( ) (ii) ((2 n + 3 n ) 1 n) (iii) 1 1 n 2 +1 n 2 +n
28 Result: Let x n 0 for all n N and let L = lim x n+1 n x n exist.
29 Result: Let x n 0 for all n N and let L = lim x n+1 n x n exist. (i) If L < 1, then x n 0. (ii) If L > 1, then (x n ) is divergent.
30 Result: Let x n 0 for all n N and let L = lim x n+1 n x n exist. (i) If L < 1, then x n 0. (ii) If L > 1, then (x n ) is divergent. Examples: (i) ( αn n! ), α R (ii) ( nk α n ), α > 1, k > 0 (iii) ( 2n n 4 )
31 Result: Let x n 0 for all n N and let L = lim x n+1 n x n exist. (i) If L < 1, then x n 0. (ii) If L > 1, then (x n ) is divergent. Examples: (i) ( αn nk n! ), α R (ii) ( α ), α > 1, k > 0 n (iii) ( 2n ) n 4 Definition: (x n ) is increasing if x n+1 x n for all n N (x n ) is decreasing if x n+1 x n for all n N (x n ) is monotonic if it is either increasing or decreasing.
32 Result: Let x n 0 for all n N and let L = lim x n+1 n x n exist. (i) If L < 1, then x n 0. (ii) If L > 1, then (x n ) is divergent. Examples: (i) ( αn nk n! ), α R (ii) ( α ), α > 1, k > 0 n (iii) ( 2n ) n 4 Definition: (x n ) is increasing if x n+1 x n for all n N (x n ) is decreasing if x n+1 x n for all n N (x n ) is monotonic if it is either increasing or decreasing. Ex. Examine whether the following sequences are monotonic. (i) (1 1 n ) (ii) (n + 1 n ) (iii) (cos nπ 3 ) (iv) ((1 + 1 n )n )
33 Definition: Let S( ) R and u R. u is an upper bound of S in R if x u for all x S.
34 Definition: Let S( ) R and u R. u is an upper bound of S in R if x u for all x S. u is the supremum (least upper bound) of S in R if (i) u is an upper bound of S in R, and (ii) u is the least among all the upper bounds of S in R, i.e. if u is any upper bound of S in R, then u u.
35 Definition: Let S( ) R and u R. u is an upper bound of S in R if x u for all x S. u is the supremum (least upper bound) of S in R if (i) u is an upper bound of S in R, and (ii) u is the least among all the upper bounds of S in R, i.e. if u is any upper bound of S in R, then u u. Lower bound and infimum (greatest lower bound) are defined similarly.
36 Result: An increasing sequence (x n ) which is bounded above converges to sup{x n : n N} A decreasing sequence (x n ) which is bounded below converges to inf{x n : n N}
37 Result: An increasing sequence (x n ) which is bounded above converges to sup{x n : n N} A decreasing sequence (x n ) which is bounded below converges to inf{x n : n N} So a monotonic sequence converges iff it is bounded.
38 Result: An increasing sequence (x n ) which is bounded above converges to sup{x n : n N} A decreasing sequence (x n ) which is bounded below converges to inf{x n : n N} So a monotonic sequence converges iff it is bounded. Example: x 1 = 1, x n+1 = 1 3 (x n + 1) for all n N. Then (x n ) is convergent and lim n x n = 1 2.
39 Result: An increasing sequence (x n ) which is bounded above converges to sup{x n : n N} A decreasing sequence (x n ) which is bounded below converges to inf{x n : n N} So a monotonic sequence converges iff it is bounded. Example: x 1 = 1, x n+1 = 1 3 (x n + 1) for all n N. Then (x n ) is convergent and lim n x n = 1 2. Ex. Let x n = 1 n n n+n for all n N. Is (x n ) convergent?
40 Result: An increasing sequence (x n ) which is bounded above converges to sup{x n : n N} A decreasing sequence (x n ) which is bounded below converges to inf{x n : n N} So a monotonic sequence converges iff it is bounded. Example: x 1 = 1, x n+1 = 1 3 (x n + 1) for all n N. Then (x n ) is convergent and lim n x n = 1 2. Ex. Let x n = 1 n n n+n for all n N. Is (x n ) convergent? Cauchy sequence: A sequence (x n ) is called a Cauchy sequence if for each ε > 0, there exists n 0 N such that x m x n < ε for all m, n n 0.
41 Result: A sequence in R is convergent iff it is a Cauchy sequence.
42 Result: A sequence in R is convergent iff it is a Cauchy sequence. Ex. Let x n = ! + 1 2! n! for all n N. Show that (x n) is convergent.
43 Result: A sequence in R is convergent iff it is a Cauchy sequence. Ex. Let x n = ! + 1 2! n! for all n N. Show that (x n) is convergent. Ex. Let (x n ) satisfy either of the following conditions: (i) x n+1 x n α n for all n N (ii) x n+2 x n+1 α x n+1 x n for all n N, where 0 < α < 1. Show that (x n ) is a Cauchy sequence.
44 Result: A sequence in R is convergent iff it is a Cauchy sequence. Ex. Let x n = ! + 1 2! n! for all n N. Show that (x n) is convergent. Ex. Let (x n ) satisfy either of the following conditions: (i) x n+1 x n α n for all n N (ii) x n+2 x n+1 α x n+1 x n for all n N, where 0 < α < 1. Show that (x n ) is a Cauchy sequence. Ex. Let x 1 = 1 and let x n+1 = 1 x n+2 for all n N. Show that (x n) is convergent and find lim n x n.
45 Subsequence: Let (x n ) be a sequence in R. If (n k ) is a sequence of positive integers such that n 1 < n 2 < n 3 <, then (x nk ) is called a subsequence of (x n ).
46 Subsequence: Let (x n ) be a sequence in R. If (n k ) is a sequence of positive integers such that n 1 < n 2 < n 3 <, then (x nk ) is called a subsequence of (x n ). Examples: Think of some divergent sequences and their convergent subsequences.
47 Subsequence: Let (x n ) be a sequence in R. If (n k ) is a sequence of positive integers such that n 1 < n 2 < n 3 <, then (x nk ) is called a subsequence of (x n ). Examples: Think of some divergent sequences and their convergent subsequences. Result: If a sequence (x n ) converges to l, then every subsequence of (x n ) must converge to l.
48 Subsequence: Let (x n ) be a sequence in R. If (n k ) is a sequence of positive integers such that n 1 < n 2 < n 3 <, then (x nk ) is called a subsequence of (x n ). Examples: Think of some divergent sequences and their convergent subsequences. Result: If a sequence (x n ) converges to l, then every subsequence of (x n ) must converge to l. So, if (x n ) has a subsequence (x nk ) such that x nk x n l. l, then
49 Subsequence: Let (x n ) be a sequence in R. If (n k ) is a sequence of positive integers such that n 1 < n 2 < n 3 <, then (x nk ) is called a subsequence of (x n ). Examples: Think of some divergent sequences and their convergent subsequences. Result: If a sequence (x n ) converges to l, then every subsequence of (x n ) must converge to l. So, if (x n ) has a subsequence (x nk ) such that x nk x n l. l, then Also, if (x n ) has two subsequences converging to two different limits, then (x n ) cannot be convergent.
50 Example: Let x n = ( 1) n (1 1 n ) for all n N. Then x n 1.
51 Example: Let x n = ( 1) n (1 1 n ) for all n N. Then x n 1. In fact, (x n ) is not convergent.
52 Example: Let x n = ( 1) n (1 1 n ) for all n N. Then x n 1. In fact, (x n ) is not convergent. Ex. Let (x n ) be a sequence such that x 2n l and x 2n 1 l. Show that x n l.
53 Example: Let x n = ( 1) n (1 1 n ) for all n N. Then x n 1. In fact, (x n ) is not convergent. Ex. Let (x n ) be a sequence such that x 2n l and x 2n 1 l. Show that x n l. Example: The sequence (1, 1 2, 1, 2 3, 1, 3 4,...) converges to 1.
54 Example: Let x n = ( 1) n (1 1 n ) for all n N. Then x n 1. In fact, (x n ) is not convergent. Ex. Let (x n ) be a sequence such that x 2n l and x 2n 1 l. Show that x n l. Example: The sequence (1, 1 2, 1, 2 3, 1, 3 4,...) converges to 1. Ex. Can you find a convergent subsequence of (( 1) n n 2 )?
55 Example: Let x n = ( 1) n (1 1 n ) for all n N. Then x n 1. In fact, (x n ) is not convergent. Ex. Let (x n ) be a sequence such that x 2n l and x 2n 1 l. Show that x n l. Example: The sequence (1, 1 2, 1, 2 3, 1, 3 4,...) converges to 1. Ex. Can you find a convergent subsequence of (( 1) n n 2 )? Bolzano-Weierstrass Theorem: Every bounded sequence in R has a convergent subsequence.
56 Example: Let x n = ( 1) n (1 1 n ) for all n N. Then x n 1. In fact, (x n ) is not convergent. Ex. Let (x n ) be a sequence such that x 2n l and x 2n 1 l. Show that x n l. Example: The sequence (1, 1 2, 1, 2 3, 1, 3 4,...) converges to 1. Ex. Can you find a convergent subsequence of (( 1) n n 2 )? Bolzano-Weierstrass Theorem: Every bounded sequence in R has a convergent subsequence. Example: If x R, then there exists a sequence (r n ) of rationals converging to x.
57 Example: Let x n = ( 1) n (1 1 n ) for all n N. Then x n 1. In fact, (x n ) is not convergent. Ex. Let (x n ) be a sequence such that x 2n l and x 2n 1 l. Show that x n l. Example: The sequence (1, 1 2, 1, 2 3, 1, 3 4,...) converges to 1. Ex. Can you find a convergent subsequence of (( 1) n n 2 )? Bolzano-Weierstrass Theorem: Every bounded sequence in R has a convergent subsequence. Example: If x R, then there exists a sequence (r n ) of rationals converging to x. Similarly, if x R, then there exists a sequence (t n ) of irrationals converging to x.
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