Assume $a_n$ is a positive sequence, and $ (1-frac {a_n} {a_{n+1}})n rightarrow a>1 $. Prove $a_n$...












0












$begingroup$


I have tried every technique I know; I can't solve this question, so I assume it's an algebraic trick. Any ideas?



Edit: [The question was written incorrectly, $a_n$ is supposed to be the divisor.]






calculus sequences-and-series







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Are you sure the assumptions are correctly stated? The current version of your statement implies that $a_n$ is eventually strictly increasing. (To see this, notice that $1-frac{a_n}{a_{n+1}}$ must be eventually positive.) So there is no chance your sequence converges to $0$. Rather, one can prove sort of the opposite: $a_n$ diverges to $infty$.
    $endgroup$
    – Sangchul Lee
    Jan 23 at 7:45












  • $begingroup$
    Look at the sequence $a_n=2(n-1)$ then your given expression converges to 2. But $a_n$ obviously goes to infinity.
    $endgroup$
    – A. P
    Jan 23 at 8:33






  • 2




    $begingroup$
    @José: I wonder why you approved the last edit. The TeX is broken, and the fraction (which was originally $frac{a_n}{a_{n+1}}$) is reversed, so that an answer is invalidated.
    $endgroup$
    – Martin R
    Jan 23 at 9:42








  • 1




    $begingroup$
    This seems like the Raabe test of series, where the sign of the terms got reversed.
    $endgroup$
    – xbh
    Jan 23 at 12:18










  • $begingroup$
    I'm sorry, it was supposed to be $a_n$ at the bottom, the Latex confused me.
    $endgroup$
    – Amit Levy
    Jan 23 at 17:16
















0












$begingroup$


I have tried every technique I know; I can't solve this question, so I assume it's an algebraic trick. Any ideas?



Edit: [The question was written incorrectly, $a_n$ is supposed to be the divisor.]






calculus sequences-and-series







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Are you sure the assumptions are correctly stated? The current version of your statement implies that $a_n$ is eventually strictly increasing. (To see this, notice that $1-frac{a_n}{a_{n+1}}$ must be eventually positive.) So there is no chance your sequence converges to $0$. Rather, one can prove sort of the opposite: $a_n$ diverges to $infty$.
    $endgroup$
    – Sangchul Lee
    Jan 23 at 7:45












  • $begingroup$
    Look at the sequence $a_n=2(n-1)$ then your given expression converges to 2. But $a_n$ obviously goes to infinity.
    $endgroup$
    – A. P
    Jan 23 at 8:33






  • 2




    $begingroup$
    @José: I wonder why you approved the last edit. The TeX is broken, and the fraction (which was originally $frac{a_n}{a_{n+1}}$) is reversed, so that an answer is invalidated.
    $endgroup$
    – Martin R
    Jan 23 at 9:42








  • 1




    $begingroup$
    This seems like the Raabe test of series, where the sign of the terms got reversed.
    $endgroup$
    – xbh
    Jan 23 at 12:18










  • $begingroup$
    I'm sorry, it was supposed to be $a_n$ at the bottom, the Latex confused me.
    $endgroup$
    – Amit Levy
    Jan 23 at 17:16














0












0








0





$begingroup$


I have tried every technique I know; I can't solve this question, so I assume it's an algebraic trick. Any ideas?



Edit: [The question was written incorrectly, $a_n$ is supposed to be the divisor.]






calculus sequences-and-series







share|cite|improve this question











$endgroup$




I have tried every technique I know; I can't solve this question, so I assume it's an algebraic trick. Any ideas?



Edit: [The question was written incorrectly, $a_n$ is supposed to be the divisor.]






calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 17:18







Amit Levy

















asked Jan 23 at 7:15









Amit LevyAmit Levy

747




747








  • 3




    $begingroup$
    Are you sure the assumptions are correctly stated? The current version of your statement implies that $a_n$ is eventually strictly increasing. (To see this, notice that $1-frac{a_n}{a_{n+1}}$ must be eventually positive.) So there is no chance your sequence converges to $0$. Rather, one can prove sort of the opposite: $a_n$ diverges to $infty$.
    $endgroup$
    – Sangchul Lee
    Jan 23 at 7:45












  • $begingroup$
    Look at the sequence $a_n=2(n-1)$ then your given expression converges to 2. But $a_n$ obviously goes to infinity.
    $endgroup$
    – A. P
    Jan 23 at 8:33






  • 2




    $begingroup$
    @José: I wonder why you approved the last edit. The TeX is broken, and the fraction (which was originally $frac{a_n}{a_{n+1}}$) is reversed, so that an answer is invalidated.
    $endgroup$
    – Martin R
    Jan 23 at 9:42








  • 1




    $begingroup$
    This seems like the Raabe test of series, where the sign of the terms got reversed.
    $endgroup$
    – xbh
    Jan 23 at 12:18










  • $begingroup$
    I'm sorry, it was supposed to be $a_n$ at the bottom, the Latex confused me.
    $endgroup$
    – Amit Levy
    Jan 23 at 17:16














  • 3




    $begingroup$
    Are you sure the assumptions are correctly stated? The current version of your statement implies that $a_n$ is eventually strictly increasing. (To see this, notice that $1-frac{a_n}{a_{n+1}}$ must be eventually positive.) So there is no chance your sequence converges to $0$. Rather, one can prove sort of the opposite: $a_n$ diverges to $infty$.
    $endgroup$
    – Sangchul Lee
    Jan 23 at 7:45












  • $begingroup$
    Look at the sequence $a_n=2(n-1)$ then your given expression converges to 2. But $a_n$ obviously goes to infinity.
    $endgroup$
    – A. P
    Jan 23 at 8:33






  • 2




    $begingroup$
    @José: I wonder why you approved the last edit. The TeX is broken, and the fraction (which was originally $frac{a_n}{a_{n+1}}$) is reversed, so that an answer is invalidated.
    $endgroup$
    – Martin R
    Jan 23 at 9:42








  • 1




    $begingroup$
    This seems like the Raabe test of series, where the sign of the terms got reversed.
    $endgroup$
    – xbh
    Jan 23 at 12:18










  • $begingroup$
    I'm sorry, it was supposed to be $a_n$ at the bottom, the Latex confused me.
    $endgroup$
    – Amit Levy
    Jan 23 at 17:16








3




3




$begingroup$
Are you sure the assumptions are correctly stated? The current version of your statement implies that $a_n$ is eventually strictly increasing. (To see this, notice that $1-frac{a_n}{a_{n+1}}$ must be eventually positive.) So there is no chance your sequence converges to $0$. Rather, one can prove sort of the opposite: $a_n$ diverges to $infty$.
$endgroup$
– Sangchul Lee
Jan 23 at 7:45






$begingroup$
Are you sure the assumptions are correctly stated? The current version of your statement implies that $a_n$ is eventually strictly increasing. (To see this, notice that $1-frac{a_n}{a_{n+1}}$ must be eventually positive.) So there is no chance your sequence converges to $0$. Rather, one can prove sort of the opposite: $a_n$ diverges to $infty$.
$endgroup$
– Sangchul Lee
Jan 23 at 7:45














$begingroup$
Look at the sequence $a_n=2(n-1)$ then your given expression converges to 2. But $a_n$ obviously goes to infinity.
$endgroup$
– A. P
Jan 23 at 8:33




$begingroup$
Look at the sequence $a_n=2(n-1)$ then your given expression converges to 2. But $a_n$ obviously goes to infinity.
$endgroup$
– A. P
Jan 23 at 8:33




2




2




$begingroup$
@José: I wonder why you approved the last edit. The TeX is broken, and the fraction (which was originally $frac{a_n}{a_{n+1}}$) is reversed, so that an answer is invalidated.
$endgroup$
– Martin R
Jan 23 at 9:42






$begingroup$
@José: I wonder why you approved the last edit. The TeX is broken, and the fraction (which was originally $frac{a_n}{a_{n+1}}$) is reversed, so that an answer is invalidated.
$endgroup$
– Martin R
Jan 23 at 9:42






1




1




$begingroup$
This seems like the Raabe test of series, where the sign of the terms got reversed.
$endgroup$
– xbh
Jan 23 at 12:18




$begingroup$
This seems like the Raabe test of series, where the sign of the terms got reversed.
$endgroup$
– xbh
Jan 23 at 12:18












$begingroup$
I'm sorry, it was supposed to be $a_n$ at the bottom, the Latex confused me.
$endgroup$
– Amit Levy
Jan 23 at 17:16




$begingroup$
I'm sorry, it was supposed to be $a_n$ at the bottom, the Latex confused me.
$endgroup$
– Amit Levy
Jan 23 at 17:16










1 Answer
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Actually $a_n to infty$!. There exist $b>1$ and $n_0$ such that $1-frac {a_n} {a_{n+1}} >frac b n$ for $n geq n_0$. This gives $frac {a_{n+1}} {a_n} >c_n$ where $c_n=frac 1 {1-frac b n}$. Iterating this gives $a_{n+n_0} >a_{n_0} prod_{k=n_0}^{n+n_0} c_k$. Hence it is enough to show that $prod_{k=n_0}^{infty} c_k=infty$. Equivalently we have to show that $sumlimits_{k=1}^{infty} log (c_k)= infty$. This follows from the divergence of harmonic series and the fact that $log(1-x)$ behaves like $x$ for $x$ near $0$.






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    $begingroup$

    Actually $a_n to infty$!. There exist $b>1$ and $n_0$ such that $1-frac {a_n} {a_{n+1}} >frac b n$ for $n geq n_0$. This gives $frac {a_{n+1}} {a_n} >c_n$ where $c_n=frac 1 {1-frac b n}$. Iterating this gives $a_{n+n_0} >a_{n_0} prod_{k=n_0}^{n+n_0} c_k$. Hence it is enough to show that $prod_{k=n_0}^{infty} c_k=infty$. Equivalently we have to show that $sumlimits_{k=1}^{infty} log (c_k)= infty$. This follows from the divergence of harmonic series and the fact that $log(1-x)$ behaves like $x$ for $x$ near $0$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Actually $a_n to infty$!. There exist $b>1$ and $n_0$ such that $1-frac {a_n} {a_{n+1}} >frac b n$ for $n geq n_0$. This gives $frac {a_{n+1}} {a_n} >c_n$ where $c_n=frac 1 {1-frac b n}$. Iterating this gives $a_{n+n_0} >a_{n_0} prod_{k=n_0}^{n+n_0} c_k$. Hence it is enough to show that $prod_{k=n_0}^{infty} c_k=infty$. Equivalently we have to show that $sumlimits_{k=1}^{infty} log (c_k)= infty$. This follows from the divergence of harmonic series and the fact that $log(1-x)$ behaves like $x$ for $x$ near $0$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Actually $a_n to infty$!. There exist $b>1$ and $n_0$ such that $1-frac {a_n} {a_{n+1}} >frac b n$ for $n geq n_0$. This gives $frac {a_{n+1}} {a_n} >c_n$ where $c_n=frac 1 {1-frac b n}$. Iterating this gives $a_{n+n_0} >a_{n_0} prod_{k=n_0}^{n+n_0} c_k$. Hence it is enough to show that $prod_{k=n_0}^{infty} c_k=infty$. Equivalently we have to show that $sumlimits_{k=1}^{infty} log (c_k)= infty$. This follows from the divergence of harmonic series and the fact that $log(1-x)$ behaves like $x$ for $x$ near $0$.






        share|cite|improve this answer











        $endgroup$



        Actually $a_n to infty$!. There exist $b>1$ and $n_0$ such that $1-frac {a_n} {a_{n+1}} >frac b n$ for $n geq n_0$. This gives $frac {a_{n+1}} {a_n} >c_n$ where $c_n=frac 1 {1-frac b n}$. Iterating this gives $a_{n+n_0} >a_{n_0} prod_{k=n_0}^{n+n_0} c_k$. Hence it is enough to show that $prod_{k=n_0}^{infty} c_k=infty$. Equivalently we have to show that $sumlimits_{k=1}^{infty} log (c_k)= infty$. This follows from the divergence of harmonic series and the fact that $log(1-x)$ behaves like $x$ for $x$ near $0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 23 at 8:06

























        answered Jan 23 at 7:46









        Kavi Rama MurthyKavi Rama Murthy

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