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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  • 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














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

















  • 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












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













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












share|cite|improve this question




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edited Jul 31 at 14:58
























asked Jul 31 at 14:52









Legend Killer

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  • 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




    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















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