Why $ sum_k=1^inftyfrac1k^1+x $ is not uniformly convergent on $ left(0,;+inftyright) $ [closed]

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Why this series of functions is not uniformly convergent on $ left(0,;+inftyright) $. beginequation
sum_k=1^inftyfrac1k^1+x
endequation

sequences-and-series convergence uniform-convergence

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asked Jul 21 at 21:22

Lithium

403

closed as off-topic by amWhy, Taroccoesbrocco, Parcly Taxel, Martin Sleziak, Jose Arnaldo Bebita Dris Jul 22 at 12:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Taroccoesbrocco, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you tried something or do you wish us to do your homework?
  – Mundron Schmidt
  Jul 21 at 21:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use the integral test for convergence
  – Michael
  Jul 21 at 21:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Uniform convergence of $sum_k=1^infty frac1k^1+x$
  – Martin Sleziak
  Jul 22 at 8:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is this question about the same series: Uniform convergence of $sum_k=1^infty frac1k^1+x$. Found using Approach0.
  – Martin Sleziak
  Jul 22 at 8:08
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Why this series of functions is not uniformly convergent on $ left(0,;+inftyright) $. beginequation
sum_k=1^inftyfrac1k^1+x
endequation

sequences-and-series convergence uniform-convergence

share|cite|improve this question

asked Jul 21 at 21:22

Lithium

403

closed as off-topic by amWhy, Taroccoesbrocco, Parcly Taxel, Martin Sleziak, Jose Arnaldo Bebita Dris Jul 22 at 12:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Taroccoesbrocco, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you tried something or do you wish us to do your homework?
  – Mundron Schmidt
  Jul 21 at 21:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use the integral test for convergence
  – Michael
  Jul 21 at 21:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Uniform convergence of $sum_k=1^infty frac1k^1+x$
  – Martin Sleziak
  Jul 22 at 8:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is this question about the same series: Uniform convergence of $sum_k=1^infty frac1k^1+x$. Found using Approach0.
  – Martin Sleziak
  Jul 22 at 8:08
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Why this series of functions is not uniformly convergent on $ left(0,;+inftyright) $. beginequation
sum_k=1^inftyfrac1k^1+x
endequation

sequences-and-series convergence uniform-convergence

share|cite|improve this question

asked Jul 21 at 21:22

Lithium

403

Why this series of functions is not uniformly convergent on $ left(0,;+inftyright) $. beginequation
sum_k=1^inftyfrac1k^1+x
endequation

sequences-and-series convergence uniform-convergence

share|cite|improve this question

asked Jul 21 at 21:22

Lithium

403

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 21:22

Lithium

403

asked Jul 21 at 21:22

Lithium

403

asked Jul 21 at 21:22

Lithium

403

403

closed as off-topic by amWhy, Taroccoesbrocco, Parcly Taxel, Martin Sleziak, Jose Arnaldo Bebita Dris Jul 22 at 12:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Taroccoesbrocco, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Taroccoesbrocco, Parcly Taxel, Martin Sleziak, Jose Arnaldo Bebita Dris Jul 22 at 12:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Taroccoesbrocco, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you tried something or do you wish us to do your homework?
  – Mundron Schmidt
  Jul 21 at 21:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use the integral test for convergence
  – Michael
  Jul 21 at 21:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Uniform convergence of $sum_k=1^infty frac1k^1+x$
  – Martin Sleziak
  Jul 22 at 8:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is this question about the same series: Uniform convergence of $sum_k=1^infty frac1k^1+x$. Found using Approach0.
  – Martin Sleziak
  Jul 22 at 8:08
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you tried something or do you wish us to do your homework?
  – Mundron Schmidt
  Jul 21 at 21:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use the integral test for convergence
  – Michael
  Jul 21 at 21:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Possible duplicate of Uniform convergence of $sum_k=1^infty frac1k^1+x$
  – Martin Sleziak
  Jul 22 at 8:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  There is this question about the same series: Uniform convergence of $sum_k=1^infty frac1k^1+x$. Found using Approach0.
  – Martin Sleziak
  Jul 22 at 8:08
  

  
  
  
  

Have you tried something or do you wish us to do your homework?
– Mundron Schmidt
Jul 21 at 21:27

Have you tried something or do you wish us to do your homework?
– Mundron Schmidt
Jul 21 at 21:27

Use the integral test for convergence
– Michael
Jul 21 at 21:47

Use the integral test for convergence
– Michael
Jul 21 at 21:47

1

1

Possible duplicate of Uniform convergence of $sum_k=1^infty frac1k^1+x$
– Martin Sleziak
Jul 22 at 8:08

Possible duplicate of Uniform convergence of $sum_k=1^infty frac1k^1+x$
– Martin Sleziak
Jul 22 at 8:08

There is this question about the same series: Uniform convergence of $sum_k=1^infty frac1k^1+x$. Found using Approach0.
– Martin Sleziak
Jul 22 at 8:08

There is this question about the same series: Uniform convergence of $sum_k=1^infty frac1k^1+x$. Found using Approach0.
– Martin Sleziak
Jul 22 at 8:08

add a comment | 

4 Answers
4

active

oldest

votes

up vote
1
down vote

Hint: What do you think happens for $xto 0^+$?

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answered Jul 21 at 21:26

Mundron Schmidt

7,1162727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I know it becomes a Harmonic series. But how can I show it is not uniformly convergent properly? For instance cauchy criterion
  – Lithium
  Jul 21 at 21:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lithium: what properties do uniformly convergent series have?
  – robjohn♦
  Jul 21 at 23:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

This is the famous Riemann $zeta$ function, shifted by 1 to the left:
https://en.wikipedia.org/wiki/Riemann_zeta_function

It is well-known that is has a singularity at 1 (in your notation due to the shift, at 0), and it is a simple pole. In other words, around 1, the $zeta$ function around 1 "looks like" $1/(x-1)$. To make it more precise, $zeta(x)-1/(x-1)$ is a smooth function on the positive half of the complex plane.

To translate all of this to your function: it has a simple pole at 0, and around 0, it looks like $1/x$, which is not uniformly continuous.

share|cite|improve this answer

answered Jul 21 at 21:30

A. Pongrácz

2,269221

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean?
  – A. Pongrácz
  Jul 21 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  any simpler method? for instance using cauchy criterion?
  – Lithium
  Jul 21 at 21:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint:

$$sup_x in (0,infty)sum_k=n+1^infty frac1k^1+x >sum_k=n+1^2n frac1k^1+1/n > fracn(2n)^1+1/n$$

share|cite|improve this answer

edited Jul 21 at 22:20

answered Jul 21 at 21:50

RRL

43.7k42260

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry did not get it. could you please explain thank you
  – Lithium
  Jul 21 at 22:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Uniform convergence requires the LHS to converge to $0$. What does the lower bound on the RHS converge to?
  – RRL
  Jul 21 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write what it means for a series to converge uniformly in your post and show it implies what I am stating.
  – RRL
  Jul 21 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why hasn't this answer received more up votes?? Here is one (+1).
  – Mark Viola
  Jul 22 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola: Thanks for being so conscientious (as usual).
  – RRL
  Jul 22 at 3:24
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

Uniform convergence means that $$sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| xrightarrowntoinfty 0$$

As @RRL suggests, for $ninmathbbN$ you can consider $x_n = frac1n$ so

beginalign
sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| &= sup_xin(0, +infty) sum_k=n+1^inftyfrac1k^1+x \
&ge sum_k=n+1^inftyfrac1k^1+frac1n\
&ge int_n+1^infty fracdtt^1+frac1n\
&= -fracnt^1/n Bigg|_n+1^infty\
&= fracnsqrt[n]n+1\
&xrightarrowntoinfty +infty
endalign

so it doesn't converge to $0$.

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answered Jul 21 at 23:35

mechanodroid

22.2k52041

add a comment | 

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Hint: What do you think happens for $xto 0^+$?

share|cite|improve this answer

answered Jul 21 at 21:26

Mundron Schmidt

7,1162727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I know it becomes a Harmonic series. But how can I show it is not uniformly convergent properly? For instance cauchy criterion
  – Lithium
  Jul 21 at 21:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lithium: what properties do uniformly convergent series have?
  – robjohn♦
  Jul 21 at 23:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint: What do you think happens for $xto 0^+$?

share|cite|improve this answer

answered Jul 21 at 21:26

Mundron Schmidt

7,1162727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I know it becomes a Harmonic series. But how can I show it is not uniformly convergent properly? For instance cauchy criterion
  – Lithium
  Jul 21 at 21:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lithium: what properties do uniformly convergent series have?
  – robjohn♦
  Jul 21 at 23:39
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint: What do you think happens for $xto 0^+$?

share|cite|improve this answer

answered Jul 21 at 21:26

Mundron Schmidt

7,1162727

Hint: What do you think happens for $xto 0^+$?

share|cite|improve this answer

answered Jul 21 at 21:26

Mundron Schmidt

7,1162727

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 21:26

Mundron Schmidt

7,1162727

answered Jul 21 at 21:26

Mundron Schmidt

7,1162727

answered Jul 21 at 21:26

Mundron Schmidt

7,1162727

7,1162727

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I know it becomes a Harmonic series. But how can I show it is not uniformly convergent properly? For instance cauchy criterion
  – Lithium
  Jul 21 at 21:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lithium: what properties do uniformly convergent series have?
  – robjohn♦
  Jul 21 at 23:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes I know it becomes a Harmonic series. But how can I show it is not uniformly convergent properly? For instance cauchy criterion
  – Lithium
  Jul 21 at 21:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lithium: what properties do uniformly convergent series have?
  – robjohn♦
  Jul 21 at 23:39
  

  
  
  
  

Yes I know it becomes a Harmonic series. But how can I show it is not uniformly convergent properly? For instance cauchy criterion
– Lithium
Jul 21 at 21:30

Yes I know it becomes a Harmonic series. But how can I show it is not uniformly convergent properly? For instance cauchy criterion
– Lithium
Jul 21 at 21:30

@Lithium: what properties do uniformly convergent series have?
– robjohn♦
Jul 21 at 23:39

@Lithium: what properties do uniformly convergent series have?
– robjohn♦
Jul 21 at 23:39

add a comment | 

up vote
1
down vote

This is the famous Riemann $zeta$ function, shifted by 1 to the left:
https://en.wikipedia.org/wiki/Riemann_zeta_function

It is well-known that is has a singularity at 1 (in your notation due to the shift, at 0), and it is a simple pole. In other words, around 1, the $zeta$ function around 1 "looks like" $1/(x-1)$. To make it more precise, $zeta(x)-1/(x-1)$ is a smooth function on the positive half of the complex plane.

To translate all of this to your function: it has a simple pole at 0, and around 0, it looks like $1/x$, which is not uniformly continuous.

share|cite|improve this answer

answered Jul 21 at 21:30

A. Pongrácz

2,269221

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean?
  – A. Pongrácz
  Jul 21 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  any simpler method? for instance using cauchy criterion?
  – Lithium
  Jul 21 at 21:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

This is the famous Riemann $zeta$ function, shifted by 1 to the left:
https://en.wikipedia.org/wiki/Riemann_zeta_function

It is well-known that is has a singularity at 1 (in your notation due to the shift, at 0), and it is a simple pole. In other words, around 1, the $zeta$ function around 1 "looks like" $1/(x-1)$. To make it more precise, $zeta(x)-1/(x-1)$ is a smooth function on the positive half of the complex plane.

To translate all of this to your function: it has a simple pole at 0, and around 0, it looks like $1/x$, which is not uniformly continuous.

share|cite|improve this answer

answered Jul 21 at 21:30

A. Pongrácz

2,269221

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean?
  – A. Pongrácz
  Jul 21 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  any simpler method? for instance using cauchy criterion?
  – Lithium
  Jul 21 at 21:34
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

This is the famous Riemann $zeta$ function, shifted by 1 to the left:
https://en.wikipedia.org/wiki/Riemann_zeta_function

It is well-known that is has a singularity at 1 (in your notation due to the shift, at 0), and it is a simple pole. In other words, around 1, the $zeta$ function around 1 "looks like" $1/(x-1)$. To make it more precise, $zeta(x)-1/(x-1)$ is a smooth function on the positive half of the complex plane.

To translate all of this to your function: it has a simple pole at 0, and around 0, it looks like $1/x$, which is not uniformly continuous.

share|cite|improve this answer

answered Jul 21 at 21:30

A. Pongrácz

2,269221

This is the famous Riemann $zeta$ function, shifted by 1 to the left:
https://en.wikipedia.org/wiki/Riemann_zeta_function

It is well-known that is has a singularity at 1 (in your notation due to the shift, at 0), and it is a simple pole. In other words, around 1, the $zeta$ function around 1 "looks like" $1/(x-1)$. To make it more precise, $zeta(x)-1/(x-1)$ is a smooth function on the positive half of the complex plane.

To translate all of this to your function: it has a simple pole at 0, and around 0, it looks like $1/x$, which is not uniformly continuous.

share|cite|improve this answer

answered Jul 21 at 21:30

A. Pongrácz

2,269221

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 21:30

A. Pongrácz

2,269221

answered Jul 21 at 21:30

A. Pongrácz

2,269221

answered Jul 21 at 21:30

A. Pongrácz

2,269221

2,269221

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean?
  – A. Pongrácz
  Jul 21 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  any simpler method? for instance using cauchy criterion?
  – Lithium
  Jul 21 at 21:34
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What do you mean?
  – A. Pongrácz
  Jul 21 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  any simpler method? for instance using cauchy criterion?
  – Lithium
  Jul 21 at 21:34
  

  
  
  
  

What do you mean?
– A. Pongrácz
Jul 21 at 21:33

What do you mean?
– A. Pongrácz
Jul 21 at 21:33

any simpler method? for instance using cauchy criterion?
– Lithium
Jul 21 at 21:34

any simpler method? for instance using cauchy criterion?
– Lithium
Jul 21 at 21:34

add a comment | 

up vote
1
down vote

Hint:

$$sup_x in (0,infty)sum_k=n+1^infty frac1k^1+x >sum_k=n+1^2n frac1k^1+1/n > fracn(2n)^1+1/n$$

share|cite|improve this answer

edited Jul 21 at 22:20

answered Jul 21 at 21:50

RRL

43.7k42260

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry did not get it. could you please explain thank you
  – Lithium
  Jul 21 at 22:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Uniform convergence requires the LHS to converge to $0$. What does the lower bound on the RHS converge to?
  – RRL
  Jul 21 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write what it means for a series to converge uniformly in your post and show it implies what I am stating.
  – RRL
  Jul 21 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why hasn't this answer received more up votes?? Here is one (+1).
  – Mark Viola
  Jul 22 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola: Thanks for being so conscientious (as usual).
  – RRL
  Jul 22 at 3:24
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

Hint:

$$sup_x in (0,infty)sum_k=n+1^infty frac1k^1+x >sum_k=n+1^2n frac1k^1+1/n > fracn(2n)^1+1/n$$

share|cite|improve this answer

edited Jul 21 at 22:20

answered Jul 21 at 21:50

RRL

43.7k42260

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry did not get it. could you please explain thank you
  – Lithium
  Jul 21 at 22:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Uniform convergence requires the LHS to converge to $0$. What does the lower bound on the RHS converge to?
  – RRL
  Jul 21 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write what it means for a series to converge uniformly in your post and show it implies what I am stating.
  – RRL
  Jul 21 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why hasn't this answer received more up votes?? Here is one (+1).
  – Mark Viola
  Jul 22 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola: Thanks for being so conscientious (as usual).
  – RRL
  Jul 22 at 3:24
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

up vote
1
down vote

Hint:

$$sup_x in (0,infty)sum_k=n+1^infty frac1k^1+x >sum_k=n+1^2n frac1k^1+1/n > fracn(2n)^1+1/n$$

share|cite|improve this answer

edited Jul 21 at 22:20

answered Jul 21 at 21:50

RRL

43.7k42260

Hint:

$$sup_x in (0,infty)sum_k=n+1^infty frac1k^1+x >sum_k=n+1^2n frac1k^1+1/n > fracn(2n)^1+1/n$$

share|cite|improve this answer

edited Jul 21 at 22:20

answered Jul 21 at 21:50

RRL

43.7k42260

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 22:20

edited Jul 21 at 22:20

edited Jul 21 at 22:20

answered Jul 21 at 21:50

RRL

43.7k42260

answered Jul 21 at 21:50

RRL

43.7k42260

answered Jul 21 at 21:50

RRL

43.7k42260

43.7k42260

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry did not get it. could you please explain thank you
  – Lithium
  Jul 21 at 22:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Uniform convergence requires the LHS to converge to $0$. What does the lower bound on the RHS converge to?
  – RRL
  Jul 21 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write what it means for a series to converge uniformly in your post and show it implies what I am stating.
  – RRL
  Jul 21 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why hasn't this answer received more up votes?? Here is one (+1).
  – Mark Viola
  Jul 22 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola: Thanks for being so conscientious (as usual).
  – RRL
  Jul 22 at 3:24
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry did not get it. could you please explain thank you
  – Lithium
  Jul 21 at 22:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Uniform convergence requires the LHS to converge to $0$. What does the lower bound on the RHS converge to?
  – RRL
  Jul 21 at 22:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Write what it means for a series to converge uniformly in your post and show it implies what I am stating.
  – RRL
  Jul 21 at 22:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why hasn't this answer received more up votes?? Here is one (+1).
  – Mark Viola
  Jul 22 at 3:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MarkViola: Thanks for being so conscientious (as usual).
  – RRL
  Jul 22 at 3:24
  

  
  
  
  

sorry did not get it. could you please explain thank you
– Lithium
Jul 21 at 22:02

sorry did not get it. could you please explain thank you
– Lithium
Jul 21 at 22:02

Uniform convergence requires the LHS to converge to $0$. What does the lower bound on the RHS converge to?
– RRL
Jul 21 at 22:17

Uniform convergence requires the LHS to converge to $0$. What does the lower bound on the RHS converge to?
– RRL
Jul 21 at 22:17

Write what it means for a series to converge uniformly in your post and show it implies what I am stating.
– RRL
Jul 21 at 22:19

Write what it means for a series to converge uniformly in your post and show it implies what I am stating.
– RRL
Jul 21 at 22:19

Why hasn't this answer received more up votes?? Here is one (+1).
– Mark Viola
Jul 22 at 3:19

Why hasn't this answer received more up votes?? Here is one (+1).
– Mark Viola
Jul 22 at 3:19

@MarkViola: Thanks for being so conscientious (as usual).
– RRL
Jul 22 at 3:24

@MarkViola: Thanks for being so conscientious (as usual).
– RRL
Jul 22 at 3:24

 | 
show 1 more comment

up vote
1
down vote

Uniform convergence means that $$sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| xrightarrowntoinfty 0$$

As @RRL suggests, for $ninmathbbN$ you can consider $x_n = frac1n$ so

beginalign
sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| &= sup_xin(0, +infty) sum_k=n+1^inftyfrac1k^1+x \
&ge sum_k=n+1^inftyfrac1k^1+frac1n\
&ge int_n+1^infty fracdtt^1+frac1n\
&= -fracnt^1/n Bigg|_n+1^infty\
&= fracnsqrt[n]n+1\
&xrightarrowntoinfty +infty
endalign

so it doesn't converge to $0$.

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answered Jul 21 at 23:35

mechanodroid

22.2k52041

add a comment | 

up vote
1
down vote

Uniform convergence means that $$sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| xrightarrowntoinfty 0$$

As @RRL suggests, for $ninmathbbN$ you can consider $x_n = frac1n$ so

beginalign
sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| &= sup_xin(0, +infty) sum_k=n+1^inftyfrac1k^1+x \
&ge sum_k=n+1^inftyfrac1k^1+frac1n\
&ge int_n+1^infty fracdtt^1+frac1n\
&= -fracnt^1/n Bigg|_n+1^infty\
&= fracnsqrt[n]n+1\
&xrightarrowntoinfty +infty
endalign

so it doesn't converge to $0$.

share|cite|improve this answer

answered Jul 21 at 23:35

mechanodroid

22.2k52041

add a comment | 

up vote
1
down vote

up vote
1
down vote

Uniform convergence means that $$sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| xrightarrowntoinfty 0$$

As @RRL suggests, for $ninmathbbN$ you can consider $x_n = frac1n$ so

beginalign
sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| &= sup_xin(0, +infty) sum_k=n+1^inftyfrac1k^1+x \
&ge sum_k=n+1^inftyfrac1k^1+frac1n\
&ge int_n+1^infty fracdtt^1+frac1n\
&= -fracnt^1/n Bigg|_n+1^infty\
&= fracnsqrt[n]n+1\
&xrightarrowntoinfty +infty
endalign

so it doesn't converge to $0$.

share|cite|improve this answer

answered Jul 21 at 23:35

mechanodroid

22.2k52041

Uniform convergence means that $$sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| xrightarrowntoinfty 0$$

As @RRL suggests, for $ninmathbbN$ you can consider $x_n = frac1n$ so

beginalign
sup_xin(0, +infty) left|sum_k=1^infty frac1k^1+x - sum_k=1^nfrac1k^1+xright| &= sup_xin(0, +infty) sum_k=n+1^inftyfrac1k^1+x \
&ge sum_k=n+1^inftyfrac1k^1+frac1n\
&ge int_n+1^infty fracdtt^1+frac1n\
&= -fracnt^1/n Bigg|_n+1^infty\
&= fracnsqrt[n]n+1\
&xrightarrowntoinfty +infty
endalign

so it doesn't converge to $0$.

share|cite|improve this answer

answered Jul 21 at 23:35

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 23:35

mechanodroid

22.2k52041

answered Jul 21 at 23:35

mechanodroid

22.2k52041

answered Jul 21 at 23:35

mechanodroid

22.2k52041

22.2k52041

add a comment | 

add a comment |