Prove or disprove that there always exists a sequence satisfying a relation with a real constant $k$.












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 17:20







Love Invariants

















asked Dec 20 '18 at 17:10









Love InvariantsLove Invariants

86115




86115












  • $begingroup$
    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










  • $begingroup$
    @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










1 Answer
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$begingroup$

You can consider the sequence
$$
x_n = k+frac 1n
$$
We clearly have $lim_{ntoinfty} x_n=k$ and
$$
lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
$$






share|cite|improve this answer









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    $begingroup$

    You can consider the sequence
    $$
    x_n = k+frac 1n
    $$
    We clearly have $lim_{ntoinfty} x_n=k$ and
    $$
    lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
    $$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You can consider the sequence
      $$
      x_n = k+frac 1n
      $$
      We clearly have $lim_{ntoinfty} x_n=k$ and
      $$
      lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
      $$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You can consider the sequence
        $$
        x_n = k+frac 1n
        $$
        We clearly have $lim_{ntoinfty} x_n=k$ and
        $$
        lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
        $$






        share|cite|improve this answer









        $endgroup$



        You can consider the sequence
        $$
        x_n = k+frac 1n
        $$
        We clearly have $lim_{ntoinfty} x_n=k$ and
        $$
        lim_{ntoinfty} frac{x_{n+1}-k}{x_n -k} = lim_{ntoinfty} frac n{n+1} = 1 = lim_{ntoinfty}frac{k+frac 1{n+1}}{k+frac 1{n}} = lim_{ntoinfty} frac{x_{n+1}}{x_n} .
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 17:24









        BigbearZzzBigbearZzz

        8,88821652




        8,88821652






























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