Does the limit of the recursive sequence $a_n+1=fracn+1frac1a_n-(n+1)$ converge to the same value regardless of whatever $a_0$ is?

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I found an interesting recursive sequence $a_n+1=fracn+1frac1a_n-(n+1)$. In playing with $a_0$ on Desmos, it seems that no matter what $a_0$ is, the sequence converges to some fixed value. Is this the case? Is this sequence the same regardless of its initial value? If so, what is the limit of this sequence?




sequences-and-series limits recursion




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  • I've added parentheses around the $n+1$ in the denominator as this looks to be your intent based on the Desmos link.
    – Michael Lugo
    Jul 18 at 19:26










  • @MichaelLugo Yes, thank you, I tend to simplify to the point of my own detriment
    – user189728
    Jul 18 at 19:28














up vote
5
down vote

favorite
1












I found an interesting recursive sequence $a_n+1=fracn+1frac1a_n-(n+1)$. In playing with $a_0$ on Desmos, it seems that no matter what $a_0$ is, the sequence converges to some fixed value. Is this the case? Is this sequence the same regardless of its initial value? If so, what is the limit of this sequence?




sequences-and-series limits recursion




share|cite|improve this question





















  • I've added parentheses around the $n+1$ in the denominator as this looks to be your intent based on the Desmos link.
    – Michael Lugo
    Jul 18 at 19:26










  • @MichaelLugo Yes, thank you, I tend to simplify to the point of my own detriment
    – user189728
    Jul 18 at 19:28












up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





I found an interesting recursive sequence $a_n+1=fracn+1frac1a_n-(n+1)$. In playing with $a_0$ on Desmos, it seems that no matter what $a_0$ is, the sequence converges to some fixed value. Is this the case? Is this sequence the same regardless of its initial value? If so, what is the limit of this sequence?




sequences-and-series limits recursion




share|cite|improve this question













I found an interesting recursive sequence $a_n+1=fracn+1frac1a_n-(n+1)$. In playing with $a_0$ on Desmos, it seems that no matter what $a_0$ is, the sequence converges to some fixed value. Is this the case? Is this sequence the same regardless of its initial value? If so, what is the limit of this sequence?





sequences-and-series limits recursion





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 18 at 19:26









Michael Lugo

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asked Jul 18 at 19:05









user189728

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  • I've added parentheses around the $n+1$ in the denominator as this looks to be your intent based on the Desmos link.
    – Michael Lugo
    Jul 18 at 19:26










  • @MichaelLugo Yes, thank you, I tend to simplify to the point of my own detriment
    – user189728
    Jul 18 at 19:28
















  • I've added parentheses around the $n+1$ in the denominator as this looks to be your intent based on the Desmos link.
    – Michael Lugo
    Jul 18 at 19:26










  • @MichaelLugo Yes, thank you, I tend to simplify to the point of my own detriment
    – user189728
    Jul 18 at 19:28















I've added parentheses around the $n+1$ in the denominator as this looks to be your intent based on the Desmos link.
– Michael Lugo
Jul 18 at 19:26




I've added parentheses around the $n+1$ in the denominator as this looks to be your intent based on the Desmos link.
– Michael Lugo
Jul 18 at 19:26












@MichaelLugo Yes, thank you, I tend to simplify to the point of my own detriment
– user189728
Jul 18 at 19:28




@MichaelLugo Yes, thank you, I tend to simplify to the point of my own detriment
– user189728
Jul 18 at 19:28










1 Answer
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The sequence $$b_n=fracn!a_n$$ solves the recursion $$b_n+1=b_n-(n+1)!$$ hence $$b_n=b_0-sum_k=1^nk!$$ which yields $$frac1a_n=frac1n!a_0-sum_k=1^nfrack!n!$$ Now, for every $ngeqslant2$,
$$1leqslantsum_k=1^nfrack!n!leqslant1+frac1n+fracn-2n(n-1)$$
Thus, for every initial condition $a_0$ not in the set of the sums $sumlimits_k=1^nk!$ for $ngeqslant0$, one has $$lim a_n=-1$$






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  • I'm admittedly not experienced with any of this, for my sake, could you bridge the gap between the first two lines for laymen?
    – user189728
    Jul 18 at 19:32










  • Dear layman, did you try to find a relation between $b_n+1$ and $b_n$?
    – Did
    Jul 18 at 19:34










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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










The sequence $$b_n=fracn!a_n$$ solves the recursion $$b_n+1=b_n-(n+1)!$$ hence $$b_n=b_0-sum_k=1^nk!$$ which yields $$frac1a_n=frac1n!a_0-sum_k=1^nfrack!n!$$ Now, for every $ngeqslant2$,
$$1leqslantsum_k=1^nfrack!n!leqslant1+frac1n+fracn-2n(n-1)$$
Thus, for every initial condition $a_0$ not in the set of the sums $sumlimits_k=1^nk!$ for $ngeqslant0$, one has $$lim a_n=-1$$






share|cite|improve this answer





















  • I'm admittedly not experienced with any of this, for my sake, could you bridge the gap between the first two lines for laymen?
    – user189728
    Jul 18 at 19:32










  • Dear layman, did you try to find a relation between $b_n+1$ and $b_n$?
    – Did
    Jul 18 at 19:34














up vote
3
down vote



accepted










The sequence $$b_n=fracn!a_n$$ solves the recursion $$b_n+1=b_n-(n+1)!$$ hence $$b_n=b_0-sum_k=1^nk!$$ which yields $$frac1a_n=frac1n!a_0-sum_k=1^nfrack!n!$$ Now, for every $ngeqslant2$,
$$1leqslantsum_k=1^nfrack!n!leqslant1+frac1n+fracn-2n(n-1)$$
Thus, for every initial condition $a_0$ not in the set of the sums $sumlimits_k=1^nk!$ for $ngeqslant0$, one has $$lim a_n=-1$$






share|cite|improve this answer





















  • I'm admittedly not experienced with any of this, for my sake, could you bridge the gap between the first two lines for laymen?
    – user189728
    Jul 18 at 19:32










  • Dear layman, did you try to find a relation between $b_n+1$ and $b_n$?
    – Did
    Jul 18 at 19:34












up vote
3
down vote



accepted







up vote
3
down vote



accepted






The sequence $$b_n=fracn!a_n$$ solves the recursion $$b_n+1=b_n-(n+1)!$$ hence $$b_n=b_0-sum_k=1^nk!$$ which yields $$frac1a_n=frac1n!a_0-sum_k=1^nfrack!n!$$ Now, for every $ngeqslant2$,
$$1leqslantsum_k=1^nfrack!n!leqslant1+frac1n+fracn-2n(n-1)$$
Thus, for every initial condition $a_0$ not in the set of the sums $sumlimits_k=1^nk!$ for $ngeqslant0$, one has $$lim a_n=-1$$






share|cite|improve this answer













The sequence $$b_n=fracn!a_n$$ solves the recursion $$b_n+1=b_n-(n+1)!$$ hence $$b_n=b_0-sum_k=1^nk!$$ which yields $$frac1a_n=frac1n!a_0-sum_k=1^nfrack!n!$$ Now, for every $ngeqslant2$,
$$1leqslantsum_k=1^nfrack!n!leqslant1+frac1n+fracn-2n(n-1)$$
Thus, for every initial condition $a_0$ not in the set of the sums $sumlimits_k=1^nk!$ for $ngeqslant0$, one has $$lim a_n=-1$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 18 at 19:30









Did

242k23208443




242k23208443











  • I'm admittedly not experienced with any of this, for my sake, could you bridge the gap between the first two lines for laymen?
    – user189728
    Jul 18 at 19:32










  • Dear layman, did you try to find a relation between $b_n+1$ and $b_n$?
    – Did
    Jul 18 at 19:34
















  • I'm admittedly not experienced with any of this, for my sake, could you bridge the gap between the first two lines for laymen?
    – user189728
    Jul 18 at 19:32










  • Dear layman, did you try to find a relation between $b_n+1$ and $b_n$?
    – Did
    Jul 18 at 19:34















I'm admittedly not experienced with any of this, for my sake, could you bridge the gap between the first two lines for laymen?
– user189728
Jul 18 at 19:32




I'm admittedly not experienced with any of this, for my sake, could you bridge the gap between the first two lines for laymen?
– user189728
Jul 18 at 19:32












Dear layman, did you try to find a relation between $b_n+1$ and $b_n$?
– Did
Jul 18 at 19:34




Dear layman, did you try to find a relation between $b_n+1$ and $b_n$?
– Did
Jul 18 at 19:34












 

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