find $limsup$ and $liminf$ of $a_n={sqrt{n} - lfloorsqrt{n}rfloor$ $ninmathbb{N}}$ [duplicate]

0

This question already has an answer here:

• 
  Supremum of $sqrt{n} -leftlfloorsqrt{n}rightrfloor $
  
  3 answers
  
  

I have to find $limsup$ and $liminf$ of $a_n={sqrt{n} - lfloorsqrt{n}rfloor : n inmathbb{N}}$

I suppose that I have to relate to subsequences .
First one is $a_{n_k}$ for all n that maintains $sqrt ninmathbb {N}$ , so the limit is $0$.

Second one $a_{n_j}$ includes the rest of the elements.I suppose it converges to 1, but dont know how to prove it.

sequences-and-series limits limsup-and-liminf

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edited Nov 21 '18 at 9:08

Chinnapparaj R

5,2951826

asked Nov 21 '18 at 8:58

Alex ˢᵅˢʰᵅ Druzina

144

marked as duplicate by Martin R, Chinnapparaj R, Community♦ Nov 21 '18 at 9:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's a subseries?
  – José Carlos Santos
  Nov 21 '18 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoséCarlosSantos I guess subsequence
  – Faustus
  Nov 21 '18 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, thank you. My english is no so great, espetially in math terminilogy
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I dont understand lim sup or lim inf.
  – Love Invariants
  Nov 21 '18 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Have a look at the link where Yves answers this question, in particular his shorter answer.Vow!:)Greetings.
  – Peter Szilas
  Nov 21 '18 at 9:26
  

  
  
  
  

add a comment | 

0

This question already has an answer here:

• 
  Supremum of $sqrt{n} -leftlfloorsqrt{n}rightrfloor $
  
  3 answers
  
  

I have to find $limsup$ and $liminf$ of $a_n={sqrt{n} - lfloorsqrt{n}rfloor : n inmathbb{N}}$

I suppose that I have to relate to subsequences .
First one is $a_{n_k}$ for all n that maintains $sqrt ninmathbb {N}$ , so the limit is $0$.

Second one $a_{n_j}$ includes the rest of the elements.I suppose it converges to 1, but dont know how to prove it.

sequences-and-series limits limsup-and-liminf

share|cite|improve this question

edited Nov 21 '18 at 9:08

Chinnapparaj R

5,2951826

asked Nov 21 '18 at 8:58

Alex ˢᵅˢʰᵅ Druzina

144

marked as duplicate by Martin R, Chinnapparaj R, Community♦ Nov 21 '18 at 9:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's a subseries?
  – José Carlos Santos
  Nov 21 '18 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoséCarlosSantos I guess subsequence
  – Faustus
  Nov 21 '18 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, thank you. My english is no so great, espetially in math terminilogy
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I dont understand lim sup or lim inf.
  – Love Invariants
  Nov 21 '18 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Have a look at the link where Yves answers this question, in particular his shorter answer.Vow!:)Greetings.
  – Peter Szilas
  Nov 21 '18 at 9:26
  

  
  
  
  

add a comment | 

0

0

0

This question already has an answer here:

• 
  Supremum of $sqrt{n} -leftlfloorsqrt{n}rightrfloor $
  
  3 answers
  
  

I have to find $limsup$ and $liminf$ of $a_n={sqrt{n} - lfloorsqrt{n}rfloor : n inmathbb{N}}$

I suppose that I have to relate to subsequences .
First one is $a_{n_k}$ for all n that maintains $sqrt ninmathbb {N}$ , so the limit is $0$.

Second one $a_{n_j}$ includes the rest of the elements.I suppose it converges to 1, but dont know how to prove it.

sequences-and-series limits limsup-and-liminf

share|cite|improve this question

edited Nov 21 '18 at 9:08

Chinnapparaj R

5,2951826

asked Nov 21 '18 at 8:58

Alex ˢᵅˢʰᵅ Druzina

144

This question already has an answer here:

• 
  Supremum of $sqrt{n} -leftlfloorsqrt{n}rightrfloor $
  
  3 answers
  
  

I have to find $limsup$ and $liminf$ of $a_n={sqrt{n} - lfloorsqrt{n}rfloor : n inmathbb{N}}$

I suppose that I have to relate to subsequences .
First one is $a_{n_k}$ for all n that maintains $sqrt ninmathbb {N}$ , so the limit is $0$.

Second one $a_{n_j}$ includes the rest of the elements.I suppose it converges to 1, but dont know how to prove it.

This question already has an answer here:

• 
  Supremum of $sqrt{n} -leftlfloorsqrt{n}rightrfloor $
  
  3 answers
  
  

sequences-and-series limits limsup-and-liminf

sequences-and-series limits limsup-and-liminf

share|cite|improve this question

edited Nov 21 '18 at 9:08

Chinnapparaj R

5,2951826

asked Nov 21 '18 at 8:58

Alex ˢᵅˢʰᵅ Druzina

144

share|cite|improve this question

edited Nov 21 '18 at 9:08

Chinnapparaj R

5,2951826

asked Nov 21 '18 at 8:58

Alex ˢᵅˢʰᵅ Druzina

144

share|cite|improve this question

share|cite|improve this question

edited Nov 21 '18 at 9:08

Chinnapparaj R

5,2951826

edited Nov 21 '18 at 9:08

Chinnapparaj R

5,2951826

edited Nov 21 '18 at 9:08

Chinnapparaj R

5,2951826

5,2951826

asked Nov 21 '18 at 8:58

Alex ˢᵅˢʰᵅ Druzina

144

asked Nov 21 '18 at 8:58

Alex ˢᵅˢʰᵅ Druzina

144

asked Nov 21 '18 at 8:58

Alex ˢᵅˢʰᵅ Druzina

144

144

marked as duplicate by Martin R, Chinnapparaj R, Community♦ Nov 21 '18 at 9:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Martin R, Chinnapparaj R, Community♦ Nov 21 '18 at 9:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's a subseries?
  – José Carlos Santos
  Nov 21 '18 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoséCarlosSantos I guess subsequence
  – Faustus
  Nov 21 '18 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, thank you. My english is no so great, espetially in math terminilogy
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I dont understand lim sup or lim inf.
  – Love Invariants
  Nov 21 '18 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Have a look at the link where Yves answers this question, in particular his shorter answer.Vow!:)Greetings.
  – Peter Szilas
  Nov 21 '18 at 9:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What's a subseries?
  – José Carlos Santos
  Nov 21 '18 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JoséCarlosSantos I guess subsequence
  – Faustus
  Nov 21 '18 at 9:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, thank you. My english is no so great, espetially in math terminilogy
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I dont understand lim sup or lim inf.
  – Love Invariants
  Nov 21 '18 at 9:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Gimusi.Have a look at the link where Yves answers this question, in particular his shorter answer.Vow!:)Greetings.
  – Peter Szilas
  Nov 21 '18 at 9:26
  

  
  
  
  

What's a subseries?
– José Carlos Santos
Nov 21 '18 at 9:00

What's a subseries?
– José Carlos Santos
Nov 21 '18 at 9:00

@JoséCarlosSantos I guess subsequence
– Faustus
Nov 21 '18 at 9:00

@JoséCarlosSantos I guess subsequence
– Faustus
Nov 21 '18 at 9:00

Yes, thank you. My english is no so great, espetially in math terminilogy
– Alex ˢᵅˢʰᵅ Druzina
Nov 21 '18 at 9:02

Yes, thank you. My english is no so great, espetially in math terminilogy
– Alex ˢᵅˢʰᵅ Druzina
Nov 21 '18 at 9:02

I dont understand lim sup or lim inf.
– Love Invariants
Nov 21 '18 at 9:07

I dont understand lim sup or lim inf.
– Love Invariants
Nov 21 '18 at 9:07

Gimusi.Have a look at the link where Yves answers this question, in particular his shorter answer.Vow!:)Greetings.
– Peter Szilas
Nov 21 '18 at 9:26

Gimusi.Have a look at the link where Yves answers this question, in particular his shorter answer.Vow!:)Greetings.
– Peter Szilas
Nov 21 '18 at 9:26

add a comment | 

1 Answer
1

active

oldest

votes

0

We have that

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor ge 0$$

and for $n=m^2 implies a_n=0$ then $liminf a_n =0$.

For $limsup$ we have

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor le 1$$

and for any $epsilon>0$ we can find $n=m^2-1$ such that $sqrt{n} - lfloorsqrt{n}rfloor=1-epsilon$.

share|cite|improve this answer

edited Nov 21 '18 at 9:11

answered Nov 21 '18 at 9:01

gimusi

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might notice that a possible duplicate target has been pointed out in the meantime. Your approach is essentially the same as math.stackexchange.com/a/2015624.
  – Martin R
  Nov 21 '18 at 9:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartinR Thanks, I see that now!
  – gimusi
  Nov 21 '18 at 9:13
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

We have that

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor ge 0$$

and for $n=m^2 implies a_n=0$ then $liminf a_n =0$.

For $limsup$ we have

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor le 1$$

and for any $epsilon>0$ we can find $n=m^2-1$ such that $sqrt{n} - lfloorsqrt{n}rfloor=1-epsilon$.

share|cite|improve this answer

edited Nov 21 '18 at 9:11

answered Nov 21 '18 at 9:01

gimusi

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might notice that a possible duplicate target has been pointed out in the meantime. Your approach is essentially the same as math.stackexchange.com/a/2015624.
  – Martin R
  Nov 21 '18 at 9:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartinR Thanks, I see that now!
  – gimusi
  Nov 21 '18 at 9:13
  

  
  
  
  

add a comment | 

0

We have that

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor ge 0$$

and for $n=m^2 implies a_n=0$ then $liminf a_n =0$.

For $limsup$ we have

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor le 1$$

and for any $epsilon>0$ we can find $n=m^2-1$ such that $sqrt{n} - lfloorsqrt{n}rfloor=1-epsilon$.

share|cite|improve this answer

edited Nov 21 '18 at 9:11

answered Nov 21 '18 at 9:01

gimusi

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might notice that a possible duplicate target has been pointed out in the meantime. Your approach is essentially the same as math.stackexchange.com/a/2015624.
  – Martin R
  Nov 21 '18 at 9:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartinR Thanks, I see that now!
  – gimusi
  Nov 21 '18 at 9:13
  

  
  
  
  

add a comment | 

0

0

0

We have that

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor ge 0$$

and for $n=m^2 implies a_n=0$ then $liminf a_n =0$.

For $limsup$ we have

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor le 1$$

and for any $epsilon>0$ we can find $n=m^2-1$ such that $sqrt{n} - lfloorsqrt{n}rfloor=1-epsilon$.

share|cite|improve this answer

edited Nov 21 '18 at 9:11

answered Nov 21 '18 at 9:01

gimusi

1

We have that

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor ge 0$$

and for $n=m^2 implies a_n=0$ then $liminf a_n =0$.

For $limsup$ we have

$$a_n={sqrt{n} - lfloorsqrt{n}rfloor le 1$$

and for any $epsilon>0$ we can find $n=m^2-1$ such that $sqrt{n} - lfloorsqrt{n}rfloor=1-epsilon$.

share|cite|improve this answer

edited Nov 21 '18 at 9:11

answered Nov 21 '18 at 9:01

gimusi

1

share|cite|improve this answer

share|cite|improve this answer

edited Nov 21 '18 at 9:11

edited Nov 21 '18 at 9:11

edited Nov 21 '18 at 9:11

answered Nov 21 '18 at 9:01

gimusi

1

answered Nov 21 '18 at 9:01

gimusi

1

answered Nov 21 '18 at 9:01

gimusi

1

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might notice that a possible duplicate target has been pointed out in the meantime. Your approach is essentially the same as math.stackexchange.com/a/2015624.
  – Martin R
  Nov 21 '18 at 9:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartinR Thanks, I see that now!
  – gimusi
  Nov 21 '18 at 9:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
  – Alex ˢᵅˢʰᵅ Druzina
  Nov 21 '18 at 9:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might notice that a possible duplicate target has been pointed out in the meantime. Your approach is essentially the same as math.stackexchange.com/a/2015624.
  – Martin R
  Nov 21 '18 at 9:12
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MartinR Thanks, I see that now!
  – gimusi
  Nov 21 '18 at 9:13
  

  
  
  
  

That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
– Alex ˢᵅˢʰᵅ Druzina
Nov 21 '18 at 9:04

That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
– Alex ˢᵅˢʰᵅ Druzina
Nov 21 '18 at 9:04

You might notice that a possible duplicate target has been pointed out in the meantime. Your approach is essentially the same as math.stackexchange.com/a/2015624.
– Martin R
Nov 21 '18 at 9:12

You might notice that a possible duplicate target has been pointed out in the meantime. Your approach is essentially the same as math.stackexchange.com/a/2015624.
– Martin R
Nov 21 '18 at 9:12

@MartinR Thanks, I see that now!
– gimusi
Nov 21 '18 at 9:13

@MartinR Thanks, I see that now!
– gimusi
Nov 21 '18 at 9:13

add a comment | 

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