Is it true that for any $g$ in $L^2$, $lim_ntoinftyint_n^n+1 g(x) dx=0$?

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Is it true that $$lim_ntoinftyint_n^n+1 g(x) dx=0,$$ for any $g$ in $L^2$? If true, how can I prove it? If not, what is the counterexample?

calculus real-analysis analysis

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edited Jul 18 at 17:42

thesmallprint

2,2191617

asked Jul 18 at 17:22

Steven Chen

116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DonAntonio He means $L^2(mathbb R)$. Constant functions aren't in there.
  – amsmath
  Jul 18 at 17:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Ah, great clearing out! Didn't realize he meant that space. Thanx.
  – DonAntonio
  Jul 18 at 17:42
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Is it true that $$lim_ntoinftyint_n^n+1 g(x) dx=0,$$ for any $g$ in $L^2$? If true, how can I prove it? If not, what is the counterexample?

calculus real-analysis analysis

share|cite|improve this question

edited Jul 18 at 17:42

thesmallprint

2,2191617

asked Jul 18 at 17:22

Steven Chen

116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DonAntonio He means $L^2(mathbb R)$. Constant functions aren't in there.
  – amsmath
  Jul 18 at 17:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Ah, great clearing out! Didn't realize he meant that space. Thanx.
  – DonAntonio
  Jul 18 at 17:42
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Is it true that $$lim_ntoinftyint_n^n+1 g(x) dx=0,$$ for any $g$ in $L^2$? If true, how can I prove it? If not, what is the counterexample?

calculus real-analysis analysis

share|cite|improve this question

edited Jul 18 at 17:42

thesmallprint

2,2191617

asked Jul 18 at 17:22

Steven Chen

116

Is it true that $$lim_ntoinftyint_n^n+1 g(x) dx=0,$$ for any $g$ in $L^2$? If true, how can I prove it? If not, what is the counterexample?

calculus real-analysis analysis

share|cite|improve this question

edited Jul 18 at 17:42

thesmallprint

2,2191617

asked Jul 18 at 17:22

Steven Chen

116

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 17:42

thesmallprint

2,2191617

edited Jul 18 at 17:42

thesmallprint

2,2191617

edited Jul 18 at 17:42

thesmallprint

2,2191617

2,2191617

asked Jul 18 at 17:22

Steven Chen

116

asked Jul 18 at 17:22

Steven Chen

116

asked Jul 18 at 17:22

Steven Chen

116

116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DonAntonio He means $L^2(mathbb R)$. Constant functions aren't in there.
  – amsmath
  Jul 18 at 17:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Ah, great clearing out! Didn't realize he meant that space. Thanx.
  – DonAntonio
  Jul 18 at 17:42
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DonAntonio He means $L^2(mathbb R)$. Constant functions aren't in there.
  – amsmath
  Jul 18 at 17:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @amsmath Ah, great clearing out! Didn't realize he meant that space. Thanx.
  – DonAntonio
  Jul 18 at 17:42
  
  

  
  
  
  

@DonAntonio He means $L^2(mathbb R)$. Constant functions aren't in there.
– amsmath
Jul 18 at 17:38

@DonAntonio He means $L^2(mathbb R)$. Constant functions aren't in there.
– amsmath
Jul 18 at 17:38

@amsmath Ah, great clearing out! Didn't realize he meant that space. Thanx.
– DonAntonio
Jul 18 at 17:42

@amsmath Ah, great clearing out! Didn't realize he meant that space. Thanx.
– DonAntonio
Jul 18 at 17:42

add a comment | 

3 Answers
3

active

oldest

votes

up vote
3
down vote



accepted

You have
$$
int_0^infty|g(x)|^2,dx = sum_n=0^inftyint_n^n+1|g(x)|^2,dx,
$$
so $int_n^n+1|g(x)|^2,dxto 0$ as $ntoinfty$. Hence, by Cauchy-Schwarz,
$$
int_n^n+1|g(x)|,dx,le,left(int_n^n+1|g(x)|^2,dxright)^1/2,to,0
$$
as $ntoinfty$.

share|cite|improve this answer

answered Jul 18 at 17:35

amsmath

1,613114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure why the Cauchy-Schwarz inequality holds in the case. Is it reasonable to restrict the integration over [n,n+1] and let n go to infinity?
  – Steven Chen
  Jul 18 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course you can use Cauchy-Schwarz. You have $1,|g|in L^2(n,n+1)$. So, $$int_n^n+1|g(x)|,dx = langle |g|,1rangle le |g|_L^2(n,n+1)|1|_L^2(n,n+1) = left(int_n^n+1|g(x)|^2,dxright)^1/2.$$
  – amsmath
  Jul 18 at 17:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When I take the limit, n will go to infinity. Doesn't that give rise to a problem in $L^2(n,n+1) $?
  – Steven Chen
  Jul 18 at 17:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StevenChen I don't understand your question. There is no problem!
  – amsmath
  Jul 18 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it. Thank you for the advice. I first misled $L^2(n,n+1)$ as $L^2(infty,infty+1)$ for n goes to infinity and thought Cauchy-Schwarz is not permitted to use here. But I found I shouldn't treat $L^2(n,n+1)$ as $L^2(infty,infty+1)$, which makes no sense to me.
  – Steven Chen
  Jul 18 at 18:02
  

  
  
  
  

add a comment | 

up vote
0
down vote

Use
$$left|int_n^n+1|g(x)|,dxright|^2leint_n^n+1|g(x)|^2,dx$$
and
$$sum_nint_n^n+1|g(x)|^2,dx<infty.$$

share|cite|improve this answer

answered Jul 18 at 17:36

Lord Shark the Unknown

85.5k951112

add a comment | 

up vote
0
down vote

Too much of true

By Holder's inequality :

$$int_n^n+1|g(x)|,dxle((n+1-n)int_n^n+1|g(x)|^2,dx)^1/2$$

$$lim_ntoinftyint_n^n+1|g(x)|,dxlelim_ntoinfty(int_n^n+1|g(x)|^2,dx)^1/2=0$$

share|cite|improve this answer

edited Jul 18 at 18:22

answered Jul 18 at 17:48

S.t.a.l.k.e.r

721320

• 
  
  
  

  
  

  
  
  
  
  
  

  
  First, what you write is wrong. Using Hölder (which is Cauchy-Schwarz here), you have to use the square root on the RHS. Also, what's the reason for taking $|n+1-n|$? Finally, the way you're suggesting was already suggested in answers that were written 13 minutes before yours.
  – amsmath
  Jul 18 at 17:53
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

You have
$$
int_0^infty|g(x)|^2,dx = sum_n=0^inftyint_n^n+1|g(x)|^2,dx,
$$
so $int_n^n+1|g(x)|^2,dxto 0$ as $ntoinfty$. Hence, by Cauchy-Schwarz,
$$
int_n^n+1|g(x)|,dx,le,left(int_n^n+1|g(x)|^2,dxright)^1/2,to,0
$$
as $ntoinfty$.

share|cite|improve this answer

answered Jul 18 at 17:35

amsmath

1,613114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure why the Cauchy-Schwarz inequality holds in the case. Is it reasonable to restrict the integration over [n,n+1] and let n go to infinity?
  – Steven Chen
  Jul 18 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course you can use Cauchy-Schwarz. You have $1,|g|in L^2(n,n+1)$. So, $$int_n^n+1|g(x)|,dx = langle |g|,1rangle le |g|_L^2(n,n+1)|1|_L^2(n,n+1) = left(int_n^n+1|g(x)|^2,dxright)^1/2.$$
  – amsmath
  Jul 18 at 17:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When I take the limit, n will go to infinity. Doesn't that give rise to a problem in $L^2(n,n+1) $?
  – Steven Chen
  Jul 18 at 17:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StevenChen I don't understand your question. There is no problem!
  – amsmath
  Jul 18 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it. Thank you for the advice. I first misled $L^2(n,n+1)$ as $L^2(infty,infty+1)$ for n goes to infinity and thought Cauchy-Schwarz is not permitted to use here. But I found I shouldn't treat $L^2(n,n+1)$ as $L^2(infty,infty+1)$, which makes no sense to me.
  – Steven Chen
  Jul 18 at 18:02
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

You have
$$
int_0^infty|g(x)|^2,dx = sum_n=0^inftyint_n^n+1|g(x)|^2,dx,
$$
so $int_n^n+1|g(x)|^2,dxto 0$ as $ntoinfty$. Hence, by Cauchy-Schwarz,
$$
int_n^n+1|g(x)|,dx,le,left(int_n^n+1|g(x)|^2,dxright)^1/2,to,0
$$
as $ntoinfty$.

share|cite|improve this answer

answered Jul 18 at 17:35

amsmath

1,613114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure why the Cauchy-Schwarz inequality holds in the case. Is it reasonable to restrict the integration over [n,n+1] and let n go to infinity?
  – Steven Chen
  Jul 18 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course you can use Cauchy-Schwarz. You have $1,|g|in L^2(n,n+1)$. So, $$int_n^n+1|g(x)|,dx = langle |g|,1rangle le |g|_L^2(n,n+1)|1|_L^2(n,n+1) = left(int_n^n+1|g(x)|^2,dxright)^1/2.$$
  – amsmath
  Jul 18 at 17:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When I take the limit, n will go to infinity. Doesn't that give rise to a problem in $L^2(n,n+1) $?
  – Steven Chen
  Jul 18 at 17:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StevenChen I don't understand your question. There is no problem!
  – amsmath
  Jul 18 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it. Thank you for the advice. I first misled $L^2(n,n+1)$ as $L^2(infty,infty+1)$ for n goes to infinity and thought Cauchy-Schwarz is not permitted to use here. But I found I shouldn't treat $L^2(n,n+1)$ as $L^2(infty,infty+1)$, which makes no sense to me.
  – Steven Chen
  Jul 18 at 18:02
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

You have
$$
int_0^infty|g(x)|^2,dx = sum_n=0^inftyint_n^n+1|g(x)|^2,dx,
$$
so $int_n^n+1|g(x)|^2,dxto 0$ as $ntoinfty$. Hence, by Cauchy-Schwarz,
$$
int_n^n+1|g(x)|,dx,le,left(int_n^n+1|g(x)|^2,dxright)^1/2,to,0
$$
as $ntoinfty$.

share|cite|improve this answer

answered Jul 18 at 17:35

amsmath

1,613114

You have
$$
int_0^infty|g(x)|^2,dx = sum_n=0^inftyint_n^n+1|g(x)|^2,dx,
$$
so $int_n^n+1|g(x)|^2,dxto 0$ as $ntoinfty$. Hence, by Cauchy-Schwarz,
$$
int_n^n+1|g(x)|,dx,le,left(int_n^n+1|g(x)|^2,dxright)^1/2,to,0
$$
as $ntoinfty$.

share|cite|improve this answer

answered Jul 18 at 17:35

amsmath

1,613114

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 17:35

amsmath

1,613114

answered Jul 18 at 17:35

amsmath

1,613114

answered Jul 18 at 17:35

amsmath

1,613114

1,613114

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure why the Cauchy-Schwarz inequality holds in the case. Is it reasonable to restrict the integration over [n,n+1] and let n go to infinity?
  – Steven Chen
  Jul 18 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course you can use Cauchy-Schwarz. You have $1,|g|in L^2(n,n+1)$. So, $$int_n^n+1|g(x)|,dx = langle |g|,1rangle le |g|_L^2(n,n+1)|1|_L^2(n,n+1) = left(int_n^n+1|g(x)|^2,dxright)^1/2.$$
  – amsmath
  Jul 18 at 17:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When I take the limit, n will go to infinity. Doesn't that give rise to a problem in $L^2(n,n+1) $?
  – Steven Chen
  Jul 18 at 17:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StevenChen I don't understand your question. There is no problem!
  – amsmath
  Jul 18 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it. Thank you for the advice. I first misled $L^2(n,n+1)$ as $L^2(infty,infty+1)$ for n goes to infinity and thought Cauchy-Schwarz is not permitted to use here. But I found I shouldn't treat $L^2(n,n+1)$ as $L^2(infty,infty+1)$, which makes no sense to me.
  – Steven Chen
  Jul 18 at 18:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure why the Cauchy-Schwarz inequality holds in the case. Is it reasonable to restrict the integration over [n,n+1] and let n go to infinity?
  – Steven Chen
  Jul 18 at 17:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course you can use Cauchy-Schwarz. You have $1,|g|in L^2(n,n+1)$. So, $$int_n^n+1|g(x)|,dx = langle |g|,1rangle le |g|_L^2(n,n+1)|1|_L^2(n,n+1) = left(int_n^n+1|g(x)|^2,dxright)^1/2.$$
  – amsmath
  Jul 18 at 17:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  When I take the limit, n will go to infinity. Doesn't that give rise to a problem in $L^2(n,n+1) $?
  – Steven Chen
  Jul 18 at 17:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StevenChen I don't understand your question. There is no problem!
  – amsmath
  Jul 18 at 17:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I got it. Thank you for the advice. I first misled $L^2(n,n+1)$ as $L^2(infty,infty+1)$ for n goes to infinity and thought Cauchy-Schwarz is not permitted to use here. But I found I shouldn't treat $L^2(n,n+1)$ as $L^2(infty,infty+1)$, which makes no sense to me.
  – Steven Chen
  Jul 18 at 18:02
  

  
  
  
  

I am not sure why the Cauchy-Schwarz inequality holds in the case. Is it reasonable to restrict the integration over [n,n+1] and let n go to infinity?
– Steven Chen
Jul 18 at 17:40

I am not sure why the Cauchy-Schwarz inequality holds in the case. Is it reasonable to restrict the integration over [n,n+1] and let n go to infinity?
– Steven Chen
Jul 18 at 17:40

Of course you can use Cauchy-Schwarz. You have $1,|g|in L^2(n,n+1)$. So, $$int_n^n+1|g(x)|,dx = langle |g|,1rangle le |g|_L^2(n,n+1)|1|_L^2(n,n+1) = left(int_n^n+1|g(x)|^2,dxright)^1/2.$$
– amsmath
Jul 18 at 17:44

Of course you can use Cauchy-Schwarz. You have $1,|g|in L^2(n,n+1)$. So, $$int_n^n+1|g(x)|,dx = langle |g|,1rangle le |g|_L^2(n,n+1)|1|_L^2(n,n+1) = left(int_n^n+1|g(x)|^2,dxright)^1/2.$$
– amsmath
Jul 18 at 17:44

When I take the limit, n will go to infinity. Doesn't that give rise to a problem in $L^2(n,n+1) $?
– Steven Chen
Jul 18 at 17:47

When I take the limit, n will go to infinity. Doesn't that give rise to a problem in $L^2(n,n+1) $?
– Steven Chen
Jul 18 at 17:47

@StevenChen I don't understand your question. There is no problem!
– amsmath
Jul 18 at 17:51

@StevenChen I don't understand your question. There is no problem!
– amsmath
Jul 18 at 17:51

I got it. Thank you for the advice. I first misled $L^2(n,n+1)$ as $L^2(infty,infty+1)$ for n goes to infinity and thought Cauchy-Schwarz is not permitted to use here. But I found I shouldn't treat $L^2(n,n+1)$ as $L^2(infty,infty+1)$, which makes no sense to me.
– Steven Chen
Jul 18 at 18:02

I got it. Thank you for the advice. I first misled $L^2(n,n+1)$ as $L^2(infty,infty+1)$ for n goes to infinity and thought Cauchy-Schwarz is not permitted to use here. But I found I shouldn't treat $L^2(n,n+1)$ as $L^2(infty,infty+1)$, which makes no sense to me.
– Steven Chen
Jul 18 at 18:02

add a comment | 

up vote
0
down vote

Use
$$left|int_n^n+1|g(x)|,dxright|^2leint_n^n+1|g(x)|^2,dx$$
and
$$sum_nint_n^n+1|g(x)|^2,dx<infty.$$

share|cite|improve this answer

answered Jul 18 at 17:36

Lord Shark the Unknown

85.5k951112

add a comment | 

up vote
0
down vote

Use
$$left|int_n^n+1|g(x)|,dxright|^2leint_n^n+1|g(x)|^2,dx$$
and
$$sum_nint_n^n+1|g(x)|^2,dx<infty.$$

share|cite|improve this answer

answered Jul 18 at 17:36

Lord Shark the Unknown

85.5k951112

add a comment | 

up vote
0
down vote

up vote
0
down vote

Use
$$left|int_n^n+1|g(x)|,dxright|^2leint_n^n+1|g(x)|^2,dx$$
and
$$sum_nint_n^n+1|g(x)|^2,dx<infty.$$

share|cite|improve this answer

answered Jul 18 at 17:36

Lord Shark the Unknown

85.5k951112

Use
$$left|int_n^n+1|g(x)|,dxright|^2leint_n^n+1|g(x)|^2,dx$$
and
$$sum_nint_n^n+1|g(x)|^2,dx<infty.$$

share|cite|improve this answer

answered Jul 18 at 17:36

Lord Shark the Unknown

85.5k951112

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 17:36

Lord Shark the Unknown

85.5k951112

answered Jul 18 at 17:36

Lord Shark the Unknown

85.5k951112

answered Jul 18 at 17:36

Lord Shark the Unknown

85.5k951112

85.5k951112

add a comment | 

add a comment | 

up vote
0
down vote

Too much of true

By Holder's inequality :

$$int_n^n+1|g(x)|,dxle((n+1-n)int_n^n+1|g(x)|^2,dx)^1/2$$

$$lim_ntoinftyint_n^n+1|g(x)|,dxlelim_ntoinfty(int_n^n+1|g(x)|^2,dx)^1/2=0$$

share|cite|improve this answer

edited Jul 18 at 18:22

answered Jul 18 at 17:48

S.t.a.l.k.e.r

721320

• 
  
  
  

  
  

  
  
  
  
  
  

  
  First, what you write is wrong. Using Hölder (which is Cauchy-Schwarz here), you have to use the square root on the RHS. Also, what's the reason for taking $|n+1-n|$? Finally, the way you're suggesting was already suggested in answers that were written 13 minutes before yours.
  – amsmath
  Jul 18 at 17:53
  

  
  
  
  

add a comment | 

up vote
0
down vote

Too much of true

By Holder's inequality :

$$int_n^n+1|g(x)|,dxle((n+1-n)int_n^n+1|g(x)|^2,dx)^1/2$$

$$lim_ntoinftyint_n^n+1|g(x)|,dxlelim_ntoinfty(int_n^n+1|g(x)|^2,dx)^1/2=0$$

share|cite|improve this answer

edited Jul 18 at 18:22

answered Jul 18 at 17:48

S.t.a.l.k.e.r

721320

• 
  
  
  

  
  

  
  
  
  
  
  

  
  First, what you write is wrong. Using Hölder (which is Cauchy-Schwarz here), you have to use the square root on the RHS. Also, what's the reason for taking $|n+1-n|$? Finally, the way you're suggesting was already suggested in answers that were written 13 minutes before yours.
  – amsmath
  Jul 18 at 17:53
  

  
  
  
  

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Too much of true

By Holder's inequality :

$$int_n^n+1|g(x)|,dxle((n+1-n)int_n^n+1|g(x)|^2,dx)^1/2$$

$$lim_ntoinftyint_n^n+1|g(x)|,dxlelim_ntoinfty(int_n^n+1|g(x)|^2,dx)^1/2=0$$

share|cite|improve this answer

edited Jul 18 at 18:22

answered Jul 18 at 17:48

S.t.a.l.k.e.r

721320

Too much of true

By Holder's inequality :

$$int_n^n+1|g(x)|,dxle((n+1-n)int_n^n+1|g(x)|^2,dx)^1/2$$

$$lim_ntoinftyint_n^n+1|g(x)|,dxlelim_ntoinfty(int_n^n+1|g(x)|^2,dx)^1/2=0$$

share|cite|improve this answer

edited Jul 18 at 18:22

answered Jul 18 at 17:48

S.t.a.l.k.e.r

721320

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 18:22

edited Jul 18 at 18:22

edited Jul 18 at 18:22

answered Jul 18 at 17:48

S.t.a.l.k.e.r

721320

answered Jul 18 at 17:48

S.t.a.l.k.e.r

721320

answered Jul 18 at 17:48

S.t.a.l.k.e.r

721320

721320

• 
  
  
  

  
  

  
  
  
  
  
  

  
  First, what you write is wrong. Using Hölder (which is Cauchy-Schwarz here), you have to use the square root on the RHS. Also, what's the reason for taking $|n+1-n|$? Finally, the way you're suggesting was already suggested in answers that were written 13 minutes before yours.
  – amsmath
  Jul 18 at 17:53
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  First, what you write is wrong. Using Hölder (which is Cauchy-Schwarz here), you have to use the square root on the RHS. Also, what's the reason for taking $|n+1-n|$? Finally, the way you're suggesting was already suggested in answers that were written 13 minutes before yours.
  – amsmath
  Jul 18 at 17:53
  

  
  
  
  

First, what you write is wrong. Using Hölder (which is Cauchy-Schwarz here), you have to use the square root on the RHS. Also, what's the reason for taking $|n+1-n|$? Finally, the way you're suggesting was already suggested in answers that were written 13 minutes before yours.
– amsmath
Jul 18 at 17:53

First, what you write is wrong. Using Hölder (which is Cauchy-Schwarz here), you have to use the square root on the RHS. Also, what's the reason for taking $|n+1-n|$? Finally, the way you're suggesting was already suggested in answers that were written 13 minutes before yours.
– amsmath
Jul 18 at 17:53

add a comment | 

 

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