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$(a_(k_n))_ninmathbb N$ converges to L?
Clash Royale CLAN TAG#URR8PPP
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The problem: Suppose $(a_n)_ninmathbb N$ is a sequence in $mathbb R$ moreover that $(a_n)_ninmathbb N$ converges to $L in mathbb R$. $(k_n)_ninmathbb N$ is a sequence in $mathbb N$, with $lim_n to inftyfrac1k_n = 0$. How can we show $(a_(k_n))_ninmathbb N$ converges to L?
My progress: Intuitively I can see why this is true moreover, after seeing a series of classes full of epsilon delta proofs of different limit theorems within sequences in general. I would bet that this will be a proof with creative uses of the epsilon delta definitions perhaps more than once in a nested fashion.
I've spent about 7-8hrs today on this question and have literally made no progress other than trying to bend the definition of convergence around like a mad man.
real-analysis sequences-and-series limits epsilon-delta
asked Aug 2 at 7:22
Florian Suess
879
add a comment |Â
up vote
1
down vote
favorite
The problem: Suppose $(a_n)_ninmathbb N$ is a sequence in $mathbb R$ moreover that $(a_n)_ninmathbb N$ converges to $L in mathbb R$. $(k_n)_ninmathbb N$ is a sequence in $mathbb N$, with $lim_n to inftyfrac1k_n = 0$. How can we show $(a_(k_n))_ninmathbb N$ converges to L?
My progress: Intuitively I can see why this is true moreover, after seeing a series of classes full of epsilon delta proofs of different limit theorems within sequences in general. I would bet that this will be a proof with creative uses of the epsilon delta definitions perhaps more than once in a nested fashion.
I've spent about 7-8hrs today on this question and have literally made no progress other than trying to bend the definition of convergence around like a mad man.
real-analysis sequences-and-series limits epsilon-delta
asked Aug 2 at 7:22
Florian Suess
879
1
A note: as @user143234 remarked in his answer, then the convergence of $a_k_n$ to $L$ follows from a general theorem in mathematical analysis, which can be found also in stadard textbooks as E. Fisher’s Intermediate real analysis. The general argument follows from the monotonicity of $limsup$, $liminf$ and from the fact that since $k_ntoinfty$ then it is possible to extract from it a strictly monotonic subsequence converging to infinity. In sum, this result relies on monotonicity and thus on order arguments.
– Daniele Tampieri
Aug 2 at 8:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The problem: Suppose $(a_n)_ninmathbb N$ is a sequence in $mathbb R$ moreover that $(a_n)_ninmathbb N$ converges to $L in mathbb R$. $(k_n)_ninmathbb N$ is a sequence in $mathbb N$, with $lim_n to inftyfrac1k_n = 0$. How can we show $(a_(k_n))_ninmathbb N$ converges to L?
My progress: Intuitively I can see why this is true moreover, after seeing a series of classes full of epsilon delta proofs of different limit theorems within sequences in general. I would bet that this will be a proof with creative uses of the epsilon delta definitions perhaps more than once in a nested fashion.
I've spent about 7-8hrs today on this question and have literally made no progress other than trying to bend the definition of convergence around like a mad man.
real-analysis sequences-and-series limits epsilon-delta
asked Aug 2 at 7:22
Florian Suess
879
The problem: Suppose $(a_n)_ninmathbb N$ is a sequence in $mathbb R$ moreover that $(a_n)_ninmathbb N$ converges to $L in mathbb R$. $(k_n)_ninmathbb N$ is a sequence in $mathbb N$, with $lim_n to inftyfrac1k_n = 0$. How can we show $(a_(k_n))_ninmathbb N$ converges to L?
My progress: Intuitively I can see why this is true moreover, after seeing a series of classes full of epsilon delta proofs of different limit theorems within sequences in general. I would bet that this will be a proof with creative uses of the epsilon delta definitions perhaps more than once in a nested fashion.
I've spent about 7-8hrs today on this question and have literally made no progress other than trying to bend the definition of convergence around like a mad man.
real-analysis sequences-and-series limits epsilon-delta
asked Aug 2 at 7:22
Florian Suess
879
asked Aug 2 at 7:22
Florian Suess
879
asked Aug 2 at 7:22
Florian Suess
879
asked Aug 2 at 7:22
Florian Suess
879
879
1
A note: as @user143234 remarked in his answer, then the convergence of $a_k_n$ to $L$ follows from a general theorem in mathematical analysis, which can be found also in stadard textbooks as E. Fisher’s Intermediate real analysis. The general argument follows from the monotonicity of $limsup$, $liminf$ and from the fact that since $k_ntoinfty$ then it is possible to extract from it a strictly monotonic subsequence converging to infinity. In sum, this result relies on monotonicity and thus on order arguments.
– Daniele Tampieri
Aug 2 at 8:56
add a comment |Â
1
A note: as @user143234 remarked in his answer, then the convergence of $a_k_n$ to $L$ follows from a general theorem in mathematical analysis, which can be found also in stadard textbooks as E. Fisher’s Intermediate real analysis. The general argument follows from the monotonicity of $limsup$, $liminf$ and from the fact that since $k_ntoinfty$ then it is possible to extract from it a strictly monotonic subsequence converging to infinity. In sum, this result relies on monotonicity and thus on order arguments.
– Daniele Tampieri
Aug 2 at 8:56
1
1
A note: as @user143234 remarked in his answer, then the convergence of $a_k_n$ to $L$ follows from a general theorem in mathematical analysis, which can be found also in stadard textbooks as E. Fisher’s Intermediate real analysis. The general argument follows from the monotonicity of $limsup$, $liminf$ and from the fact that since $k_ntoinfty$ then it is possible to extract from it a strictly monotonic subsequence converging to infinity. In sum, this result relies on monotonicity and thus on order arguments.
– Daniele Tampieri
Aug 2 at 8:56
A note: as @user143234 remarked in his answer, then the convergence of $a_k_n$ to $L$ follows from a general theorem in mathematical analysis, which can be found also in stadard textbooks as E. Fisher’s Intermediate real analysis. The general argument follows from the monotonicity of $limsup$, $liminf$ and from the fact that since $k_ntoinfty$ then it is possible to extract from it a strictly monotonic subsequence converging to infinity. In sum, this result relies on monotonicity and thus on order arguments.
– Daniele Tampieri
Aug 2 at 8:56
add a comment |Â
2 Answers
2
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oldest
votes
up vote
5
down vote
accepted
Let $epsilon >0$. There exists $m$ such that $|a_n-L| <epsilon$ for $n >m$. There exists $p$ such that $frac 1 k_n < frac 1 m$ for $n>p$. If $n >p$ then $|a_k_n -L| < epsilon$.
answered Aug 2 at 7:31
Kavi Rama Murthy
19.2k2829
add a comment |Â
up vote
2
down vote
If $lim_n frac1k_n to 0$, then it is obvious that $lim_n k_n to infty$ (you can use simple $epsilon-delta$ techniques to prove this). If a sequence converges, then every subsequence converges to the same limit (Prove: If a sequence converges, then every subsequence converges to the same limit.). $a_k_n$ is just a subsequence, so the result follows.
answered Aug 2 at 7:30
user143234
1298
2
This may be a technicality, but I think it's worth pointing out for these type of exercises: $(k_n)_n geq 1$ need not be increasing and therefore $a_k_n$ is not always well defined.
– Guido A.
Aug 2 at 7:33
1
My answer shows that $a_k_n$ converges to $L$ whether or not it is a subsequence of $a_n$.
– Kavi Rama Murthy
Aug 2 at 7:38
1
The sequence $k_n$ contains an increasing subsequence, so this argument can still be saved.
– Dirk Liebhold
Aug 2 at 7:41
3
I think convergence for a sub-sequence does not imply convergence for the general sequence to the same limit
– user143234
Aug 2 at 7:47
1
What I am saying that $k_n$ need not be strictly increasing to $infty$. It can even decrease in-between but the hypothesis tells us that $k_n to infty$ and this is crucial. In choosing $p$ I have used the fact that $k_n to infty$. You can use the term subsequence only when $k_n$ is strictly increasing.
– Kavi Rama Murthy
Aug 2 at 8:09
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Let $epsilon >0$. There exists $m$ such that $|a_n-L| <epsilon$ for $n >m$. There exists $p$ such that $frac 1 k_n < frac 1 m$ for $n>p$. If $n >p$ then $|a_k_n -L| < epsilon$.
answered Aug 2 at 7:31
Kavi Rama Murthy
19.2k2829
add a comment |Â
up vote
5
down vote
accepted
Let $epsilon >0$. There exists $m$ such that $|a_n-L| <epsilon$ for $n >m$. There exists $p$ such that $frac 1 k_n < frac 1 m$ for $n>p$. If $n >p$ then $|a_k_n -L| < epsilon$.
answered Aug 2 at 7:31
Kavi Rama Murthy
19.2k2829
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Let $epsilon >0$. There exists $m$ such that $|a_n-L| <epsilon$ for $n >m$. There exists $p$ such that $frac 1 k_n < frac 1 m$ for $n>p$. If $n >p$ then $|a_k_n -L| < epsilon$.
answered Aug 2 at 7:31
Kavi Rama Murthy
19.2k2829
Let $epsilon >0$. There exists $m$ such that $|a_n-L| <epsilon$ for $n >m$. There exists $p$ such that $frac 1 k_n < frac 1 m$ for $n>p$. If $n >p$ then $|a_k_n -L| < epsilon$.
answered Aug 2 at 7:31
Kavi Rama Murthy
19.2k2829
answered Aug 2 at 7:31
Kavi Rama Murthy
19.2k2829
answered Aug 2 at 7:31
Kavi Rama Murthy
19.2k2829
answered Aug 2 at 7:31
Kavi Rama Murthy
19.2k2829
19.2k2829
add a comment |Â
add a comment |Â
up vote
2
down vote
If $lim_n frac1k_n to 0$, then it is obvious that $lim_n k_n to infty$ (you can use simple $epsilon-delta$ techniques to prove this). If a sequence converges, then every subsequence converges to the same limit (Prove: If a sequence converges, then every subsequence converges to the same limit.). $a_k_n$ is just a subsequence, so the result follows.
answered Aug 2 at 7:30
user143234
1298
2
This may be a technicality, but I think it's worth pointing out for these type of exercises: $(k_n)_n geq 1$ need not be increasing and therefore $a_k_n$ is not always well defined.
– Guido A.
Aug 2 at 7:33
1
My answer shows that $a_k_n$ converges to $L$ whether or not it is a subsequence of $a_n$.
– Kavi Rama Murthy
Aug 2 at 7:38
1
The sequence $k_n$ contains an increasing subsequence, so this argument can still be saved.
– Dirk Liebhold
Aug 2 at 7:41
3
I think convergence for a sub-sequence does not imply convergence for the general sequence to the same limit
– user143234
Aug 2 at 7:47
1
What I am saying that $k_n$ need not be strictly increasing to $infty$. It can even decrease in-between but the hypothesis tells us that $k_n to infty$ and this is crucial. In choosing $p$ I have used the fact that $k_n to infty$. You can use the term subsequence only when $k_n$ is strictly increasing.
– Kavi Rama Murthy
Aug 2 at 8:09
 |Â
show 4 more comments
up vote
2
down vote
If $lim_n frac1k_n to 0$, then it is obvious that $lim_n k_n to infty$ (you can use simple $epsilon-delta$ techniques to prove this). If a sequence converges, then every subsequence converges to the same limit (Prove: If a sequence converges, then every subsequence converges to the same limit.). $a_k_n$ is just a subsequence, so the result follows.
answered Aug 2 at 7:30
user143234
1298
2
This may be a technicality, but I think it's worth pointing out for these type of exercises: $(k_n)_n geq 1$ need not be increasing and therefore $a_k_n$ is not always well defined.
– Guido A.
Aug 2 at 7:33
1
My answer shows that $a_k_n$ converges to $L$ whether or not it is a subsequence of $a_n$.
– Kavi Rama Murthy
Aug 2 at 7:38
1
The sequence $k_n$ contains an increasing subsequence, so this argument can still be saved.
– Dirk Liebhold
Aug 2 at 7:41
3
I think convergence for a sub-sequence does not imply convergence for the general sequence to the same limit
– user143234
Aug 2 at 7:47
1
What I am saying that $k_n$ need not be strictly increasing to $infty$. It can even decrease in-between but the hypothesis tells us that $k_n to infty$ and this is crucial. In choosing $p$ I have used the fact that $k_n to infty$. You can use the term subsequence only when $k_n$ is strictly increasing.
– Kavi Rama Murthy
Aug 2 at 8:09
 |Â
show 4 more comments
up vote
2
down vote
up vote
2
down vote
If $lim_n frac1k_n to 0$, then it is obvious that $lim_n k_n to infty$ (you can use simple $epsilon-delta$ techniques to prove this). If a sequence converges, then every subsequence converges to the same limit (Prove: If a sequence converges, then every subsequence converges to the same limit.). $a_k_n$ is just a subsequence, so the result follows.
answered Aug 2 at 7:30
user143234
1298
If $lim_n frac1k_n to 0$, then it is obvious that $lim_n k_n to infty$ (you can use simple $epsilon-delta$ techniques to prove this). If a sequence converges, then every subsequence converges to the same limit (Prove: If a sequence converges, then every subsequence converges to the same limit.). $a_k_n$ is just a subsequence, so the result follows.
answered Aug 2 at 7:30
user143234
1298
answered Aug 2 at 7:30
user143234
1298
answered Aug 2 at 7:30
user143234
1298
answered Aug 2 at 7:30
user143234
1298
1298
2
This may be a technicality, but I think it's worth pointing out for these type of exercises: $(k_n)_n geq 1$ need not be increasing and therefore $a_k_n$ is not always well defined.
– Guido A.
Aug 2 at 7:33
1
My answer shows that $a_k_n$ converges to $L$ whether or not it is a subsequence of $a_n$.
– Kavi Rama Murthy
Aug 2 at 7:38
1
The sequence $k_n$ contains an increasing subsequence, so this argument can still be saved.
– Dirk Liebhold
Aug 2 at 7:41
3
I think convergence for a sub-sequence does not imply convergence for the general sequence to the same limit
– user143234
Aug 2 at 7:47
1
What I am saying that $k_n$ need not be strictly increasing to $infty$. It can even decrease in-between but the hypothesis tells us that $k_n to infty$ and this is crucial. In choosing $p$ I have used the fact that $k_n to infty$. You can use the term subsequence only when $k_n$ is strictly increasing.
– Kavi Rama Murthy
Aug 2 at 8:09
 |Â
show 4 more comments
2
This may be a technicality, but I think it's worth pointing out for these type of exercises: $(k_n)_n geq 1$ need not be increasing and therefore $a_k_n$ is not always well defined.
– Guido A.
Aug 2 at 7:33
1
My answer shows that $a_k_n$ converges to $L$ whether or not it is a subsequence of $a_n$.
– Kavi Rama Murthy
Aug 2 at 7:38
1
The sequence $k_n$ contains an increasing subsequence, so this argument can still be saved.
– Dirk Liebhold
Aug 2 at 7:41
3
I think convergence for a sub-sequence does not imply convergence for the general sequence to the same limit
– user143234
Aug 2 at 7:47
1
What I am saying that $k_n$ need not be strictly increasing to $infty$. It can even decrease in-between but the hypothesis tells us that $k_n to infty$ and this is crucial. In choosing $p$ I have used the fact that $k_n to infty$. You can use the term subsequence only when $k_n$ is strictly increasing.
– Kavi Rama Murthy
Aug 2 at 8:09
2
2
This may be a technicality, but I think it's worth pointing out for these type of exercises: $(k_n)_n geq 1$ need not be increasing and therefore $a_k_n$ is not always well defined.
– Guido A.
Aug 2 at 7:33
This may be a technicality, but I think it's worth pointing out for these type of exercises: $(k_n)_n geq 1$ need not be increasing and therefore $a_k_n$ is not always well defined.
– Guido A.
Aug 2 at 7:33
1
1
My answer shows that $a_k_n$ converges to $L$ whether or not it is a subsequence of $a_n$.
– Kavi Rama Murthy
Aug 2 at 7:38
My answer shows that $a_k_n$ converges to $L$ whether or not it is a subsequence of $a_n$.
– Kavi Rama Murthy
Aug 2 at 7:38
1
1
The sequence $k_n$ contains an increasing subsequence, so this argument can still be saved.
– Dirk Liebhold
Aug 2 at 7:41
The sequence $k_n$ contains an increasing subsequence, so this argument can still be saved.
– Dirk Liebhold
Aug 2 at 7:41
3
3
I think convergence for a sub-sequence does not imply convergence for the general sequence to the same limit
– user143234
Aug 2 at 7:47
I think convergence for a sub-sequence does not imply convergence for the general sequence to the same limit
– user143234
Aug 2 at 7:47
1
1
What I am saying that $k_n$ need not be strictly increasing to $infty$. It can even decrease in-between but the hypothesis tells us that $k_n to infty$ and this is crucial. In choosing $p$ I have used the fact that $k_n to infty$. You can use the term subsequence only when $k_n$ is strictly increasing.
– Kavi Rama Murthy
Aug 2 at 8:09
What I am saying that $k_n$ need not be strictly increasing to $infty$. It can even decrease in-between but the hypothesis tells us that $k_n to infty$ and this is crucial. In choosing $p$ I have used the fact that $k_n to infty$. You can use the term subsequence only when $k_n$ is strictly increasing.
– Kavi Rama Murthy
Aug 2 at 8:09
 |Â
show 4 more comments
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1
A note: as @user143234 remarked in his answer, then the convergence of $a_k_n$ to $L$ follows from a general theorem in mathematical analysis, which can be found also in stadard textbooks as E. Fisher’s Intermediate real analysis. The general argument follows from the monotonicity of $limsup$, $liminf$ and from the fact that since $k_ntoinfty$ then it is possible to extract from it a strictly monotonic subsequence converging to infinity. In sum, this result relies on monotonicity and thus on order arguments.
– Daniele Tampieri
Aug 2 at 8:56