Proving divergence of the fraction of the square root
$begingroup$
I saw this question somewhere and was having a hard time proving it so I would love to get some help.
Prove using $varepsilon$ and $N$ that the following sequence is divergent
$$(a_n)=sqrt n - lfloor sqrt n rfloor$$
$$lim_{ntoinfty} (a_n) neq L$$
I have managed to prove it for $ L neq 0 $ but can't manage to do so for $L=0$.
sequences-and-series divergent-series epsilon-delta
edited Jan 14 '16 at 18:55
gebruiker
5,00851963
asked Nov 15 '15 at 18:21
user290361user290361
112
$endgroup$
add a comment |
$begingroup$
I saw this question somewhere and was having a hard time proving it so I would love to get some help.
Prove using $varepsilon$ and $N$ that the following sequence is divergent
$$(a_n)=sqrt n - lfloor sqrt n rfloor$$
$$lim_{ntoinfty} (a_n) neq L$$
I have managed to prove it for $ L neq 0 $ but can't manage to do so for $L=0$.
sequences-and-series divergent-series epsilon-delta
edited Jan 14 '16 at 18:55
gebruiker
5,00851963
asked Nov 15 '15 at 18:21
user290361user290361
112
$endgroup$
1
$begingroup$
Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
$endgroup$
– Wojowu
Nov 15 '15 at 21:10
$begingroup$
didn't really understand,can't really see the connection between the 2 sequences
$endgroup$
– user290361
Nov 15 '15 at 21:32
add a comment |
$begingroup$
I saw this question somewhere and was having a hard time proving it so I would love to get some help.
Prove using $varepsilon$ and $N$ that the following sequence is divergent
$$(a_n)=sqrt n - lfloor sqrt n rfloor$$
$$lim_{ntoinfty} (a_n) neq L$$
I have managed to prove it for $ L neq 0 $ but can't manage to do so for $L=0$.
sequences-and-series divergent-series epsilon-delta
edited Jan 14 '16 at 18:55
gebruiker
5,00851963
asked Nov 15 '15 at 18:21
user290361user290361
112
$endgroup$
I saw this question somewhere and was having a hard time proving it so I would love to get some help.
Prove using $varepsilon$ and $N$ that the following sequence is divergent
$$(a_n)=sqrt n - lfloor sqrt n rfloor$$
$$lim_{ntoinfty} (a_n) neq L$$
I have managed to prove it for $ L neq 0 $ but can't manage to do so for $L=0$.
sequences-and-series divergent-series epsilon-delta
sequences-and-series divergent-series epsilon-delta
edited Jan 14 '16 at 18:55
gebruiker
5,00851963
asked Nov 15 '15 at 18:21
user290361user290361
112
edited Jan 14 '16 at 18:55
gebruiker
5,00851963
asked Nov 15 '15 at 18:21
user290361user290361
112
edited Jan 14 '16 at 18:55
gebruiker
5,00851963
edited Jan 14 '16 at 18:55
gebruiker
5,00851963
edited Jan 14 '16 at 18:55
gebruiker
5,00851963
5,00851963
asked Nov 15 '15 at 18:21
user290361user290361
112
asked Nov 15 '15 at 18:21
user290361user290361
112
asked Nov 15 '15 at 18:21
user290361user290361
112
112
1
$begingroup$
Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
$endgroup$
– Wojowu
Nov 15 '15 at 21:10
$begingroup$
didn't really understand,can't really see the connection between the 2 sequences
$endgroup$
– user290361
Nov 15 '15 at 21:32
add a comment |
1
$begingroup$
Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
$endgroup$
– Wojowu
Nov 15 '15 at 21:10
$begingroup$
didn't really understand,can't really see the connection between the 2 sequences
$endgroup$
– user290361
Nov 15 '15 at 21:32
1
1
$begingroup$
Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
$endgroup$
– Wojowu
Nov 15 '15 at 21:10
$begingroup$
Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
$endgroup$
– Wojowu
Nov 15 '15 at 21:10
$begingroup$
didn't really understand,can't really see the connection between the 2 sequences
$endgroup$
– user290361
Nov 15 '15 at 21:32
$begingroup$
didn't really understand,can't really see the connection between the 2 sequences
$endgroup$
– user290361
Nov 15 '15 at 21:32
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.
We may always assume without loss of generality that $N>1$.
Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.
$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$
This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.
If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.
(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )
Since no $L$ will do, we must conclude that $(a_n)$ diverges.
answered Jan 14 '16 at 18:54
gebruikergebruiker
5,00851963
$endgroup$
$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22
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@matan129 Should work now
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– gebruiker
Dec 1 '18 at 15:20
add a comment |
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1 Answer
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$begingroup$
If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.
We may always assume without loss of generality that $N>1$.
Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.
$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$
This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.
If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.
(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )
Since no $L$ will do, we must conclude that $(a_n)$ diverges.
answered Jan 14 '16 at 18:54
gebruikergebruiker
5,00851963
$endgroup$
$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22
$begingroup$
@matan129 Should work now
$endgroup$
– gebruiker
Dec 1 '18 at 15:20
add a comment |
$begingroup$
If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.
We may always assume without loss of generality that $N>1$.
Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.
$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$
This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.
If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.
(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )
Since no $L$ will do, we must conclude that $(a_n)$ diverges.
answered Jan 14 '16 at 18:54
gebruikergebruiker
5,00851963
$endgroup$
$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22
$begingroup$
@matan129 Should work now
$endgroup$
– gebruiker
Dec 1 '18 at 15:20
add a comment |
$begingroup$
If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.
We may always assume without loss of generality that $N>1$.
Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.
$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$
This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.
If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.
(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )
Since no $L$ will do, we must conclude that $(a_n)$ diverges.
answered Jan 14 '16 at 18:54
gebruikergebruiker
5,00851963
$endgroup$
If $limlimits_{ntoinfty} (a_n)=L$, then for every $varepsilon>0$ there is an $N>0$, such that $vert a_n-Lvert <varepsilon$ if $n>N$.
We may always assume without loss of generality that $N>1$.
Now $Lneq0$ is impossible. We can see this by choosing $0<varepsilon<vert Lvert$ and $n=N^2$.
$L=0$ is also impossible. We can see this by choosing $$m=beta+2+2sqrt{2beta},quadtext{where }beta=2k^2text{, for some odd $k>N$.}$$
This way, we have $N<minmathbb N$ and $a_m=a_{beta} + a_2$.
If $(a_n)to 0$ then we must also have that $(a_m)to 0$. But if the subsequence $(a_m)to 0$ then $(a_ {beta})to -a_{2}$, whereas it should converge to $0$. This is a contradiction and so $L=0$ is not a limit.
(If you like things simple, than take Wojowu's advice form the comments and just show that $(a_{n^2-1})$ converges to $1$. )
Since no $L$ will do, we must conclude that $(a_n)$ diverges.
answered Jan 14 '16 at 18:54
gebruikergebruiker
5,00851963
edited Dec 1 '18 at 14:47
answered Jan 14 '16 at 18:54
gebruikergebruiker
5,00851963
answered Jan 14 '16 at 18:54
gebruikergebruiker
5,00851963
answered Jan 14 '16 at 18:54
gebruikergebruiker
5,00851963
5,00851963
$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22
$begingroup$
@matan129 Should work now
$endgroup$
– gebruiker
Dec 1 '18 at 15:20
add a comment |
$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22
$begingroup$
@matan129 Should work now
$endgroup$
– gebruiker
Dec 1 '18 at 15:20
$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22
$begingroup$
You're right. I will edit.
$endgroup$
– gebruiker
Dec 1 '18 at 14:22
$begingroup$
@matan129 Should work now
$endgroup$
– gebruiker
Dec 1 '18 at 15:20
$begingroup$
@matan129 Should work now
$endgroup$
– gebruiker
Dec 1 '18 at 15:20
add a comment |
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1
$begingroup$
Hint: look at subsequence $a_{n^2-1}$ and show it converges to $1$.
$endgroup$
– Wojowu
Nov 15 '15 at 21:10
$begingroup$
didn't really understand,can't really see the connection between the 2 sequences
$endgroup$
– user290361
Nov 15 '15 at 21:32