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











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  • 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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marked as duplicate by Martin R, Chinnapparaj R, Community♦ Nov 21 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 at 9:00










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










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












  • I dont understand lim sup or lim inf.
    – Love Invariants
    Nov 21 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 at 9:26















up vote
0
down vote

favorite













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















marked as duplicate by Martin R, Chinnapparaj R, Community♦ Nov 21 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 at 9:00










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










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












  • I dont understand lim sup or lim inf.
    – Love Invariants
    Nov 21 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 at 9:26













up vote
0
down vote

favorite









up vote
0
down vote

favorite












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
















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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 at 9:08









Chinnapparaj R

4,7671825




4,7671825










asked Nov 21 at 8:58









Alex ˢᵅˢʰᵅ Druzina

144




144




marked as duplicate by Martin R, Chinnapparaj R, Community♦ Nov 21 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 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 at 9:00










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










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












  • I dont understand lim sup or lim inf.
    – Love Invariants
    Nov 21 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 at 9:26


















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










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










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












  • I dont understand lim sup or lim inf.
    – Love Invariants
    Nov 21 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 at 9:26
















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




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












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




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












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






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














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




I dont understand lim sup or lim inf.
– Love Invariants
Nov 21 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 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 at 9:26










1 Answer
1






active

oldest

votes

















up vote
0
down vote













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























  • That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
    – Alex ˢᵅˢʰᵅ Druzina
    Nov 21 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 at 9:12












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


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













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























  • That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
    – Alex ˢᵅˢʰᵅ Druzina
    Nov 21 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 at 9:12












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















up vote
0
down vote













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























  • That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
    – Alex ˢᵅˢʰᵅ Druzina
    Nov 21 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 at 9:12












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













up vote
0
down vote










up vote
0
down vote









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














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














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 9:11

























answered Nov 21 at 9:01









gimusi

91.2k74495




91.2k74495












  • That is exacly what Im asking. I have already found lim inf but, has no idea about lim sup
    – Alex ˢᵅˢʰᵅ Druzina
    Nov 21 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 at 9:12












  • @MartinR Thanks, I see that now!
    – gimusi
    Nov 21 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 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 at 9:12












  • @MartinR Thanks, I see that now!
    – gimusi
    Nov 21 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 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 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 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 at 9:12














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




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



-

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