Sequences | Mathematics Optional Notes for UPSC PDF Download

 Page 1


Edurev123 
2. Sequences 
2.1 Show that a bounded infinite subset of?R must have a limit point. 
(2009 : 15 Marks) 
Solution: 
Let ?? be an infinite and bounded set. There exist and interval [?? ,?? ] such that ?? ?[?? ,?? ]. 
We define a set ?? as follows : 
?? ??? : It exceeds at the most a finite number of member of the set ?? . 
Thus, while ?? ??? and ?? ??? . 
Also, ?? is bounded above in as much as ?? is an upper bounded of the same. Let ?? the 
least upper bound of 
?? . Surely it exists by the order completeness properly of R. 
We show that ?? is a limit point of ?? 
Consider a neighbourhood say ?? of ?? , 
?? ?(?? ,?? )??? 
Now the number ' ?? ' which is less than the least upper bound ?? of the set is not an 
upper bound of ?? . Thus there exist a number, ?? (sequence) of ?? such that 
?? <?? =?? ,?? ??? 
Also ?? being a member of ?? exceeds at not a finite number of member of ?? : It follower 
and that the number ' ?? ' also exceeds at the most a finite a finite number of member of 
?? . 
Again the number ?? which is greater than ?? is an upper bound of ?? without being a 
member of ?? . Thus must exceed an infinite number of member of ?? . 
It follows that 
(i) ' ?? ' exceeds at the most a finite number of member of ?? . 
(ii) ' ?? ' exceeds an infinite number of members of ?? . 
Thus, [?? ,?? ] contains an infinite number of member of ?? . So, that ?? is the limit of point of 
?? . 
Page 2


Edurev123 
2. Sequences 
2.1 Show that a bounded infinite subset of?R must have a limit point. 
(2009 : 15 Marks) 
Solution: 
Let ?? be an infinite and bounded set. There exist and interval [?? ,?? ] such that ?? ?[?? ,?? ]. 
We define a set ?? as follows : 
?? ??? : It exceeds at the most a finite number of member of the set ?? . 
Thus, while ?? ??? and ?? ??? . 
Also, ?? is bounded above in as much as ?? is an upper bounded of the same. Let ?? the 
least upper bound of 
?? . Surely it exists by the order completeness properly of R. 
We show that ?? is a limit point of ?? 
Consider a neighbourhood say ?? of ?? , 
?? ?(?? ,?? )??? 
Now the number ' ?? ' which is less than the least upper bound ?? of the set is not an 
upper bound of ?? . Thus there exist a number, ?? (sequence) of ?? such that 
?? <?? =?? ,?? ??? 
Also ?? being a member of ?? exceeds at not a finite number of member of ?? : It follower 
and that the number ' ?? ' also exceeds at the most a finite a finite number of member of 
?? . 
Again the number ?? which is greater than ?? is an upper bound of ?? without being a 
member of ?? . Thus must exceed an infinite number of member of ?? . 
It follows that 
(i) ' ?? ' exceeds at the most a finite number of member of ?? . 
(ii) ' ?? ' exceeds an infinite number of members of ?? . 
Thus, [?? ,?? ] contains an infinite number of member of ?? . So, that ?? is the limit of point of 
?? . 
2.2 Discuss the convergence of the sequence {?? ?? } where ?? ?? =
?????? ?
????
?? ?? . 
(2010 : 12 Marks) 
Solution: 
Given: 
 Sequence {?? ?? }={
1
8
,0,-
1
8
,0,
1
8
,0,-
1
8
,….} 
So, the given sequence {?? ?? } assumes 3 values viz., 0,-
1
8
 and 
1
8
 and is oscillatory in 
nature. 
?{?? ?? } does not converge. 
2.3 Define {?? ?? } by ?? ?? =?? and ?? ?? +?? =v?? +?? ?? for ?? >?? . Show that the sequence 
converges to 
?? +v????
?? . 
(2010 : 12 Marks) 
Solution: 
 Given: ?? 1
=5 and ?? ?? +1
=v4+?? ?? for ?? >1
??
?? 2
?=v4+?? 1
=v4+5=v9=3
?? 3
?=v4+?? 2
=v4+3=v7
 
Let ?? =1 
???????????????????????????????????????????????? 2
<?? 1
 
? True for ?? =1. 
Let it is also true for ?? ??? . 
?? ?? +1
?<?? ?? ?? ?? +1
+4 ?<?? ?? +4
v?? ?? +1
+4 ?<v?? ?? +4
?? ?? +2
?<?? ?? +1
 
? True for ?? +1 also. 
So, by mathematical induction it is true for all ?? ??? . 
?{?? ?? } is monotonically decreasing sequence. 
Now, 
Page 3


Edurev123 
2. Sequences 
2.1 Show that a bounded infinite subset of?R must have a limit point. 
(2009 : 15 Marks) 
Solution: 
Let ?? be an infinite and bounded set. There exist and interval [?? ,?? ] such that ?? ?[?? ,?? ]. 
We define a set ?? as follows : 
?? ??? : It exceeds at the most a finite number of member of the set ?? . 
Thus, while ?? ??? and ?? ??? . 
Also, ?? is bounded above in as much as ?? is an upper bounded of the same. Let ?? the 
least upper bound of 
?? . Surely it exists by the order completeness properly of R. 
We show that ?? is a limit point of ?? 
Consider a neighbourhood say ?? of ?? , 
?? ?(?? ,?? )??? 
Now the number ' ?? ' which is less than the least upper bound ?? of the set is not an 
upper bound of ?? . Thus there exist a number, ?? (sequence) of ?? such that 
?? <?? =?? ,?? ??? 
Also ?? being a member of ?? exceeds at not a finite number of member of ?? : It follower 
and that the number ' ?? ' also exceeds at the most a finite a finite number of member of 
?? . 
Again the number ?? which is greater than ?? is an upper bound of ?? without being a 
member of ?? . Thus must exceed an infinite number of member of ?? . 
It follows that 
(i) ' ?? ' exceeds at the most a finite number of member of ?? . 
(ii) ' ?? ' exceeds an infinite number of members of ?? . 
Thus, [?? ,?? ] contains an infinite number of member of ?? . So, that ?? is the limit of point of 
?? . 
2.2 Discuss the convergence of the sequence {?? ?? } where ?? ?? =
?????? ?
????
?? ?? . 
(2010 : 12 Marks) 
Solution: 
Given: 
 Sequence {?? ?? }={
1
8
,0,-
1
8
,0,
1
8
,0,-
1
8
,….} 
So, the given sequence {?? ?? } assumes 3 values viz., 0,-
1
8
 and 
1
8
 and is oscillatory in 
nature. 
?{?? ?? } does not converge. 
2.3 Define {?? ?? } by ?? ?? =?? and ?? ?? +?? =v?? +?? ?? for ?? >?? . Show that the sequence 
converges to 
?? +v????
?? . 
(2010 : 12 Marks) 
Solution: 
 Given: ?? 1
=5 and ?? ?? +1
=v4+?? ?? for ?? >1
??
?? 2
?=v4+?? 1
=v4+5=v9=3
?? 3
?=v4+?? 2
=v4+3=v7
 
Let ?? =1 
???????????????????????????????????????????????? 2
<?? 1
 
? True for ?? =1. 
Let it is also true for ?? ??? . 
?? ?? +1
?<?? ?? ?? ?? +1
+4 ?<?? ?? +4
v?? ?? +1
+4 ?<v?? ?? +4
?? ?? +2
?<?? ?? +1
 
? True for ?? +1 also. 
So, by mathematical induction it is true for all ?? ??? . 
?{?? ?? } is monotonically decreasing sequence. 
Now, 
?? 1
=5>2
?? 2
=3>2
?? 3
=v7>2
??
?? ?? +1
=v4+?? ?? >2 as ?? ?? >0
 
?{?? ?? } is bounded below. 
As {?? ?? } is monotonically decreasing and bounded below, ? it is convergent. 
Let it converges to ?? . (?? >0) . 
????? ?? +1
=v4+?? ?? ???lim
?? ?8
??? ?? +1= lim
?? ?8
?v4+?? ?? ????? =v4+?? ????? 2
=4+?? ????? 2
-?? -4=0
????? =
1±v1+16
2
=
1±v17
2
 Now, ??? >0
????? =
1+v17
2
 
2.4 Let ?? ?? (?? )=?? ?? on -?? <?? =?? for ?? =?? ,?? ,….. Find the limit function. is the 
convergence uniform? Justify your answer. 
(2010: 15 marks) 
Solution: 
Given : 
?? ?? (?? )=?? ?? 
If ?? ?1 : 
(?? ) = lim
?? ?8
??? ?? (?? )= lim
?? ?8
??? ?? =0 as |?? |<1
? |?? ?? (?? )-?? (?? )| =|?? ?? -0|=|?? ?? |
 At sup. |?? ?? (?? )-?? (?? )|,
?? ????
(?? ?? ) =0
? ?? ?? ?? -1
=0 at ?? =0
 At ?? =0, sup. |?? ?? (?? )-?? (?? )| =0.
 
So, limit function is 0 and is uniformly convergent. 
At ?? =1, 
Page 4


Edurev123 
2. Sequences 
2.1 Show that a bounded infinite subset of?R must have a limit point. 
(2009 : 15 Marks) 
Solution: 
Let ?? be an infinite and bounded set. There exist and interval [?? ,?? ] such that ?? ?[?? ,?? ]. 
We define a set ?? as follows : 
?? ??? : It exceeds at the most a finite number of member of the set ?? . 
Thus, while ?? ??? and ?? ??? . 
Also, ?? is bounded above in as much as ?? is an upper bounded of the same. Let ?? the 
least upper bound of 
?? . Surely it exists by the order completeness properly of R. 
We show that ?? is a limit point of ?? 
Consider a neighbourhood say ?? of ?? , 
?? ?(?? ,?? )??? 
Now the number ' ?? ' which is less than the least upper bound ?? of the set is not an 
upper bound of ?? . Thus there exist a number, ?? (sequence) of ?? such that 
?? <?? =?? ,?? ??? 
Also ?? being a member of ?? exceeds at not a finite number of member of ?? : It follower 
and that the number ' ?? ' also exceeds at the most a finite a finite number of member of 
?? . 
Again the number ?? which is greater than ?? is an upper bound of ?? without being a 
member of ?? . Thus must exceed an infinite number of member of ?? . 
It follows that 
(i) ' ?? ' exceeds at the most a finite number of member of ?? . 
(ii) ' ?? ' exceeds an infinite number of members of ?? . 
Thus, [?? ,?? ] contains an infinite number of member of ?? . So, that ?? is the limit of point of 
?? . 
2.2 Discuss the convergence of the sequence {?? ?? } where ?? ?? =
?????? ?
????
?? ?? . 
(2010 : 12 Marks) 
Solution: 
Given: 
 Sequence {?? ?? }={
1
8
,0,-
1
8
,0,
1
8
,0,-
1
8
,….} 
So, the given sequence {?? ?? } assumes 3 values viz., 0,-
1
8
 and 
1
8
 and is oscillatory in 
nature. 
?{?? ?? } does not converge. 
2.3 Define {?? ?? } by ?? ?? =?? and ?? ?? +?? =v?? +?? ?? for ?? >?? . Show that the sequence 
converges to 
?? +v????
?? . 
(2010 : 12 Marks) 
Solution: 
 Given: ?? 1
=5 and ?? ?? +1
=v4+?? ?? for ?? >1
??
?? 2
?=v4+?? 1
=v4+5=v9=3
?? 3
?=v4+?? 2
=v4+3=v7
 
Let ?? =1 
???????????????????????????????????????????????? 2
<?? 1
 
? True for ?? =1. 
Let it is also true for ?? ??? . 
?? ?? +1
?<?? ?? ?? ?? +1
+4 ?<?? ?? +4
v?? ?? +1
+4 ?<v?? ?? +4
?? ?? +2
?<?? ?? +1
 
? True for ?? +1 also. 
So, by mathematical induction it is true for all ?? ??? . 
?{?? ?? } is monotonically decreasing sequence. 
Now, 
?? 1
=5>2
?? 2
=3>2
?? 3
=v7>2
??
?? ?? +1
=v4+?? ?? >2 as ?? ?? >0
 
?{?? ?? } is bounded below. 
As {?? ?? } is monotonically decreasing and bounded below, ? it is convergent. 
Let it converges to ?? . (?? >0) . 
????? ?? +1
=v4+?? ?? ???lim
?? ?8
??? ?? +1= lim
?? ?8
?v4+?? ?? ????? =v4+?? ????? 2
=4+?? ????? 2
-?? -4=0
????? =
1±v1+16
2
=
1±v17
2
 Now, ??? >0
????? =
1+v17
2
 
2.4 Let ?? ?? (?? )=?? ?? on -?? <?? =?? for ?? =?? ,?? ,….. Find the limit function. is the 
convergence uniform? Justify your answer. 
(2010: 15 marks) 
Solution: 
Given : 
?? ?? (?? )=?? ?? 
If ?? ?1 : 
(?? ) = lim
?? ?8
??? ?? (?? )= lim
?? ?8
??? ?? =0 as |?? |<1
? |?? ?? (?? )-?? (?? )| =|?? ?? -0|=|?? ?? |
 At sup. |?? ?? (?? )-?? (?? )|,
?? ????
(?? ?? ) =0
? ?? ?? ?? -1
=0 at ?? =0
 At ?? =0, sup. |?? ?? (?? )-?? (?? )| =0.
 
So, limit function is 0 and is uniformly convergent. 
At ?? =1, 
?? ?? (?? )=1
?? ?8 as ?? ?8 
? At ?? ?8 
?? ?? (?? )={
0 if ?? ?(-1,1)
1 if ?? =1
 
? In the interval (-1,1],?? ?? (?? ) is discontinuous at ?? =1 when ?? ?8. 
So, ?? ?? (?? ) is not uniformly convergent on (-1,1] 
2.5 Let ?? ?? (?? )=???? (?? -?? )?? ,?? ?[?? ,?? ]. Examine the uniform convergence of {?? ?? (?? )} 
on [?? ,?? ]. 
(2011 : 15 Marks) 
Solution: 
Given                                      
?? ?? (?? )=???? (1-?? )
?? ,?? ?[0,1] 
At ?? =0,??????????????????????????????????lim
?? ?8
??? ?? (?? )= lim
?? ?8
?0=0 
At ?? =1,                                lim
?? ?8
??? ?? (?? )= lim
?? ?8
?0=0 
For 0<?? <1, we have          
-1?<-?? <0
0?<1-?? <1
 
???????????????????????????????????????????????????lim
?? ?8
??? ?? (?? )= lim
?? ?8
????? (1-?? )
?? =0=?? (?? ) (say) 
??????????????????????????????????????????????????lim
?? ?8
??? ?? (?? )=0=?? (?? )??? ?[0,1] 
Again, define                           
?? ?? = sup
?? ?[0,1]
?|?? ?? (?? )-?? (?? )| 
Then, define                                   
?? ?? = sup
?? ?[0,1]
?|?? ?? (?? )-?? (?? )| 
Then,                                            {?? ?? }? f uniformly ?lim
?? ?8
??? ?? =0 
Now,                                                  ?? ?? =sup
?? ?[0,?? ]
?|???? (?? -?? )
?? | 
Let 
???????????????????????????????????????????????????????????
?? (?? )?=???? (1-?? )
?? '
(?? )?=?? (1-?? )
?? -?? 2
?? (1-?? )
?? -1
 
Page 5


Edurev123 
2. Sequences 
2.1 Show that a bounded infinite subset of?R must have a limit point. 
(2009 : 15 Marks) 
Solution: 
Let ?? be an infinite and bounded set. There exist and interval [?? ,?? ] such that ?? ?[?? ,?? ]. 
We define a set ?? as follows : 
?? ??? : It exceeds at the most a finite number of member of the set ?? . 
Thus, while ?? ??? and ?? ??? . 
Also, ?? is bounded above in as much as ?? is an upper bounded of the same. Let ?? the 
least upper bound of 
?? . Surely it exists by the order completeness properly of R. 
We show that ?? is a limit point of ?? 
Consider a neighbourhood say ?? of ?? , 
?? ?(?? ,?? )??? 
Now the number ' ?? ' which is less than the least upper bound ?? of the set is not an 
upper bound of ?? . Thus there exist a number, ?? (sequence) of ?? such that 
?? <?? =?? ,?? ??? 
Also ?? being a member of ?? exceeds at not a finite number of member of ?? : It follower 
and that the number ' ?? ' also exceeds at the most a finite a finite number of member of 
?? . 
Again the number ?? which is greater than ?? is an upper bound of ?? without being a 
member of ?? . Thus must exceed an infinite number of member of ?? . 
It follows that 
(i) ' ?? ' exceeds at the most a finite number of member of ?? . 
(ii) ' ?? ' exceeds an infinite number of members of ?? . 
Thus, [?? ,?? ] contains an infinite number of member of ?? . So, that ?? is the limit of point of 
?? . 
2.2 Discuss the convergence of the sequence {?? ?? } where ?? ?? =
?????? ?
????
?? ?? . 
(2010 : 12 Marks) 
Solution: 
Given: 
 Sequence {?? ?? }={
1
8
,0,-
1
8
,0,
1
8
,0,-
1
8
,….} 
So, the given sequence {?? ?? } assumes 3 values viz., 0,-
1
8
 and 
1
8
 and is oscillatory in 
nature. 
?{?? ?? } does not converge. 
2.3 Define {?? ?? } by ?? ?? =?? and ?? ?? +?? =v?? +?? ?? for ?? >?? . Show that the sequence 
converges to 
?? +v????
?? . 
(2010 : 12 Marks) 
Solution: 
 Given: ?? 1
=5 and ?? ?? +1
=v4+?? ?? for ?? >1
??
?? 2
?=v4+?? 1
=v4+5=v9=3
?? 3
?=v4+?? 2
=v4+3=v7
 
Let ?? =1 
???????????????????????????????????????????????? 2
<?? 1
 
? True for ?? =1. 
Let it is also true for ?? ??? . 
?? ?? +1
?<?? ?? ?? ?? +1
+4 ?<?? ?? +4
v?? ?? +1
+4 ?<v?? ?? +4
?? ?? +2
?<?? ?? +1
 
? True for ?? +1 also. 
So, by mathematical induction it is true for all ?? ??? . 
?{?? ?? } is monotonically decreasing sequence. 
Now, 
?? 1
=5>2
?? 2
=3>2
?? 3
=v7>2
??
?? ?? +1
=v4+?? ?? >2 as ?? ?? >0
 
?{?? ?? } is bounded below. 
As {?? ?? } is monotonically decreasing and bounded below, ? it is convergent. 
Let it converges to ?? . (?? >0) . 
????? ?? +1
=v4+?? ?? ???lim
?? ?8
??? ?? +1= lim
?? ?8
?v4+?? ?? ????? =v4+?? ????? 2
=4+?? ????? 2
-?? -4=0
????? =
1±v1+16
2
=
1±v17
2
 Now, ??? >0
????? =
1+v17
2
 
2.4 Let ?? ?? (?? )=?? ?? on -?? <?? =?? for ?? =?? ,?? ,….. Find the limit function. is the 
convergence uniform? Justify your answer. 
(2010: 15 marks) 
Solution: 
Given : 
?? ?? (?? )=?? ?? 
If ?? ?1 : 
(?? ) = lim
?? ?8
??? ?? (?? )= lim
?? ?8
??? ?? =0 as |?? |<1
? |?? ?? (?? )-?? (?? )| =|?? ?? -0|=|?? ?? |
 At sup. |?? ?? (?? )-?? (?? )|,
?? ????
(?? ?? ) =0
? ?? ?? ?? -1
=0 at ?? =0
 At ?? =0, sup. |?? ?? (?? )-?? (?? )| =0.
 
So, limit function is 0 and is uniformly convergent. 
At ?? =1, 
?? ?? (?? )=1
?? ?8 as ?? ?8 
? At ?? ?8 
?? ?? (?? )={
0 if ?? ?(-1,1)
1 if ?? =1
 
? In the interval (-1,1],?? ?? (?? ) is discontinuous at ?? =1 when ?? ?8. 
So, ?? ?? (?? ) is not uniformly convergent on (-1,1] 
2.5 Let ?? ?? (?? )=???? (?? -?? )?? ,?? ?[?? ,?? ]. Examine the uniform convergence of {?? ?? (?? )} 
on [?? ,?? ]. 
(2011 : 15 Marks) 
Solution: 
Given                                      
?? ?? (?? )=???? (1-?? )
?? ,?? ?[0,1] 
At ?? =0,??????????????????????????????????lim
?? ?8
??? ?? (?? )= lim
?? ?8
?0=0 
At ?? =1,                                lim
?? ?8
??? ?? (?? )= lim
?? ?8
?0=0 
For 0<?? <1, we have          
-1?<-?? <0
0?<1-?? <1
 
???????????????????????????????????????????????????lim
?? ?8
??? ?? (?? )= lim
?? ?8
????? (1-?? )
?? =0=?? (?? ) (say) 
??????????????????????????????????????????????????lim
?? ?8
??? ?? (?? )=0=?? (?? )??? ?[0,1] 
Again, define                           
?? ?? = sup
?? ?[0,1]
?|?? ?? (?? )-?? (?? )| 
Then, define                                   
?? ?? = sup
?? ?[0,1]
?|?? ?? (?? )-?? (?? )| 
Then,                                            {?? ?? }? f uniformly ?lim
?? ?8
??? ?? =0 
Now,                                                  ?? ?? =sup
?? ?[0,?? ]
?|???? (?? -?? )
?? | 
Let 
???????????????????????????????????????????????????????????
?? (?? )?=???? (1-?? )
?? '
(?? )?=?? (1-?? )
?? -?? 2
?? (1-?? )
?? -1
 
??????????????????????????????????????????????????????????????????????=?? (1-?? )
?? -1
(1-?? -???? ) 
????????????????????????????????????????????????????????????? '
(?? )=0
? 
?
 
??????????????????????? (1-?? )
?? -1
(1-?? -???? )=0 
                                                            ?? =
1
?? +1
 
??????????????????????????????????????????????????????????????? ?? =sup
?? ?[0,1]
?|???? (1-?? )
?? | 
????????????????????????????????????????????????????????????????????=|?? ·
1
?? +1
·(1-
1
?? +1
)
?? | 
?????????????????????????????????????????????????????????????????????=(
?? ?? +1
)
?? +1
 
????????????????????????????????????????????????lim
?? ?8
??? ?? =lim
?? ?8
?(
?? ?? +1
)
?? +1
 
?????????????????????????????????????????????????????????????????????=lim
?? ?8
?(
?? ?? +1
)
?? ·(
?? ?? +1
) 
?????????????????????????????????????????????????????????????????????=lim
?? ?8
?(
1
1+
1
?? )
?? ·(
1
1+
1
?? ) 
=
1
?? ·1=
1
?? ?0 
?[?? ?? (?? )] is not uniformly convergent on [0,1]. 
2.6 Two sequences {?? ?? } and {?? ?? } are defined inductively by the following : 
?? ?? ?=
?? ?? ,?? ?? =?? and ?? ?? =v?? ?? +?? ?? +?? ,?? =?? ,?? ,?? ?? ?? ?? ?=
?? ?? (
?? ?? ?? +
?? ?? ?? -?? ),?? =?? ,?? ,?? ,….
 
Prove that ?? ?? -?? <?? ?? <?? ?? <?? ?? -?? ,?? =?? ,?? ,?? ,…. and deduce that both the 
sequences comes to the same limit ?? , where 
?? ?? <?? <?? . 
(2016 : 10 Marks) 
Solution: 
Given: