Prove or disprove that there always exists a sequence satisfying a relation with a real constant $k$.
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Question:
Prove or disprove for a real number $k$ there always exists a sequence $x_n$ where $lim_{nto infty}x_n=k$ and $$lim_{nto infty} frac{x_{n+1}-k}{x_n-k}=lim_{ntoinfty}frac{x_{n+1}}{x_n}$$
Here I'm interested in a proof. This will not have a counter example as all
possible sequences are to be considered.
Tell me if I can do something to improve the question.
Post Script:
This question is a mere work of curiosity and is inspired by a question put up on this website. It is not duplicate and is slightly different from the original one.
Anyways here is the link
real-analysis sequences-and-series limits
asked Dec 20 '18 at 17:10
Love InvariantsLove Invariants
89015
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add a comment |
$begingroup$
Question:
Prove or disprove for a real number $k$ there always exists a sequence $x_n$ where $lim_{nto infty}x_n=k$ and $$lim_{nto infty} frac{x_{n+1}-k}{x_n-k}=lim_{ntoinfty}frac{x_{n+1}}{x_n}$$
Here I'm interested in a proof. This will not have a counter example as all
possible sequences are to be considered.
Tell me if I can do something to improve the question.
Post Script:
This question is a mere work of curiosity and is inspired by a question put up on this website. It is not duplicate and is slightly different from the original one.
Anyways here is the link
real-analysis sequences-and-series limits
asked Dec 20 '18 at 17:10
Love InvariantsLove Invariants
89015
$endgroup$
$begingroup$
Did you mean to say $lim_{nto infty} x_n=k$? Also, what is $l$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:12
$begingroup$
@BigbearZzz Here series is defined by the constant $k$ and sorry for that $l$
$endgroup$
– Love Invariants
Dec 20 '18 at 17:16
$begingroup$
There's a part where you wrote just $lim_{nto infty}=k$. Shouldn't it be $lim_{nto infty}x_n=k$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:17
$begingroup$
Oh yep...........
$endgroup$
– Love Invariants
Dec 20 '18 at 17:20
$begingroup$
$$x_n=k+ frac{1}{n}$$ works.
$endgroup$
– Crostul
Dec 20 '18 at 17:23
add a comment |
$begingroup$
Question:
Prove or disprove for a real number $k$ there always exists a sequence $x_n$ where $lim_{nto infty}x_n=k$ and $$lim_{nto infty} frac{x_{n+1}-k}{x_n-k}=lim_{ntoinfty}frac{x_{n+1}}{x_n}$$
Here I'm interested in a proof. This will not have a counter example as all
possible sequences are to be considered.
Tell me if I can do something to improve the question.
Post Script:
This question is a mere work of curiosity and is inspired by a question put up on this website. It is not duplicate and is slightly different from the original one.
Anyways here is the link
real-analysis sequences-and-series limits
asked Dec 20 '18 at 17:10
Love InvariantsLove Invariants
89015
$endgroup$
Question:
Prove or disprove for a real number $k$ there always exists a sequence $x_n$ where $lim_{nto infty}x_n=k$ and $$lim_{nto infty} frac{x_{n+1}-k}{x_n-k}=lim_{ntoinfty}frac{x_{n+1}}{x_n}$$
Here I'm interested in a proof. This will not have a counter example as all
possible sequences are to be considered.
Tell me if I can do something to improve the question.
Post Script:
This question is a mere work of curiosity and is inspired by a question put up on this website. It is not duplicate and is slightly different from the original one.
Anyways here is the link
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
asked Dec 20 '18 at 17:10
Love InvariantsLove Invariants
89015
asked Dec 20 '18 at 17:10
Love InvariantsLove Invariants
89015
edited Dec 20 '18 at 17:20
Love Invariants
asked Dec 20 '18 at 17:10
Love InvariantsLove Invariants
89015
asked Dec 20 '18 at 17:10
Love InvariantsLove Invariants
89015
asked Dec 20 '18 at 17:10
Love InvariantsLove Invariants
89015
89015
$begingroup$
Did you mean to say $lim_{nto infty} x_n=k$? Also, what is $l$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:12
$begingroup$
@BigbearZzz Here series is defined by the constant $k$ and sorry for that $l$
$endgroup$
– Love Invariants
Dec 20 '18 at 17:16
$begingroup$
There's a part where you wrote just $lim_{nto infty}=k$. Shouldn't it be $lim_{nto infty}x_n=k$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:17
$begingroup$
Oh yep...........
$endgroup$
– Love Invariants
Dec 20 '18 at 17:20
$begingroup$
$$x_n=k+ frac{1}{n}$$ works.
$endgroup$
– Crostul
Dec 20 '18 at 17:23
add a comment |
$begingroup$
Did you mean to say $lim_{nto infty} x_n=k$? Also, what is $l$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:12
$begingroup$
@BigbearZzz Here series is defined by the constant $k$ and sorry for that $l$
$endgroup$
– Love Invariants
Dec 20 '18 at 17:16
$begingroup$
There's a part where you wrote just $lim_{nto infty}=k$. Shouldn't it be $lim_{nto infty}x_n=k$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:17
$begingroup$
Oh yep...........
$endgroup$
– Love Invariants
Dec 20 '18 at 17:20
$begingroup$
$$x_n=k+ frac{1}{n}$$ works.
$endgroup$
– Crostul
Dec 20 '18 at 17:23
$begingroup$
Did you mean to say $lim_{nto infty} x_n=k$? Also, what is $l$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:12
$begingroup$
Did you mean to say $lim_{nto infty} x_n=k$? Also, what is $l$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:12
$begingroup$
@BigbearZzz Here series is defined by the constant $k$ and sorry for that $l$
$endgroup$
– Love Invariants
Dec 20 '18 at 17:16
$begingroup$
@BigbearZzz Here series is defined by the constant $k$ and sorry for that $l$
$endgroup$
– Love Invariants
Dec 20 '18 at 17:16
$begingroup$
There's a part where you wrote just $lim_{nto infty}=k$. Shouldn't it be $lim_{nto infty}x_n=k$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:17
$begingroup$
There's a part where you wrote just $lim_{nto infty}=k$. Shouldn't it be $lim_{nto infty}x_n=k$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:17
$begingroup$
Oh yep...........
$endgroup$
– Love Invariants
Dec 20 '18 at 17:20
$begingroup$
Oh yep...........
$endgroup$
– Love Invariants
Dec 20 '18 at 17:20
$begingroup$
$$x_n=k+ frac{1}{n}$$ works.
$endgroup$
– Crostul
Dec 20 '18 at 17:23
$begingroup$
$$x_n=k+ frac{1}{n}$$ works.
$endgroup$
– Crostul
Dec 20 '18 at 17:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can consider the sequence
$$
x_n = k+frac 1n
$$
We clearly have $lim_{ntoinfty} x_n=k$ and
$$
lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
$$
answered Dec 20 '18 at 17:24
BigbearZzzBigbearZzz
9,04221653
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
You can consider the sequence
$$
x_n = k+frac 1n
$$
We clearly have $lim_{ntoinfty} x_n=k$ and
$$
lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
$$
answered Dec 20 '18 at 17:24
BigbearZzzBigbearZzz
9,04221653
$endgroup$
add a comment |
$begingroup$
You can consider the sequence
$$
x_n = k+frac 1n
$$
We clearly have $lim_{ntoinfty} x_n=k$ and
$$
lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
$$
answered Dec 20 '18 at 17:24
BigbearZzzBigbearZzz
9,04221653
$endgroup$
add a comment |
$begingroup$
You can consider the sequence
$$
x_n = k+frac 1n
$$
We clearly have $lim_{ntoinfty} x_n=k$ and
$$
lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
$$
answered Dec 20 '18 at 17:24
BigbearZzzBigbearZzz
9,04221653
$endgroup$
You can consider the sequence
$$
x_n = k+frac 1n
$$
We clearly have $lim_{ntoinfty} x_n=k$ and
$$
lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
$$
answered Dec 20 '18 at 17:24
BigbearZzzBigbearZzz
9,04221653
answered Dec 20 '18 at 17:24
BigbearZzzBigbearZzz
9,04221653
answered Dec 20 '18 at 17:24
BigbearZzzBigbearZzz
9,04221653
answered Dec 20 '18 at 17:24
BigbearZzzBigbearZzz
9,04221653
9,04221653
add a comment |
add a comment |
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$begingroup$
Did you mean to say $lim_{nto infty} x_n=k$? Also, what is $l$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:12
$begingroup$
@BigbearZzz Here series is defined by the constant $k$ and sorry for that $l$
$endgroup$
– Love Invariants
Dec 20 '18 at 17:16
$begingroup$
There's a part where you wrote just $lim_{nto infty}=k$. Shouldn't it be $lim_{nto infty}x_n=k$?
$endgroup$
– BigbearZzz
Dec 20 '18 at 17:17
$begingroup$
Oh yep...........
$endgroup$
– Love Invariants
Dec 20 '18 at 17:20
$begingroup$
$$x_n=k+ frac{1}{n}$$ works.
$endgroup$
– Crostul
Dec 20 '18 at 17:23