Boundedness criterion for a sequence

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Let $x_1=0$.
$x_n+1=ax_n+frac1n$ where $a>0$.
Prove that $x_n$ is bounded iff $0<a<1$.

In the if part,
we have $x_n+1-x_n<frac1n$
From this we get $x_n<sum_k=1^n-1 frac1k$.
But I cannot show that the sequence is uniformly bounded.
No ideas for the only if part.

real-analysis sequences-and-series convergence

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edited Jul 31 at 14:58

asked Jul 31 at 14:52

Legend Killer

1,500523

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In the if part you have not used the fact that $0<alpha<1$. Moreover, $sum_k=1^n-1k^-1$ is not bounded.
  – Julián Aguirre
  Jul 31 at 15:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint: prove that your recurrence has the closed form $$ x_n = frac1n-1 + fracan-2+fraca^2n-3+dots+fraca^n-21 $$
  – Mike Earnest
  Jul 31 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  We have $x_2=1.$ If $ageq 1$ then use induction on $n geq 2:$ If $x_ngeq S(n-1)=sum_j=1^n-1(1/j)$ then $x_n+1=$ $ax_n+1/ngeq x_n+1/ngeq $ $S(n-1)+1/n=S(n).$
  – DanielWainfleet
  Jul 31 at 15:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MikeEarnest. Very good. Solves everything.
  – DanielWainfleet
  Jul 31 at 15:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Let $x_1=0$.
$x_n+1=ax_n+frac1n$ where $a>0$.
Prove that $x_n$ is bounded iff $0<a<1$.

In the if part,
we have $x_n+1-x_n<frac1n$
From this we get $x_n<sum_k=1^n-1 frac1k$.
But I cannot show that the sequence is uniformly bounded.
No ideas for the only if part.

real-analysis sequences-and-series convergence

share|cite|improve this question

edited Jul 31 at 14:58

asked Jul 31 at 14:52

Legend Killer

1,500523

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In the if part you have not used the fact that $0<alpha<1$. Moreover, $sum_k=1^n-1k^-1$ is not bounded.
  – Julián Aguirre
  Jul 31 at 15:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint: prove that your recurrence has the closed form $$ x_n = frac1n-1 + fracan-2+fraca^2n-3+dots+fraca^n-21 $$
  – Mike Earnest
  Jul 31 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  We have $x_2=1.$ If $ageq 1$ then use induction on $n geq 2:$ If $x_ngeq S(n-1)=sum_j=1^n-1(1/j)$ then $x_n+1=$ $ax_n+1/ngeq x_n+1/ngeq $ $S(n-1)+1/n=S(n).$
  – DanielWainfleet
  Jul 31 at 15:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MikeEarnest. Very good. Solves everything.
  – DanielWainfleet
  Jul 31 at 15:36
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $x_1=0$.
$x_n+1=ax_n+frac1n$ where $a>0$.
Prove that $x_n$ is bounded iff $0<a<1$.

In the if part,
we have $x_n+1-x_n<frac1n$
From this we get $x_n<sum_k=1^n-1 frac1k$.
But I cannot show that the sequence is uniformly bounded.
No ideas for the only if part.

real-analysis sequences-and-series convergence

share|cite|improve this question

edited Jul 31 at 14:58

asked Jul 31 at 14:52

Legend Killer

1,500523

Let $x_1=0$.
$x_n+1=ax_n+frac1n$ where $a>0$.
Prove that $x_n$ is bounded iff $0<a<1$.

In the if part,
we have $x_n+1-x_n<frac1n$
From this we get $x_n<sum_k=1^n-1 frac1k$.
But I cannot show that the sequence is uniformly bounded.
No ideas for the only if part.

real-analysis sequences-and-series convergence

share|cite|improve this question

edited Jul 31 at 14:58

asked Jul 31 at 14:52

Legend Killer

1,500523

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 14:58

edited Jul 31 at 14:58

edited Jul 31 at 14:58

asked Jul 31 at 14:52

Legend Killer

1,500523

asked Jul 31 at 14:52

Legend Killer

1,500523

asked Jul 31 at 14:52

Legend Killer

1,500523

1,500523

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In the if part you have not used the fact that $0<alpha<1$. Moreover, $sum_k=1^n-1k^-1$ is not bounded.
  – Julián Aguirre
  Jul 31 at 15:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint: prove that your recurrence has the closed form $$ x_n = frac1n-1 + fracan-2+fraca^2n-3+dots+fraca^n-21 $$
  – Mike Earnest
  Jul 31 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  We have $x_2=1.$ If $ageq 1$ then use induction on $n geq 2:$ If $x_ngeq S(n-1)=sum_j=1^n-1(1/j)$ then $x_n+1=$ $ax_n+1/ngeq x_n+1/ngeq $ $S(n-1)+1/n=S(n).$
  – DanielWainfleet
  Jul 31 at 15:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MikeEarnest. Very good. Solves everything.
  – DanielWainfleet
  Jul 31 at 15:36
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In the if part you have not used the fact that $0<alpha<1$. Moreover, $sum_k=1^n-1k^-1$ is not bounded.
  – Julián Aguirre
  Jul 31 at 15:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint: prove that your recurrence has the closed form $$ x_n = frac1n-1 + fracan-2+fraca^2n-3+dots+fraca^n-21 $$
  – Mike Earnest
  Jul 31 at 15:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  We have $x_2=1.$ If $ageq 1$ then use induction on $n geq 2:$ If $x_ngeq S(n-1)=sum_j=1^n-1(1/j)$ then $x_n+1=$ $ax_n+1/ngeq x_n+1/ngeq $ $S(n-1)+1/n=S(n).$
  – DanielWainfleet
  Jul 31 at 15:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MikeEarnest. Very good. Solves everything.
  – DanielWainfleet
  Jul 31 at 15:36
  

  
  
  
  

1

1

In the if part you have not used the fact that $0<alpha<1$. Moreover, $sum_k=1^n-1k^-1$ is not bounded.
– Julián Aguirre
Jul 31 at 15:13

In the if part you have not used the fact that $0<alpha<1$. Moreover, $sum_k=1^n-1k^-1$ is not bounded.
– Julián Aguirre
Jul 31 at 15:13

Hint: prove that your recurrence has the closed form $$ x_n = frac1n-1 + fracan-2+fraca^2n-3+dots+fraca^n-21 $$
– Mike Earnest
Jul 31 at 15:31

Hint: prove that your recurrence has the closed form $$ x_n = frac1n-1 + fracan-2+fraca^2n-3+dots+fraca^n-21 $$
– Mike Earnest
Jul 31 at 15:31

We have $x_2=1.$ If $ageq 1$ then use induction on $n geq 2:$ If $x_ngeq S(n-1)=sum_j=1^n-1(1/j)$ then $x_n+1=$ $ax_n+1/ngeq x_n+1/ngeq $ $S(n-1)+1/n=S(n).$
– DanielWainfleet
Jul 31 at 15:32

We have $x_2=1.$ If $ageq 1$ then use induction on $n geq 2:$ If $x_ngeq S(n-1)=sum_j=1^n-1(1/j)$ then $x_n+1=$ $ax_n+1/ngeq x_n+1/ngeq $ $S(n-1)+1/n=S(n).$
– DanielWainfleet
Jul 31 at 15:32

@MikeEarnest. Very good. Solves everything.
– DanielWainfleet
Jul 31 at 15:36

@MikeEarnest. Very good. Solves everything.
– DanielWainfleet
Jul 31 at 15:36

add a comment | 

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