Comprehensive Solutions: Convergence and Sequences in Math Chap 3

This document presents detailed solutions to exercises from Chapter 3 of a mathematics textbook, focusing on the concepts of convergence and divergence of sequences and series. The solutions cover problems related to determining the convergence of sequences using definitions and theorems, finding upper and lower limits of sequences, and applying convergence tests such as the comparison test, integral test, and ratio test to determine the convergence or divergence of infinite series. Additionally, the document includes solutions for finding the radius of convergence and interval of convergence for power series. The exercises also address the properties of Cauchy sequences and their relationship to convergent sequences, providing a comprehensive overview of key concepts in real analysis. Desklib offers a wide range of solved assignments and past papers for students.

Math
Chap. 3, Exercises: 3
Given that S1= √2 Sn +1= √ 2+ √ Sn n=1,2,3,4,5 … S1= √ 2 ∴ S1<2 S2= √ 2+S1 < √ 2+2=2
S3= √ 2+ S2 < √ 4=2∴ Sn <2
On the other hand
S1= √ 2 ∴ S1> 0S2= √2+ S1 > √2=S1 S3= √ 2+ S2 > √ 2+S1=S2 ∴ Sn >Sn−1
Combining Sn <2and Sn > Sn−1for all values of n, shows that the sequence converges; the
value 2
Chap. 3, Exercises: 4
The upper and the lower limit of the sequence defined by:
S1=0; S2 m = S2 m −1
2 ∧S2 m +1=1
2 +S2 m
By observing the sequence, we see that m=1,2,3,4 ,… ..S2 m=−1
2m + 1
2 S2 m+ 1=−1
2m +1
By induction on m, m=1, then S2= S1
2 =0=1
2 − 1
21
Suppose now S2 m=−1
2m + 1
2 for some integer m ≥1, thus we have, S2 (m +1 )=S2m +2= S (2 m−1 )
2
¿ 1
2 (S2m + 1
2 ) ¿ 1
2 ( 1− 1
2m )¿ 1
2 − 1
2m+1

Thus, S2 m=−1
2m + 1
2 for all m, consequently, the results obtained from the above argument
is:
lim
n → ∞
inf Sn= 1
2 ∧lim
n → ∞
¿ Sn=1
Chap. 3, Exercises: 6
Convergence and divergence of ∑ an
a) ∑ an=∑ √n+1− √n
By use of comparison test and let bn= 1
√ 2 n+1
lim
n → ∞
√ n+1− √ n
1
√ 2 n+1
= 1
√ 2
By the limit comparison test criteria ∑ an diverges
b) ∑ an=∑ √ n+1− √ n
n
Using Integral test∫
1
∞
√n+1− √n
n dn=−ln ( 3−2 √ 2 )−2 √2−2
Therefore, by integral test criteria ∑ an converges
c) ∑ an=∑ ( n
√n−1 )n
Using ratio test:
lim
n → ∞
an+1
an
= ( n +1
√ n+1−1 )
n+1
( n
√ n−1 )
n =∞
L>1 by the ratio test
Thus, using the ratio test criteria ∑ an diverges

d) ∑ an=∑ 1
1+ zn for complex value of z
Using ratio test:
lim
n → ∞
an+1
an
=
lim
n →∞
1
1+ zn+ 1
1
1+ zn
¿ lim
n → ∞ (| 1+ zn
1+ zn+ 1 |)= 1
z
L<1 by the ratio test
Thus, using the ratio test criteria ∑ an converges
Chap. 3, Exercises: 9.
Radius of convergence
a) ∑ n3 zn
Using ratio test:
lim
n → ∞
an+1
an
=lim
n → ∞ (| ( n+1 ) 3 zn +1
n3 zn |) ;|z|
The sum converges for L<1, now solving for |z|< 1, we get that the radius of convergence
R=1
And the interval of convergence (−1 , 1)
b) ∑ 2n
n ! zn
Using ratio test:

lim
n → ∞
an+1
an
=lim
n → ∞
(| 2n+1
(n+1)! zn+1
2n
n ! zn |) ; 0
The sum converges for L<1, now solving for |z|< 0, the sum converges for all z: we get
that the radius of convergence R=∞
And the interval of convergence (−∞, ∞)
c) ∑ 2n
n2 zn
Using ratio test:
lim
n → ∞
an+1
an
=lim
n → ∞
(| 2n+1
(n+1)2 zn +1
2n
n2 zn |) ;2∨z∨¿
The sum converges for L<1, now solving for 2|z|< 1, we get that the radius of
convergence R=1
2
And the interval of convergence [ −1
2 , 1
2 ]
d) ∑ n3
3n zn
Using ratio test:

lim
n → ∞
an+1
an
=lim
n → ∞
(| (n+1)3
3n+1 zn +1
n3
3n zn |);| z
3 |
The sum converges for L<1, now solving for| z
3 |<1, we get that the radius of convergence
R=3
And the interval of convergence (−3 , 3)
Chap. 3, Exercises: 20
Given ε > 0, as { pni
} converges to p, then a positive integer N x exists such that
i> Nx :d ( pni
, p)<ε. Because { pn } is Cauchy, then there exists a positive integer N y :d ( pm , p ) <ε
for all m ≥ N y∧n ≥ N y
By choosing N ≥ N ysuch that ni =N for some i≥ N x and then for all n ≥ N y
d ( pn , p ) ≤ d ( pn , pN ) + d ( pN , p ) <ε +ε=2 ε

Chap. 3, Exercises: 24.

Chap. 3, Exercise 22