Mixing
Sequences that diverge, limit stuff
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I have a quadratic based sequence,
say its of the general form : $y = ax^2 + bx +c$
To put a concrete example the sequence : $a(n) = n^2 +5n +6$
$lima(n)toinfty$ when $ntoinfty$
I am trying to think of an Epsilon-N way to prove this.
I am not sure how to start to break down the quadratic, would one factor it in product form and manipulate that, or would one use some kind of Triangle Inequality on the quadratic.
Hope to get some clarification on this.
sequences-and-series limits limits-without-lhopital
edited Jul 22 at 4:55
prog_SAHIL
823217
asked Jul 22 at 4:37
Palu
2952620
add a comment |Â
up vote
0
down vote
favorite
I have a quadratic based sequence,
say its of the general form : $y = ax^2 + bx +c$
To put a concrete example the sequence : $a(n) = n^2 +5n +6$
$lima(n)toinfty$ when $ntoinfty$
I am trying to think of an Epsilon-N way to prove this.
I am not sure how to start to break down the quadratic, would one factor it in product form and manipulate that, or would one use some kind of Triangle Inequality on the quadratic.
Hope to get some clarification on this.
sequences-and-series limits limits-without-lhopital
edited Jul 22 at 4:55
prog_SAHIL
823217
asked Jul 22 at 4:37
Palu
2952620
You won't actually have an $varepsilon$ value, but factor out $n^2$ and look at $a(n)=n^2(1+frac5n+frac6n^2)geq n^2$.
– Clayton
Jul 22 at 4:57
Hi Clayton, so I see that when you factored out that way, one factor being the squared term the other factor would tend to --> 1 as n--> infinity. that is clever. But hence can we just do an Epsilon-N proof (I know that there is no epsilon, i just call it that) on the quadratic part n^2.
– Palu
Jul 22 at 5:06
So we choose some N, such that n >N, a(n) > M [for any (large, positive) number M] so therefore N^2 > M, therefore N > root(M). Is this sufficient to do this just for treating this : a(n) as a(n) == n^2.
– Palu
Jul 22 at 5:10
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a quadratic based sequence,
say its of the general form : $y = ax^2 + bx +c$
To put a concrete example the sequence : $a(n) = n^2 +5n +6$
$lima(n)toinfty$ when $ntoinfty$
I am trying to think of an Epsilon-N way to prove this.
I am not sure how to start to break down the quadratic, would one factor it in product form and manipulate that, or would one use some kind of Triangle Inequality on the quadratic.
Hope to get some clarification on this.
sequences-and-series limits limits-without-lhopital
edited Jul 22 at 4:55
prog_SAHIL
823217
asked Jul 22 at 4:37
Palu
2952620
I have a quadratic based sequence,
say its of the general form : $y = ax^2 + bx +c$
To put a concrete example the sequence : $a(n) = n^2 +5n +6$
$lima(n)toinfty$ when $ntoinfty$
I am trying to think of an Epsilon-N way to prove this.
I am not sure how to start to break down the quadratic, would one factor it in product form and manipulate that, or would one use some kind of Triangle Inequality on the quadratic.
Hope to get some clarification on this.
sequences-and-series limits limits-without-lhopital
edited Jul 22 at 4:55
prog_SAHIL
823217
asked Jul 22 at 4:37
Palu
2952620
edited Jul 22 at 4:55
prog_SAHIL
823217
edited Jul 22 at 4:55
prog_SAHIL
823217
edited Jul 22 at 4:55
prog_SAHIL
823217
823217
asked Jul 22 at 4:37
Palu
2952620
asked Jul 22 at 4:37
Palu
2952620
asked Jul 22 at 4:37
Palu
2952620
2952620
You won't actually have an $varepsilon$ value, but factor out $n^2$ and look at $a(n)=n^2(1+frac5n+frac6n^2)geq n^2$.
– Clayton
Jul 22 at 4:57
Hi Clayton, so I see that when you factored out that way, one factor being the squared term the other factor would tend to --> 1 as n--> infinity. that is clever. But hence can we just do an Epsilon-N proof (I know that there is no epsilon, i just call it that) on the quadratic part n^2.
– Palu
Jul 22 at 5:06
So we choose some N, such that n >N, a(n) > M [for any (large, positive) number M] so therefore N^2 > M, therefore N > root(M). Is this sufficient to do this just for treating this : a(n) as a(n) == n^2.
– Palu
Jul 22 at 5:10
add a comment |Â
You won't actually have an $varepsilon$ value, but factor out $n^2$ and look at $a(n)=n^2(1+frac5n+frac6n^2)geq n^2$.
– Clayton
Jul 22 at 4:57
Hi Clayton, so I see that when you factored out that way, one factor being the squared term the other factor would tend to --> 1 as n--> infinity. that is clever. But hence can we just do an Epsilon-N proof (I know that there is no epsilon, i just call it that) on the quadratic part n^2.
– Palu
Jul 22 at 5:06
So we choose some N, such that n >N, a(n) > M [for any (large, positive) number M] so therefore N^2 > M, therefore N > root(M). Is this sufficient to do this just for treating this : a(n) as a(n) == n^2.
– Palu
Jul 22 at 5:10
You won't actually have an $varepsilon$ value, but factor out $n^2$ and look at $a(n)=n^2(1+frac5n+frac6n^2)geq n^2$.
– Clayton
Jul 22 at 4:57
You won't actually have an $varepsilon$ value, but factor out $n^2$ and look at $a(n)=n^2(1+frac5n+frac6n^2)geq n^2$.
– Clayton
Jul 22 at 4:57
Hi Clayton, so I see that when you factored out that way, one factor being the squared term the other factor would tend to --> 1 as n--> infinity. that is clever. But hence can we just do an Epsilon-N proof (I know that there is no epsilon, i just call it that) on the quadratic part n^2.
– Palu
Jul 22 at 5:06
Hi Clayton, so I see that when you factored out that way, one factor being the squared term the other factor would tend to --> 1 as n--> infinity. that is clever. But hence can we just do an Epsilon-N proof (I know that there is no epsilon, i just call it that) on the quadratic part n^2.
– Palu
Jul 22 at 5:06
So we choose some N, such that n >N, a(n) > M [for any (large, positive) number M] so therefore N^2 > M, therefore N > root(M). Is this sufficient to do this just for treating this : a(n) as a(n) == n^2.
– Palu
Jul 22 at 5:10
So we choose some N, such that n >N, a(n) > M [for any (large, positive) number M] so therefore N^2 > M, therefore N > root(M). Is this sufficient to do this just for treating this : a(n) as a(n) == n^2.
– Palu
Jul 22 at 5:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
We notice that for positive $n$ we have $$a(n)> n^2$$Now for each $M$ we can choose $N=maxM$ and thus for $n>N$ we get $$a(n)>n^2ge N^2ge |M|^2ge |M|ge M$$
answered Jul 22 at 5:30
Holo
4,2512629
Hi Holo, I think I understand, based on your analysis and Clayton. Via the factoring from Clayton, we are able to set up an inequality to use such as a(n) >n^2. And if we want a(n) > M, for each n > N. which means for each choice of n, we always want that a(n) > M.
– Palu
Jul 22 at 5:36
Yes, basically we notice that $a(n)>b(n)$ and then show that for all $M$ we have a point such that after that point we have $b(n)>M$ for all $n$, that implies $a(n)>M$. to remind you the definition of limit diverge to infinity: $$forall Mexists Nforall nquad n>Nimplies a(n)>M$$(note that this definition sometimes being called $M-N$ definition as analogy to the $varepsilon-delta$ and $varepsilon-N$ definitions)
– Holo
Jul 22 at 5:44
Hi Holo, thanks for the recap of the different definitions of these precise limit definitions, that is really helpful in general. I will award your answer as THE answer to this question. Regards, Palu
– Palu
Jul 22 at 5:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
We notice that for positive $n$ we have $$a(n)> n^2$$Now for each $M$ we can choose $N=maxM$ and thus for $n>N$ we get $$a(n)>n^2ge N^2ge |M|^2ge |M|ge M$$
answered Jul 22 at 5:30
Holo
4,2512629
Hi Holo, I think I understand, based on your analysis and Clayton. Via the factoring from Clayton, we are able to set up an inequality to use such as a(n) >n^2. And if we want a(n) > M, for each n > N. which means for each choice of n, we always want that a(n) > M.
– Palu
Jul 22 at 5:36
Yes, basically we notice that $a(n)>b(n)$ and then show that for all $M$ we have a point such that after that point we have $b(n)>M$ for all $n$, that implies $a(n)>M$. to remind you the definition of limit diverge to infinity: $$forall Mexists Nforall nquad n>Nimplies a(n)>M$$(note that this definition sometimes being called $M-N$ definition as analogy to the $varepsilon-delta$ and $varepsilon-N$ definitions)
– Holo
Jul 22 at 5:44
Hi Holo, thanks for the recap of the different definitions of these precise limit definitions, that is really helpful in general. I will award your answer as THE answer to this question. Regards, Palu
– Palu
Jul 22 at 5:48
add a comment |Â
up vote
2
down vote
accepted
We notice that for positive $n$ we have $$a(n)> n^2$$Now for each $M$ we can choose $N=maxM$ and thus for $n>N$ we get $$a(n)>n^2ge N^2ge |M|^2ge |M|ge M$$
answered Jul 22 at 5:30
Holo
4,2512629
Hi Holo, I think I understand, based on your analysis and Clayton. Via the factoring from Clayton, we are able to set up an inequality to use such as a(n) >n^2. And if we want a(n) > M, for each n > N. which means for each choice of n, we always want that a(n) > M.
– Palu
Jul 22 at 5:36
Yes, basically we notice that $a(n)>b(n)$ and then show that for all $M$ we have a point such that after that point we have $b(n)>M$ for all $n$, that implies $a(n)>M$. to remind you the definition of limit diverge to infinity: $$forall Mexists Nforall nquad n>Nimplies a(n)>M$$(note that this definition sometimes being called $M-N$ definition as analogy to the $varepsilon-delta$ and $varepsilon-N$ definitions)
– Holo
Jul 22 at 5:44
Hi Holo, thanks for the recap of the different definitions of these precise limit definitions, that is really helpful in general. I will award your answer as THE answer to this question. Regards, Palu
– Palu
Jul 22 at 5:48
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
We notice that for positive $n$ we have $$a(n)> n^2$$Now for each $M$ we can choose $N=maxM$ and thus for $n>N$ we get $$a(n)>n^2ge N^2ge |M|^2ge |M|ge M$$
answered Jul 22 at 5:30
Holo
4,2512629
We notice that for positive $n$ we have $$a(n)> n^2$$Now for each $M$ we can choose $N=maxM$ and thus for $n>N$ we get $$a(n)>n^2ge N^2ge |M|^2ge |M|ge M$$
answered Jul 22 at 5:30
Holo
4,2512629
answered Jul 22 at 5:30
Holo
4,2512629
answered Jul 22 at 5:30
Holo
4,2512629
answered Jul 22 at 5:30
Holo
4,2512629
4,2512629
Hi Holo, I think I understand, based on your analysis and Clayton. Via the factoring from Clayton, we are able to set up an inequality to use such as a(n) >n^2. And if we want a(n) > M, for each n > N. which means for each choice of n, we always want that a(n) > M.
– Palu
Jul 22 at 5:36
Yes, basically we notice that $a(n)>b(n)$ and then show that for all $M$ we have a point such that after that point we have $b(n)>M$ for all $n$, that implies $a(n)>M$. to remind you the definition of limit diverge to infinity: $$forall Mexists Nforall nquad n>Nimplies a(n)>M$$(note that this definition sometimes being called $M-N$ definition as analogy to the $varepsilon-delta$ and $varepsilon-N$ definitions)
– Holo
Jul 22 at 5:44
Hi Holo, thanks for the recap of the different definitions of these precise limit definitions, that is really helpful in general. I will award your answer as THE answer to this question. Regards, Palu
– Palu
Jul 22 at 5:48
add a comment |Â
Hi Holo, I think I understand, based on your analysis and Clayton. Via the factoring from Clayton, we are able to set up an inequality to use such as a(n) >n^2. And if we want a(n) > M, for each n > N. which means for each choice of n, we always want that a(n) > M.
– Palu
Jul 22 at 5:36
Yes, basically we notice that $a(n)>b(n)$ and then show that for all $M$ we have a point such that after that point we have $b(n)>M$ for all $n$, that implies $a(n)>M$. to remind you the definition of limit diverge to infinity: $$forall Mexists Nforall nquad n>Nimplies a(n)>M$$(note that this definition sometimes being called $M-N$ definition as analogy to the $varepsilon-delta$ and $varepsilon-N$ definitions)
– Holo
Jul 22 at 5:44
Hi Holo, thanks for the recap of the different definitions of these precise limit definitions, that is really helpful in general. I will award your answer as THE answer to this question. Regards, Palu
– Palu
Jul 22 at 5:48
Hi Holo, I think I understand, based on your analysis and Clayton. Via the factoring from Clayton, we are able to set up an inequality to use such as a(n) >n^2. And if we want a(n) > M, for each n > N. which means for each choice of n, we always want that a(n) > M.
– Palu
Jul 22 at 5:36
Hi Holo, I think I understand, based on your analysis and Clayton. Via the factoring from Clayton, we are able to set up an inequality to use such as a(n) >n^2. And if we want a(n) > M, for each n > N. which means for each choice of n, we always want that a(n) > M.
– Palu
Jul 22 at 5:36
Yes, basically we notice that $a(n)>b(n)$ and then show that for all $M$ we have a point such that after that point we have $b(n)>M$ for all $n$, that implies $a(n)>M$. to remind you the definition of limit diverge to infinity: $$forall Mexists Nforall nquad n>Nimplies a(n)>M$$(note that this definition sometimes being called $M-N$ definition as analogy to the $varepsilon-delta$ and $varepsilon-N$ definitions)
– Holo
Jul 22 at 5:44
Yes, basically we notice that $a(n)>b(n)$ and then show that for all $M$ we have a point such that after that point we have $b(n)>M$ for all $n$, that implies $a(n)>M$. to remind you the definition of limit diverge to infinity: $$forall Mexists Nforall nquad n>Nimplies a(n)>M$$(note that this definition sometimes being called $M-N$ definition as analogy to the $varepsilon-delta$ and $varepsilon-N$ definitions)
– Holo
Jul 22 at 5:44
Hi Holo, thanks for the recap of the different definitions of these precise limit definitions, that is really helpful in general. I will award your answer as THE answer to this question. Regards, Palu
– Palu
Jul 22 at 5:48
Hi Holo, thanks for the recap of the different definitions of these precise limit definitions, that is really helpful in general. I will award your answer as THE answer to this question. Regards, Palu
– Palu
Jul 22 at 5:48
add a comment |Â
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You won't actually have an $varepsilon$ value, but factor out $n^2$ and look at $a(n)=n^2(1+frac5n+frac6n^2)geq n^2$.
– Clayton
Jul 22 at 4:57
Hi Clayton, so I see that when you factored out that way, one factor being the squared term the other factor would tend to --> 1 as n--> infinity. that is clever. But hence can we just do an Epsilon-N proof (I know that there is no epsilon, i just call it that) on the quadratic part n^2.
– Palu
Jul 22 at 5:06
So we choose some N, such that n >N, a(n) > M [for any (large, positive) number M] so therefore N^2 > M, therefore N > root(M). Is this sufficient to do this just for treating this : a(n) as a(n) == n^2.
– Palu
Jul 22 at 5:10