Chap 02 Real Analysis: Sequences And Series

Chapter 2 – Sequences and Series Subject: Real Analysis Level: M.Sc. Source: Syed Gul Shah (Chairman, Department of Mathematics, UoS Sargodha) Collected & Composed by: Atiq ur Rehman([email protected]), http://www.mathcity.org Sequence A sequence is a function whose domain of definition is the set of natural numbers. Or it can also be defined as an ordered set. Notation: An infinite sequence is denoted as ∞

{S n } or n=1

e.g.

i)

{Sn : n ∈ ¥} or {S1, S2 , S3 ,...........} or simply as {Sn }

{n} = {1, 2,3,..........}

1  1 1  ii)   = 1, , ,............... n  2 3  n+1 iii) {(−1) } = {1, −1,1, −1,.............} Subsequence It is a sequence whose terms are contained in given sequence. ∞

∞

A subsequence of {S n } is usually written as {S nk } . n=1

Increasing Sequence A sequence {Sn } is said to be an increasing sequence if S n +1 ≥ S n ∀ n ≥ 1 . Decreasing Sequence A sequence {Sn } is said to be an decreasing sequence if S n +1 ≤ S n ∀ n ≥ 1 . Monotonic Sequence A sequence {Sn } is said to be monotonic sequence if it is either increasing or decreasing. S {Sn } is monotonically increasing if Sn+1 − Sn ≥ 0 or n+1 ≥ 1, ∀ n ≥ 1 Sn S {Sn } is monotonically decreasing if Sn − Sn+1 ≥ 0 or n ≥ 1, ∀ n ≥ 1 S n+1 Strictly Increasing or Decreasing {Sn } is called strictly increasing or decreasing according as S n+1 > S n or S n+1 < S n ∀ n ≥ 1 . Bernoulli’s Inequality Let p ∈ ¡ , p ≥ −1 and p ≠ 0 then for n ≥ 2 we have

(1 + p )

n

> 1 + np

Proof: We shall use mathematical induction to prove this inequality. If n = 2 L.H.S = (1 + p) 2 = 1 + 2 p + p 2 R.H.S = 1 + 2 p ⇒ L.H .S > R.H .S

Sequences and Series

-2-

i.e. condition I of mathematical induction is satisfied. Suppose (1 + p ) > 1 + kp .................. (i ) k

Now

(1 + p )

k +1

where k ≥ 2

= (1 + p )(1 + p ) > (1 + p )(1 + kp )

using (i)

= 1 + kp + p + kp 2 = 1 + (k + 1) p + kp 2 ≥ 1 + (k + 1) p

ignoring kp 2 ≥ 0

k

⇒ (1 + p ) > 1 + ( k + 1) p Since the truth for n = k implies the truth for n = k + 1 therefore condition II of n mathematical induction is satisfied. Hence we conclude that (1 + p ) > 1 + np . k +1

Example  1 Let S n = 1 +   n

n

where n ≥ 1

To prove that this sequence is an increasing sequence, we use p = Bernoulli’s inequality to have n 1  n  1 − 2  > 1 − 2 n  n  n

  1  1   1 ⇒  1 − 1 +   > 1 − n   n  n   1− n

1− n

 1  1  n −1  n  ⇒ 1 +  > 1 −  =   =   n  n  n   n −1 ⇒ Sn > S n −1 ∀ n ≥1 which shows that {Sn } is increasing sequence. n

n−1

1   = 1 +   n −1 

Example n+1

 1 Let tn = 1 +  ; n ≥1  n then the sequence is decreasing sequence. 1 We use p = 2 in Bernoulli’s inequality. n −1 n n 1   ……….. (i) 1 + 2  >1+ 2 n −1  n −1 where 1 n2  n  n  = 2 = 1+ 2   n − 1 n − 1  n − 1  n + 1  1  n + 1   n   ⇒ 1 + 2  =  …………… (ii)  n − 1  n   n − 1  n n 1   n   Now tn−1 = 1 +  =  n − 1    n −1  1  n + 1   =  1 + 2     n − 1  n  

n

from (ii)

n −1

−1 , n ≥ 2 in n2

Sequences and Series

1   n +1  = 1 + 2     n −1  n  n n  n + 1   > 1 + 2    n − 1  n  n

-3-

n

from (i)

n n 1  1  n + 1  > 1 +  Q 2 > 2=  n −1 n n  n  n  n+1  n +1 =  = tn  n  i.e. tn−1 > t n Hence the given sequence is decreasing sequence. n

Bounded Sequence A sequence {Sn } is said to be bounded if there exists a positive real number λ such that Sn < λ ∀ n ∈ ¥ If S and s are the supremum and infimum of elements forming the bounded sequence {Sn } we write S = sup S n and s = inf S n All the elements of the sequence S n such that Sn < λ ∀ n ∈ ¥ lie with in the strip { y : − λ < y < λ} . But the elements of the unbounded sequence can not be contained in any strip of a finite width. Examples (i) (ii)

 (−1)n  {U n} =   is a bounded sequence  n 

{Vn } = {sin nx} is also bounded sequence. Its supremum is 1 and infimum is −1 .

(iii) The geometric sequence {ar n −1} , r > 1 is an unbounded above sequence. It is bounded below by a. nπ   (iv)  tan  is an unbounded sequence. 2  

Convergence of the Sequence A sequence {Sn } of real numbers is said to convergent to limit ‘s’ as n → ∞ , if for every positive real number ε > 0 , however small, there exists a positive integer n0 , depending uponε , such that Sn − s < ε ∀ n > n0 . Theorem A convergent sequence of real number has one and only one limit (i.e. Limit of the sequence is unique.) Proof: Suppose {Sn } converges to two limits s and t, where s ≠ t . Put ε =

s−t 2

then there exits two positive integers n1 and n2 such that

∀ n > n1 Sn − s < ε and Sn − t < ε ∀ n > n2 ⇒ Sn − s < ε and Sn − t < ε hold simultaneously ∀ n > max(n1 , n2 ) . Thus for all n > max(n1 , n2 ) we have s − t = s − S n + Sn − t

Sequences and Series

-4-

≤ Sn − s + S n − t < ε + ε = 2ε  s−t  ⇒ s − t < 2   2  ⇒ s −t < s −t Which is impossible, therefore the limit of the sequence is unique. Note: If {Sn } converges to s then all of its infinite subsequence converge to s. Cauchy Sequence A sequence { xn } of real number is said to be a Cauchy sequence if for given positive real number ε , ∃ a positive integer n0 (ε ) such that xn − xm < ε ∀ m, n > n0 Theorem A Cauchy sequence of real numbers is bounded. Proof Let {Sn } be a Cauchy sequence. Take ε = 1 , then there exits a positive integers n0 such that Sn − Sm < 1 ∀ m, n > n0 . Fix m = n0 + 1 then Sn = S n − S n0 +1 + S n0 +1 ≤ S n − S n0 +1 + S n0 +1 < 1 + S n0 +1

<λ

∀ n > n0 ∀ n > 1 , and λ = 1 + S n0 +1

( n0 changes as ε changes)

Hence we conclude that {Sn } is a Cauchy sequence, which is bounded one. Note: (i) Convergent sequence is bounded. (ii) The converse of the above theorem does not hold. i.e. every bounded sequence is not Cauchy. Consider the sequence {Sn } where Sn = (−1)n , n ≥ 1 . It is bounded sequence because (−1) n = 1 < 2

∀ n ≥1

But it is not a Cauchy sequence if it is then for ε = 1 we should be able to find a positive integer n0 such that Sn − Sm < 1 for all m, n > n0 But with m = 2k + 1 , n = 2k + 2 when 2 k + 1 > n0 , we arrive at Sn − S m = (−1)2 n+2 − (−1) 2 k +1

= 1 + 1 = 2 < 1 is absurd. Hence {Sn } is not a Cauchy sequence. Also this sequence is not a convergent sequence. (it is an oscillatory sequence) ……………………………

Sequences and Series

-5-

Divergent Sequence A {Sn } is said to be divergent if it is not convergent or it is unbounded.

{ n } is divergent, it is unbounded. (ii) {(−1) } tends to 1 or -1 according as n is even or odd. It oscillates finitely. (iii) {(−1) n} is a divergent sequence. It oscillates infinitely. 2

e.g.

n

n

Note: If two subsequence of a sequence converges to two different limits then the sequence itself is a divergent. Theorem If S n < U n < t n ∀ n ≥ n0 and if both the {Sn } and {tn } converge to same limits as s, then the sequence {U n } also converges to s. Proof Since the sequence {Sn } and {tn } converge to the same limit s, therefore, for given ε > 0 there exists two positive integers n1 , n2 > n0 such that Sn − s < ε ∀ n > n1

tn − s < ε ∀ n > n2 i.e. s − ε < S n < s + ε ∀ n > n1 s − ε < tn < s + ε ∀ n > n2 Since we have given Sn < U n < tn ∀ n > n0 ∴ s − ε < Sn < U n < tn < s + ε ∀ n > max(n0 , n1 , n2 ) ⇒ s − ε < Un < s + ε ∀ n > max(n0 , n1 , n2 ) Un − s < ε i.e. ∀ n > max(n0 , n1 , n2 ) i.e. lim U n = s n→∞

Example 1 n

Show that lim n = 1 n→∞

Solution Using Bernoulli’s Inequality n 1  n  ≥ n ≥1 1 +  ≥1+ n n  Also 2 n

1   1    1 1 + = +       > n   n    2

1   ⇒ 1 ≤ n < 1 +  n  1 n

n

( n)

2 n

1 n

> n ≥1

2

1   ⇒ lim1 ≤ lim n < lim 1 +  n→∞ n→∞ n→∞ n  1 n

∀ n.

2

1 n

⇒ 1 ≤ lim n < 1 n→∞

1 n

i.e. lim n = 1 . n→∞

……………………..

Sequences and Series

Example  1 1 1  Show that lim  + + ............ + =0 2 2 n→∞ ( n + 1) (n + 2) (2 n)2   Solution We have  1 1 1  Sn =  + + ............ +  2 (n + 2)2 (2n) 2   (n + 1) and n n < S < n (2n) 2 n2 1 1 ⇒ < Sn < 4n n 1 1 ⇒ lim < lim S n < lim n→∞ 4n n→∞ n→∞ n ⇒ 0 < lim S n < 0 n→∞

⇒ lim S n = 0 n→∞

Theorem If the sequence {Sn } converges to s then ∃ a positive integer n 1 such that Sn > s . 2 Proof 1 We fix ε = s > 0 2 ⇒ ∃ a positive integer n1 such that Sn − s < ε for n > n1 1 ⇒ Sn − s < s 2 Now 1 1 s = s − s 2 2 < s − Sn − s ≤ s + ( S n − s ) ⇒

1 s < Sn 2

Theorem Let a and b be fixed real numbers if {Sn } and {tn } converge to s and t respectively, then (i) {aSn + btn } converges to as + bt. (ii) {Sn tn } converges to st. S  s (iii)  n  converges to , provided tn ≠ 0 ∀ n and t ≠ 0 . t  tn  Proof Since {Sn } and {tn } converge to s and t respectively, ∴ Sn − s < ε ∀ n > n1 ∈ ¥

-6-

Sequences and Series

-7-

tn − t < ε ∀ n > n2 ∈ ¥ Also ∃ λ > 0 such that Sn < λ ∀ n > 1 ( Q {Sn } is bounded ) (i) We have ( aSn + btn ) − ( as + bt ) = a ( Sn − s ) + b ( tn − t ) ≤ a ( S n − s ) + b ( tn − t )

(ii)

< a ε+ bε ∀ n > max (n1 , n2 ) = ε1 Where ε1 = a ε + b ε a certain number. This implies {aSn + btn } converges to as + bt. Sn tn − st = Sn tn − Sn t + Sn t − s t = S n ( tn − t ) + t ( S n − s )

≤ S n ⋅ ( tn − t ) + t ⋅ ( S n − s )

< λε + t ε ∀ n > max (n1 , n2 ) = ε2 where ε 2 = λ ε + t ε a certain number. This implies {Sn tn } converges to st. (iii)

1 1 t − tn − = tn t tn t tn − t ε < 1 tn t 2 t t ε = = ε3 1 t 2 2 =

∀ n > max (n1 , n2 ) where ε 3 =

ε 1 2

1 1 This implies   converges to . t  tn  S   1 s 1 Hence  n  =  Sn ⋅  converges to s ⋅ = . tn  t t  tn  

t

2

1 Q tn > t 2 a certain number.

( from (ii) )

Theorem For each irrational number x, there exists a sequence {rn } of distinct rational numbers such that lim rn = x . n→∞

Proof Since x and x + 1 are two different real numbers Q ∃ a rational number r1 such that x < r1 < x + 1 Similarly ∃ a rational number r2 ≠ r1 such that 1  x < r2 < min  r1 , x +  < x + 1 2  Continuing in this manner we have 1  x < r3 < min  r2 , x +  < x + 1 3  1  x < r4 < min  r3 , x +  < x + 1 4  …………………………….......... …………………………….......... ……………………………..........

1  x < rn < min  rn−1 , x +  < x + 1 n 

Sequences and Series

-8-

This implies that ∃ a sequence {rn } of the distinct rational number such that 1 1 x − < x < rn < x + n n Since 1 1   lim  x −  = lim  x +  = x n→∞ n  n→∞  n  Therefore lim rn = x n→∞

Theorem Let a sequence {Sn } be a bounded sequence. (i) If {Sn } is monotonically increasing then it converges to its supremum. (ii) If {Sn } is monotonically decreasing then it converges to its infimum. Proof Let S = sup S n and s = inf S n Take ε > 0 (i) Since S = sup S n ∴ ∃ Sn0 such that S − ε < Sn0

Since {Sn } is ↑ ∴ S − ε < Sn0 < Sn < S < S + ε ⇒ S − ε < Sn < S + ε ⇒ Sn − S < ε ⇒ lim S n = S

( ↑ stands for monotonically increasing )

for n > n0 for n > n0 for n > n0

n→∞

(ii)

Since s = inf S n ∴ ∃ Sn1 such that Sn1 < s + ε Since {Sn } is ↓ . ∴ s − ε < s < Sn < Sn1 < s + ε

( ↓ stands for monotonically decreasing )

for n > n1

⇒ s − ε < Sn < s + ε for n > n1 ⇒ Sn − s < ε for n > n1 Thus lim S n = s n→∞

Note A monotonic sequence can not oscillate infinitely. Example:  1 n  Consider {S n } = 1 +    n   As shown earlier it is an increasing sequence 2n 1   Take S 2 n = 1 +   2n  n

1   Then S 2 n = 1 +   2n  n n 1 1 1   2n   ⇒ = ⇒ = 1 −   S 2 n  2n + 1  S 2 n  2n + 1  Using Bernoulli’s Inequality we have

Sequences and Series

⇒

-9n

1  n Q  1 −  ≥ 1− 2n + 1  2n + 1 

1 n n 1 ≥ 1− > 1− = 2n + 1 2n 2 S2 n

⇒ S2n < 2 ∀ n = 1,2,3,.......... ⇒ S2n < 4 ∀ n = 1,2,3,.......... ⇒ Sn < S2 n < 4 ∀ n = 1,2,3,.......... Which show that the sequence {Sn } is bounded one. Hence {Sn } is a convergent sequence the number to which it converges is its supremum, which is denoted by ‘e’ and 2 < e < 3 . Recurrence Relation A sequence is said to be defined recursively or by recurrence relation if the general term is given as a relation of its preceding and succeeding terms in the sequence together with some initial condition. Example Let t1 > 0 and let {tn } be defined by tn+1 > 2 − ⇒ tn > 0

Also

∀ n ≥1 1 tn − tn+1 = tn − 2 + tn

1 ; n ≥1 tn

( t − 1) > 0 t 2 − 2tn + 1 = n = n tn tn ⇒ tn > tn+1 ∀ n ≥1 This implies that tn is monotonically decreasing. Since tn > 1 ∀ n ≥1 ⇒ t n is bounded below ⇒ tn is convergent. Let us suppose lim t n = t 2

n→∞

lim tn+1 = lim tn

Then

n→∞

n→∞

 1 ⇒ lim  2 −  = lim tn n→∞ t n  n→∞  1 2t − 1 ⇒ 2− =t ⇒ =t t t 2 ⇒ ( t − 1) = 0 ⇒ t =1

⇒ 2t − 1 = t 2

Example

⇒ t 2 − 2t + 1 = 0

Let {Sn } be defined by S n +1 = S n + b ; n ≥ 1 and S1 = a > b . It is clear that S n > 0 ∀ n ≥ 1 and S 2 > S1 and

Sn2+1 − Sn2 = ( Sn + b ) − ( Sn−1 + b ) = S n − S n−1 ⇒ ( Sn+1 + Sn )( Sn +1 − Sn ) = Sn − Sn−1 S − S n −1 ⇒ S n+1 − Sn = n S n+1 + S n Since S n +1 + S n > 0 ∀ n ≥ 1 Therefore S n +1 − S n and S n − S n−1 have the same sign. i.e. S n+1 > S n if and only if S n > S n−1 and S n+1 < S n if and only if S n < S n−1 .

Sequences and Series

- 10 -

But we know that S 2 > S1 therefore S3 > S 2 , S 4 > S3 , and so on. This implies the sequence is an increasing sequence. Also S n2+1 − S n2 =

(

Sn + b

)

2

− S n2 = Sn + b − S n2

= − ( S n2 − S n − b )

Since S n > 0 ∀ n ≥ 1 , therefore S n is the root (+ive) of the

Sn2 − Sn − b = 0 1 + 1 + 4b For equation ax 2 + bx + c = 0 2 −b The product of roots is αβ = c a the other root of equation is therefore α c i.e. the other root β = Since S n+1 > S n ∀ n ≥ 1 aα b  Also − ( S n − α )  S n +  = Sn2+1 − S n2 > 0 α  b ∴ S n + > 0 or − ( Sn − α ) ≥ 0 α ⇒ Sn < α ∀ n ≥ 1 which shows that S n is bounded and hence it is convergent. Suppose lim S n = s Take this value of S n as α where α =

n→∞

Then lim( S n+1 )2 = lim( S n + b) n→∞

n→∞

⇒ s = s+b 2

Which shows that α =

⇒ s2 − s − b = 0 1 + 1 + 4b is the limit of the sequence. 2

Theorem Every Cauchy sequence of real numbers has a convergent subsequence. Proof Suppose {Sn } is a Cauchy sequence. Let ε > 0 then ∃ a positive integer n0 ≥ 1 such that ε Snk − Snk −1 < k ∀ nk , nk −1 , k = 1,2,3,......... 2 Put bk = S n1 − S n0 + S n2 − S n1 + ............ + S nk − S nk −1 ⇒ bk

( = (S ≤ (S <

n1

− S n0

n1

− S n0

) ( ) ( ) ) + ( S − S ) + ............ + ( S − S ) ) + ( S − S ) + ............ + ( S − S ) n2

n1

n2

ε ε ε + 2 + ............ + k 2 2 2

nk

n1

nk −1

nk

(

 1 1 − 1k 1 2 1 1 2 = ε  + 2 + ............ + k  = ε  1  1− 2 2  2 2  ⇒ bk < ε ∀ k ≥1 ⇒ {bk } is convergent Q bk = Snk − Sn0 ∴ Snk = bk + Sn0

nk −1

)  = ε 1 − 1   

 

 2k 

{ } which is a subsequence of {S } is

Where Sn0 is a certain fix number therefore S nk convergent.

n

Sequences and Series

- 11 -

Theorem (Cauchy’s General Principle for Convergence) A sequence of real number is convergent if and only if it is a Cauchy sequence. Proof Necessary Condition Let {Sn } be a convergent sequence, which converges to s . Then for given ε > 0 ∃ a positive integer n0 , such that ε Sn − s < ∀ n > n0 2 Now for n > m > n0 Sn − S m = S n − s + S m − s

≤ Sn − s + Sm − s ε ε < + =ε 2 2 Which shows that {Sn } is a Cauchy sequence. Sufficient Condition Let us suppose that {Sn } is a Cauchy sequence then for ε > 0 , ∃ a positive integer m1 such that ε Sn − S m < ∀ n, m > m1 ……….. (i) 2 Since {Sn } is a Cauchy sequence

{ } converging to s (say).

therefore it has a subsequence S nk

⇒ ∃ a positive integer m2 such that ε S nk − s < ∀ n > m2 ……….. (ii) 2 Now Sn − s = S n − S nk + S nk − s ≤ S n − S nk + S nk − s

ε ε + =ε ∀ n > max(m1 , m2 ) 2 2 which shows that {Sn } is a convergent sequence. <

Example Let {Sn } be define by 0 < a < S1 < S 2 < b and also S n +1 = Sn ⋅ S n −1 , Here S n > 0 , ∀ n ≥ 1 and a < S1 < b Let for some k > 2 a < Sk < b then a 2 < a Sk < Sk Sk −1 = ( Sk +1 )2 < b2 i.e. a 2 < Sk2+1 < b2 ⇒ a < S k +1 < b ⇒ a < Sn < b ∀ n ∈ ¥ S a Q n > Sn +1 b S a ∴ n +1 > +1 Sn +1 b

n > 2 ………. (i)

Q Sn +1 = Sn S n−1

Sequences and Series

S n + S n+1 a+b > S n +1 b S + S n+1 a+b ⇒ n > Sn b ⇒

S n +1 is replace by S n ∴ S n < S n+1

Sn2+1 − Sn2 = Sn ⋅ Sn−1 − Sn2 Q Sn +1 = Sn S n−1 = Sn ( Sn−1 − Sn ) Sn ⇒ S n+1 − Sn = S n−1 − S n S n + S n+1 b < S n−1 − S n a+b b ⇒ S n+1 − Sn < Sn − S n−1 Q S n−1 − S n = S n − S n−1 a+b 2  b  <  S n−1 − S n−2 a+b 3  b  <  S n − 2 − S n −3 a+b ……………………… ……………………… ……………………… n −1  b  <  (b − a ) a+b b Take r = <1 a+b Then for n > m we have Sn − Sm = Sn − Sn −1 + Sn −1 − Sn −2 + .............. + Sm+1 − Sm ≤ Sn − Sn−1 + Sn−1 − Sn−2 + .............. + Sm+1 − Sm

And

< ( r n −2 + r n −3 + ............... + r m−1 ) (b − a )

=ε This implies that {Sn } is a Cauchy sequence, therefore it is convergent. Example Let {tn } be defined by 1 1 1 tn = 1 + + + ............... + 2 3 n For m, n ∈ ¥ , n > m we have 1 1 1 tn − tm = + + ............. + m +1 m + 2 n 1 m > ( n − m) = 1 − n n In particular if n = 2m then 1 tn − tm > 2 This implies that {tn } is not a Cauchy sequence therefore it is divergent. 7⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅8

- 12 -

Sequences and Series

- 13 -

Theorem (nested intervals) Suppose that {I n } is a sequence of the closed interval such that I n = [ an , bn ] ,

I n+1 ⊂ I n ∀ n ≥ 1 , and ( bn − an ) → 0 as n → ∞ then I I n contains one and only one point. Proof Since I n+1 ⊂ I n ∴ a1 < a2 < a3 < ............. < an−1 < an < bn < bn−1 < .......... < b3 < b2 < b1 {an } is increasing sequence, bounded above by b1 and bounded below by a1 . And {bn } is decreasing sequence bounded below by a1 and bounded above by b1 . ⇒ {an } and {bn } both are convergent. Suppose {an } converges to a and {bn } converges to b. But

and

a − b = a − an + an − bn + bn − b ≤ an − a + an − bn + bn − b → 0 as n → ∞ . ⇒ a=b an < a < bn ∀ n ≥ 1 .

Theorem (Bolzano-Weierstrass theorem) Every bounded sequence has a convergent subsequence. Proof Let {Sn } be a bounded sequence. Take a1 = inf S n and b1 = sup S n Then a1 < S n < b1 ∀ n ≥ 1 . Now bisect interval [ a1 , b1 ] such that at least one of the two sub-intervals contains infinite numbers of terms of the sequence. Denote this sub-interval by [ a2 , b2 ] . If both the sub-intervals contain infinite number of terms of the sequence then choose the one on the right hand. Then clearly a1 ≤ a2 < b2 ≤ b1 . Suppose there exist a subinterval [ ak , bk ] such that a1 ≤ a2 ≤ ........... ≤ ak < bk ≤ ........... ≤ b2 ≤ b1 1 ⇒ ( bk − ak ) = k ( b1 − a1 ) 2 Bisect the interval [ ak , bk ] in the same manner and choose [ ak +1 , bk +1 ] to have a1 ≤ a2 ≤ ........... ≤ ak ≤ ak +1 < bk +1 ≤ bk ≤ ........... ≤ b2 ≤ b1 1 and bk +1 − ak +1 = k +1 ( b1 − a1 ) 2 This implies that we obtain a sequence of interval [ an , bn ] such that 1 bn − an = n (b1 − a1 ) → 0 as n → ∞ . 2 ⇒ we have a unique point s such that s = I [ an , bn ] there are infinitely many terms of the sequence whose length is ε > 0 that contain s. For ε = 1 there are infinitely many values of n such that Sn − s < 1 Let n1 be one of such value then S n1 − s < 1

Sequences and Series

- 14 -

Again choose n2 > n1 such that 1 2 Continuing in this manner we find a sequence S nk S n2 − s <

{ } for each positive integer k such

that nk < nk +1 and S nk − s <

{ }

Hence there is a subsequence S nk

1 ∀ k = 1, 2,3,............ k which converges to s.

Limit Inferior of the sequence Suppose {Sn } is bounded then we define limit inferior of {Sn } as follow lim ( inf S n ) = lim U k where U k = inf {Sn : n ≥ k } n→∞

n→∞

If S n is bounded below then lim ( inf S n ) = − ∞ n→∞

Limit Superior of the sequence Suppose {Sn } is bounded above then we define limit superior of {Sn } as follow lim ( sup S n ) = lim Vk where Vk = inf {Sn : n ≥ k } n→∞

n→∞

If S n is not bounded above then we have lim ( sup S n ) = + ∞ n→∞

Note: (i) A bounded sequence has unique limit inferior and superior (ii) Let {Sn } contains all the rational numbers, then every real number is a subsequencial limit then limit superior of S n is + ∞ and limit inferior of S n is −∞  1 (iii) Let {S n } = (−1)n 1 +   n then limit superior of S n is 1 and limit inferior of S n is −1 . (iv) Let U k = inf {Sn : n ≥ k } 1  1   1     = inf 1 +  cos kπ , 1 +  cos( k + 1)π ,  1 +  cos(k + 2)π ,................  k +1  k +2  k    1  if k is odd  1 + k  cos kπ   = 1 + 1  cos(k + 1)π if k is even  k + 1  ⇒ lim ( inf Sn ) = lim U k = −1 n→∞

Also

n→∞

Vk = sup{Sn : n ≥ k }

 1  1 +  k + 1  cos(k + 1)π  =  1 + 1  cos kπ  k  ⇒ lim ( inf Sn ) = lim Vk = 1 n→∞

if k is odd if k is even

n→∞

ï⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ð

Sequences and Series

Theorem If {Sn } is a convergent sequence then lim S n = lim ( inf S n ) = lim ( sup S n ) n→∞

n→∞

n→∞

Proof Let lim S n = s then for a real number ε > 0 , ∃ a positive integer n0 such that n→∞

Sn − s < ε i.e. s − ε < Sn < s + ε If Vk = sup{Sn : n ≥ k } Then s − ε < Vn < s + ε ⇒ s − ε < lim Vn < s + ε k →∞

∀ n ≥ n0 ……….. (i) ∀ n ≥ n0 ∀ k ≥ n0 ∀ k ≥ n0 …………. (ii)

from (i) and (ii) we have s = lim sup {S n } k →∞

We can have the same result for limit inferior of {Sn } by taking

U k = inf {Sn : n ≥ k }

ï⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ð

- 15 -

Sequences and Series

- 16 -

Infinite Series Given a sequence {an } , we use the notation

∞

∑a i =1

n

or simply

∑a

n

to denotes the

sum a1 + a2 + a3 + .............. and called a infinite series or just series. n

The numbers S n = ∑ ak are called the partial sum of the series. k =1

If the sequence {Sn } converges to s, we say that the series converges and write ∞

∑a n=1

n

= s , the number s is called the sum of the series but it should be clearly

understood that the ‘s’ is the limit of the sequence of sums and is not obtained simply by addition. If the sequence {Sn } diverges then the series is said to be diverge. Note: The behaviors of the series remain unchanged by addition or deletion of the certain terms Theorem ∞

If

∑a n=1

converges then lim an = 0 .

n

n→∞

Proof Let S n = a1 + a2 + a3 + .......... + an lim S n = s = ∑ an

Take

n→∞

Since an = Sn − S n −1 Therefore lim an = lim ( S n − S n−1 ) n→∞

n→∞

= lim S n − lim S n−1 n→∞

=s−s=0

n→∞

Note: The converse of the above theorem is false Example ∞

1

∑n.

Consider the series

n=1

We know that the sequence {Sn } where S n = 1 + ∞

therefore

1

∑n n=1

1 1 1 + + ............. + is divergent 2 3 n

is divergent series, although lim an = 0 .

This implies that if

n→∞

lim an ≠ 0 , then n→∞

∑a

n

is divergent.

It is know as basic divergent test. Theorem (General Principle of Convergence) A series ∑ an is convergent if and only if for any real number ε > 0 , there exists a positive integer n0 such that ∞

∑a

i = m +1

i

<ε

∀ n > m > n0

Proof Let S n = a1 + a2 + a3 + ............. + an then {Sn } is convergent if and only if for ε > 0 ∃ a positive integer n0 such that

Sequences and Series

Sn − Sm < ε ∞

∑a

⇒

i

i = m +1

- 17 -

∀ n > m > n0

= Sn − Sm < ε

Example If

∞

∑x

x < 1 then

n

1 1− x

=

n= 0

∞

∑x

And if x ≥ 1 then

n

is divergent.

n =0

Theorem Let ∑ an be an infinite series of non-negative terms and let {Sn } be a sequence of

its partial sums then ∑ an is convergent if {Sn } is bounded and it diverges if {Sn } is unbounded. Proof Since an ≥ 0 ∀ n ≥ 0 S n = S n−1 + an > S n−1 ∀ n≥0 therefore the sequence {Sn } is monotonic increasing and hence it is converges if {Sn } is bounded and it will diverge if it is unbounded. Hence we conclude that

{Sn } is unbounded.

∑a

is convergent if {Sn } is bounded and it divergent if

n

Theorem (Comparison Test) Suppose ∑ an and ∑ bn are infinite series such that an > 0 , bn > 0 ∀ n . Also

suppose that for a fixed positive number λ and positive integer k , an < λ bn ∀ n ≥ k Then

∑a

Proof Suppose

n

converges if

∑b

n

is converges and

∑b

n

is diverges if

∑b

is convergent and an < λ bn ∀ n ≥ k …………. (i) then for any positive number ε > 0 there exists n0 such that n ε bi < n > m > n0 ∑ λ i = m +1 from (i) n

n

∑ ⇒ ∑a Now suppose ∑ a ⇒

i =m +1

ai < λ n

n

∑ b <ε

i = m +1

i

,

n > m > n0

is convergent.

is divergent then {Sn } is unbounded. ⇒ ∃ a real number β > 0 such that n

n

∑ b >λβ

i = m +1

,

i

n>m

from (i) n

1 n ⇒ ∑ bi > ∑ ai > β λ i =m+1 i =m +1 ⇒ ∑ bn is convergent.

,

n>m

∑a

n

is diverges.

Sequences and Series

- 18 -

Example We know that

1

∑n

is divergent and

n≥ n ∀ n ≥1 1 1 ⇒ ≤ n n 1 is divergent as ⇒ ∑ n

1

∑n

is divergent.

Example The series Let

1

∑n

Sn = 1 +

If α > 1 then

α

is convergent if α > 1 and diverges if α ≤ 1 .

1 1 1 + + .................. + 2α 3α nα

Sn < S2n

and

1 1 < nα (n − 1)α

 1 1 1 1  Now S 2 n = 1 + α + α + α ............ + 3 4 (2n)α   2   1 1 1 1 1 1 1  = 1 + α + α + ............ + + + + + ............ + 5 (2n − 1)α   2α 4α 6α (2n)α   3   1  1 1 1 1 1 1  = 1 + α + α + ............ + + α 1 + α + α + ............ + α  α  5 (2n − 1)  2  2 3 ( n)   3   1 1 1 1 Sn < 1 + α + α + ............ + + α  4 (2n − 2)  2α  2 replacing 3 by 2, 5 by 4 and so on.

1  1 1  1 1 + + ............ + + Sn 2α  2α (n − 1)α  2α 1 1 1 1 =1 + α Sn −1 + α S n =1 + α S 2 n + α S2 n Q Sn −1 < S n < S 2 n 2 2 2 2 2 =1 + α S 2 n 2 1 ⇒ S 2 n < 1 + α −1 S 2 n 2  2α −1 − 1  1  2α −1  ⇒ 1 − α −1  S 2 n < 1 ⇒  α −1  S 2 n < 1 ⇒ S 2 n < α −1 2 −1  2   2  2α −1 i.e. S n < S 2 n < α −1 2 −1 1 ⇒ {Sn } is bounded and also monotonic. Hence we conclude that ∑ α is n convergent when α > 1 . If α ≤ 1 then nα ≤ n ∀ n ≥ 1 1 1 ⇒ α ≥ ∀ n ≥1 n n 1 1 Q ∑ is divergent therefore ∑ α is divergent when α ≤ 1. n n =1 +

Sequences and Series

- 19 -

Theorem an = λ ≠ 0 then the series n→∞ b n

Let an > 0 , bn > 0 and lim

∑a

n

and

∑b

n

behave alike.

Proof an =λ n→∞ b n

Since

lim

an − λ < ε ∀ n ≥ n0 . bn λ Use ε = 2 a λ ⇒ n −λ < ∀ n ≥ n0 . 2 bn λ a λ ⇒ λ− < n <λ+ 2 bn 2 λ an 3λ ⇒ < < 2 bn 2 then we got 3λ 2 an < bn and bn < an 2 λ Hence by comparison test we conclude that together. ⇒

∑a

n

and

∑b

n

converge or diverge

Example x diverges or converges consider n 1 x 1 an = sin 2 and take bn = 3 n n n an x = n 2 sin 2 then bn n To check

1

∑ n sin

2

2

x x  sin sin   n = x2 n =  1 x    n2  n  Applying limit as n → ∞ 2 2 x x   sin  sin  an 2 2 n n = x 2 (1) = x 2 lim = lim x   = x  lim  n→∞ b n→∞ n→∞ x x n     n   n   ⇒ ∑ an and ∑ bn have the similar behavior ∀ finite values of x except x = 0. 2

Since

1

∑n

3

is convergent series therefore the given series is also convergent for

finite values of x except x = 0. ï⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ð

Sequences and Series

- 20 -

Theorem ( Cauchy Condensation Test ) Let an ≥ 0 , an > an +1 ∀ n ≥ 1 , then the series ∑ an and diverges together. Proof Let us suppose S n = a1 + a2 + a3 + ............... + an

∑2

n −1

a2n −1 converges or

tn = a1 + 2a2 + 2 2 a22 + ................ + 2 n−1 a2n−1 .

and

Q an ≥ 0 and n < 2n−1 < 2n − 1 ∴ Sn < S2n−1 < S2n −1 for n > 2 then S2n −1 = a1 + a2 + a3 + ...... + a2n −1

( + a ) + ....... + ( a

= a1 + ( a2 + a3 ) + ( a4 + a5 + a6 + a7 ) + ....... + a2n−1 + a2n −1+1 + a2n −1+ 2 + ..... + a2n −1 < a1 + ( a2 + a2 ) + ( a4 + a4 + a4

4

< a1 + 2 a2 + 22 a4 + ....... + 2 n−1 a2n −1 = tn ⇒ S n < tn ⇒ Sn < tn < 2S2n …………. (i) Now consider S2n = a1 + a2 + a3 + ............... + a2n

2n −1

+ a2n −1 + a2n−1 + ..... + a2n−1

)

(

)

= a1 + a2 + ( a3 + a4 ) + ( a5 + a6 + a7 + a8 ) + ....... + a2n−1+1 + a2n−1+ 2 + a2n−1+3 + .... + a2n

(

)

1 > a1 + a2 + ( a4 + a4 ) + ( a8 + a8 + a8 + a8 ) + ...... + a2n + a2n + a2n + ..... + a2n 2 1 = a1 + a2 + 2 a4 + 2 2 a8 + ................ + 2 n−1 a2n 2 1 = a1 + 2 a2 + 2 3a4 + 23 a8 + ................ + 2 n a2n 2 1 ⇒ S 2 n > t n ………… (ii) 2 ⇒ 2 S 2 n > tn From (i) and (ii) we see that the sequence S n and tn are either both bounded or both

(

)

unbounded, implies that

∑a

n

and

∑2

n −1

a2n −1 converges or diverges together.

Example Consider the series

1

∑n

p

1 ≠0 n→∞ n p therefore the series diverges when p ≤ 0 . If p > 0 then the condensation test is applicable and we are lead to the series ∞ ∞ 1 1 k 2 k p = ∑ kp −k ∑ (2 ) k =0 k =0 2 If p ≤ 0 then lim

∞

=∑ =

k =0 ∞

1 2( p −1) k

∑2

∞

 1  = ∑  ( p −1)   k =0  2

k

(1− p ) k

k =0

1− p

Now 2

< 1 iff 1 − p < 0 i.e. when p > 1

)

Sequences and Series

And the result follows by comparing this series with the geometric series having common ratio less than one. The series diverges when 21− p = 1 ( i.e. when p = 1 ) The series is also divergent if 0 < p < 1. Example If p > 1 ,

∞

∑ n (ln n) 1

n= 2

p

converges and

If p ≤ 1 the series is divergent.  1  ∴   decreases  n ln n  and we can use the condensation test to the above series. 1 We have an = p n ( ln n ) 1 1 ⇒ 2 n a2n = ⇒ a2n = p p ( n ln 2 ) 2 n ln 2 n

Q {ln n} is increasing

⇒

(

)

we have the series 1 1 1 ∑ 2n a2n = ∑ (n ln 2) p = ( ln 2 ) p ∑ n p

which converges when p > 1 and diverges when p ≤ 1 . Example Consider

1

∑ ln n

 1  Since {ln n} is increasing there   decreases.  ln n  And we can apply the condensation test to check the behavior of the series 1 1 Q an = ∴ a2n = ln n ln 2 n 2n 2n n n ⇒ 2 a2n = so 2 a2n = ln 2n n ln 2 n 2 1 since > ∀ n ≥1 n n 1 and ∑ is diverges therefore the given series is also diverges. n ï⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ð

- 21 -

Sequences and Series

- 22 -

Alternating Series A series in which successive terms have opposite signs is called an alternating series. (−1)n +1 1 1 1 e.g. ∑ = 1 − + − + ............... is an alternating series. n 2 3 4 Theorem (Alternating Series Test or Leibniz Test) Let {an } be a decreasing sequence of positive numbers such that lim an = 0 then the n→∞

∞

alternating series

∑ (−1)

n+1

n=1

an = a1 − a2 + a3 − a4 + ................ converges.

Proof Looking at the odd numbered partial sums of this series we find that S 2 n +1 = (a1 − a2 ) + (a3 − a4 ) + (a5 − a6 ) + ........... + (a2 n−1 − a2 n ) + a2 n+1 Since {an } is decreasing therefore all the terms in the parenthesis are non-negative ⇒ S 2 n +1 > 0 ∀ n Moreover S 2 n +3 = S 2 n+1 − a2 n+2 + a2 n+3 = S2 n +1 − ( a2 n +2 − a2 n+3 ) Since a2 n+ 2 − a2 n +3 ≥ 0 therefore S 2 n +3 ≤ S 2 n +1 Hence the sequence of odd numbered partial sum is decreasing and is bounded below by zero. (as it has +ive terms) It is therefore convergent. Thus S 2 n+1 converges to some limit l (say). Now consider the even numbered partial sum. We find that S 2 n +2 = S2 n+1 − a2 n+ 2 and lim S 2 n+ 2 = lim ( S 2 n+1 − a2 n +2 ) n→∞

n→∞

= lim S 2 n+1 − lim a2 n+2 n→∞

=l −0 =l

n→∞

Q lim an = 0 n→∞

so that the even partial sum is also convergent to l . ⇒ both sequences of odd and even partial sums converge to the same limit. Hence we conclude that the corresponding series is convergent. Absolute Convergence ∑ an is said to converge absolutely if

∑a

n

converges.

Theorem An absolutely convergent series is convergent. Proof: If ∑ an is convergent then for a real number ε > 0 , ∃ a positive integer n0 such that n

∑a

i = m +1

⇒ the series

<

i

n

∑

∑a

i = m +1 n

ai < ε

∀ n, m > n0

is convergent. (Cauchy Criterion has been used)

Note The converse of the above theorem does not hold. (−1)n +1 1 e.g. ∑ is convergent but ∑ is divergent. n n

Sequences and Series

- 23 -

Theorem (The Root Test) Let Then

lim Sup an

∑a

1

n

n→∞

n

=p

converges absolutely if p < 1 and it diverges if p > 1 .

Proof Let p < 1 then we can find the positive number ε > 0 such that p + ε < 1 ⇒ an

1

n

< p +ε < 1

∀ n > n0

⇒ an < ( p + ε ) < 1 n

∑ ( p + ε ) is convergent because it is a geometric series with ∴ ∑ a is convergent ⇒ ∑ a converges absolutely.

Q

n

r < 1.

n

n

Now let p > 1 then we can find a number ε1 > 0 such that p − ε1 > 1 . ⇒ an

1

n

> p +ε > 1

⇒ an > 1 for infinitely many values of n. ⇒ lim an ≠ 0 n→∞

⇒

∑a

n

is divergent.

Note:

The above test give no information when p = 1 . 1 1 e.g. Consider the series ∑ and ∑ 2 . n n 1 For each of these series p = 1 , but ∑ is divergent and n

1

∑n

2

is convergent.

Theorem (Ratio Test) The series ∑ an (i)

Converges if lim Sup n→∞

(ii) Diverges if

an+1 <1 an

an +1 > 1 for n ≥ n0 , where n0 is some fixed integer. an

Proof

If (i) holds we can find β < 1 and integer N such that an +1 < β for n ≥ N an

In particular a N +1 < β aN

⇒ aN +1 < β aN ⇒ aN +2 < β aN +1 < β 2 aN ⇒ a N +3 < β 3 a N ……………………. ……………………. ……………………. ⇒ a N + p < β p aN

Sequences and Series

⇒ an < β n − N aN

∑β

n

we put N + p = n .

an < aN β − N β n for n ≥ N .

i.e.

Q

- 24 -

is convergent because it is geometric series with common ration < 1 .

Therefore Now if

∑a

n

is convergent (by comparison test)

an+1 ≥ an then lim an ≠ 0

for n ≥ n0

n→∞

⇒

∑a

n

is divergent.

Note The knowledge

an +1 = 1 implies nothing about the convergent or divergent of series. an

Example n+1  n  n   − Consider the series ∑ an with an =    n + 1  n + 1    n Q <1 ∴ an > 0 ∀ n . n +1

Also

( an )

1 n

n+1  n  n   =  −    n + 1  n + 1  

−1

n  n +1   n   =   1 −     n    n + 1  

−1

−n  1   1  = 1 +  1 − 1 +    n    n  

−n  n +1   n +1  =   1 −     n    n  

−1

  1 − n   1 = lim 1 +  lim 1 − 1 +   n→∞  n  n→∞   n   = 1 ⋅ 1 − e 

⇒ the series is divergent.

 1 = 1 −   e

−1

−1

−n  1   1  lim an = lim 1 +  1 − 1 +   n→∞ n→∞  n    n  

−1 −1

−n

−1

−1

 e − 1 =   e 

−1

 e  =   e − 1 

>1

Theorem (Dirichlet) Suppose that {Sn } , S n = a1 + a2 + a3 + ............. + an is bounded. Let { bn } be positive term decreasing sequence such that lim bn = 0 , then n→∞

Proof Q {Sn } is bounded ∴ ∃ a positive number λ such that ∀ n ≥ 1. Sn < λ Then

ai bi = ( Si − Si −1 ) bi for i ≥ 2 = Si bi − Si−1 bi = Si bi − Si−1 bi + Si bi +1 − Si bi +1

∑a

n

bn is convergent.

Sequences and Series

- 25 -

= Si ( bi − bi+1 ) − Si −1 bi + Si bi +1 n

∑

⇒

i = m +1

ai bi =

n

∑ S (b − b ) − ( S i

i = m +1

i +1

i

m

bm+1 − S n bn +1 )

{ bn } is decreasing

Q

n

∑ab

∴

i

i = m +1

n

∑ S (b − b ) − S

=

i

i = m+1

i

i +1

i

m

n

∑ { S (b − b )} +

<

i

i = m +1 n

i +1

i

bm+1 + Sn bn+1 S m bm+1 + S n bn +1

∑ {λ (b − b )} + λ b

<

i

i = m +1

i +1

Q Si < λ

+ λ bn+1

m +1

 n  = λ  ∑ ( bi − bi+1 ) + bm+1 + bn +1   i =m+1  = λ ( ( bm+1 − bn+1 ) + bm+1 + bn +1 ) = 2λ ( bm+1 ) ⇒

n

∑ab

i = m +1

i

∑a

⇒ The

n

Theorem Suppose that then

∑a

n

where ε = 2λ ( bm+1 ) a certain number

<ε

i

bn is convergent. ( We have use Cauchy Criterion here. )

∑a

n

is convergent and that { bn } is monotonic convergent sequence

bn is also convergent.

Proof Suppose { bn } is decreasing and it converges to b . Put cn = bn − b ⇒ cn ≥ 0 and lim cn = 0 Q

∑a

n→∞

is convergent

n

∴ {Sn } , S n = a1 + a2 + a3 + ................. + an is convergent ⇒ It is bounded ⇒ ∑ an cn is bounded. Q an bn = an cn + an b and ∴

∑a

n

∑a

n

cn and

∑a

n

b are convergent.

bn is convergent.

Now if { bn } is increasing and converges to b then we shall put cn = b − bn . Example 1

∑ (n ln n)

α

is convergent if α > 1 and divergent if α ≤ 1.

To see this we proceed as follows 1 an = (n ln n)α Take bn = 2 a2n = n

=

2n

=

( 2n ln 2n )

α

2n 2nα nα ( ln 2 )

α

(2 =

2n

n

n ln 2 )

α

1 2nα −n nα ( ln 2 )

α

Sequences and Series

- 26 -

(α −1) n

1

=

( ln 2 )

α

1   2 ⋅ α n

1 1 Since ∑ α is convergent when α > 1 and   n 2 converges to 0. Therefore ∑ bn is convergent

(α −1) n

∑ a is also convergent. Now ∑ b is divergent for α ≤ 1 therefore ∑ a

is decreasing for α > 1 and it

⇒

n

n

n

diverges for α ≤ 1.

Example To check We have an = Take

∑n

α

1 is convergent or divergent. ln n

1 n ln n α

bn = 2 a2n n

2n = (2 n )α (ln 2 n )

2n = nα 2 (n ln 2) (α −1) n

1 2(1−α ) n = ⋅ n ln 2 1 Q ∑ is divergent although n

∑ b is divergent. ⇒ ∑ a is divergent.

therefore

1 1  2  = ⋅ n ln 2

 1 n (α −1)     is decreasing, tending to zero for α > 1  2  

n

n

The series also divergent if α ≤ 1. i.e. it is always divergent. References:

(1) Lectures (2003-04) Prof. Syyed Gull Shah

Chairman, Department of Mathematics. University of Sargodha, Sargodha.

(2) Book Principles of Mathematical Analysis Walter Rudin (McGraw-Hill, Inc.)

Made by: Atiq ur Rehman ([email protected]) Available online at http://www.mathcity.org in PDF Format. Page Setup: Legal ( 8′′ 1 2 × 14′′ ) Printed: October 20, 2004. Updated: October 11, 2005