Find maximum of succession $a_n+1=fracn^2+n+42n^2+1a_n$ knowing $a_1=1$

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I want to find out what's the maximum of the following sequence:
$$
left{
beginarrayc
a_1=1 \
a_n+1=fracn^2+n+42n^2+1a_n \
endarray
right.
$$
I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3dots$ but we couldn't find the maximum. Please note that $ngeq 0$. Any hints?
sequences-and-series
 |Â
show 1 more comment
up vote
1
down vote
favorite
I want to find out what's the maximum of the following sequence:
$$
left{
beginarrayc
a_1=1 \
a_n+1=fracn^2+n+42n^2+1a_n \
endarray
right.
$$
I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3dots$ but we couldn't find the maximum. Please note that $ngeq 0$. Any hints?
sequences-and-series
1
There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
– José Carlos Santos
Sep 10 at 9:59
@JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
– Cesare
Sep 10 at 10:01
1
So, you are sure that the definition of $a_n+1$ does not use $a_n$.
– José Carlos Santos
Sep 10 at 10:02
It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
– Claude Leibovici
Sep 10 at 10:03
@JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
– Cesare
Sep 10 at 10:05
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to find out what's the maximum of the following sequence:
$$
left{
beginarrayc
a_1=1 \
a_n+1=fracn^2+n+42n^2+1a_n \
endarray
right.
$$
I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3dots$ but we couldn't find the maximum. Please note that $ngeq 0$. Any hints?
sequences-and-series
I want to find out what's the maximum of the following sequence:
$$
left{
beginarrayc
a_1=1 \
a_n+1=fracn^2+n+42n^2+1a_n \
endarray
right.
$$
I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3dots$ but we couldn't find the maximum. Please note that $ngeq 0$. Any hints?
sequences-and-series
sequences-and-series
edited Sep 10 at 11:29
asked Sep 10 at 9:58
Cesare
734410
734410
1
There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
– José Carlos Santos
Sep 10 at 9:59
@JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
– Cesare
Sep 10 at 10:01
1
So, you are sure that the definition of $a_n+1$ does not use $a_n$.
– José Carlos Santos
Sep 10 at 10:02
It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
– Claude Leibovici
Sep 10 at 10:03
@JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
– Cesare
Sep 10 at 10:05
 |Â
show 1 more comment
1
There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
– José Carlos Santos
Sep 10 at 9:59
@JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
– Cesare
Sep 10 at 10:01
1
So, you are sure that the definition of $a_n+1$ does not use $a_n$.
– José Carlos Santos
Sep 10 at 10:02
It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
– Claude Leibovici
Sep 10 at 10:03
@JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
– Cesare
Sep 10 at 10:05
1
1
There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
– José Carlos Santos
Sep 10 at 9:59
There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
– José Carlos Santos
Sep 10 at 9:59
@JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
– Cesare
Sep 10 at 10:01
@JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
– Cesare
Sep 10 at 10:01
1
1
So, you are sure that the definition of $a_n+1$ does not use $a_n$.
– José Carlos Santos
Sep 10 at 10:02
So, you are sure that the definition of $a_n+1$ does not use $a_n$.
– José Carlos Santos
Sep 10 at 10:02
It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
– Claude Leibovici
Sep 10 at 10:03
It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
– Claude Leibovici
Sep 10 at 10:03
@JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
– Cesare
Sep 10 at 10:05
@JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
– Cesare
Sep 10 at 10:05
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.
My textbook says that the extreme value of the function (maximum) is $frac 2019$.
– Cesare
Sep 10 at 10:16
@Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
– José Carlos Santos
Sep 10 at 10:19
@Winther I saw that. Why do you think otherwise?
– José Carlos Santos
Sep 10 at 10:20
@Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
– José Carlos Santos
Sep 10 at 10:25
Therefore $a_4$ should be less than $frac 209$?
– Cesare
Sep 10 at 10:27
 |Â
show 3 more comments
up vote
1
down vote
In general for a recursion of the form:
$$
a_n+1 = f(n)a_n, a_1 > 0.
$$
if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.
in this case:
$f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.
My textbook says that the extreme value of the function (maximum) is $frac 2019$.
– Cesare
Sep 10 at 10:16
@Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
– José Carlos Santos
Sep 10 at 10:19
@Winther I saw that. Why do you think otherwise?
– José Carlos Santos
Sep 10 at 10:20
@Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
– José Carlos Santos
Sep 10 at 10:25
Therefore $a_4$ should be less than $frac 209$?
– Cesare
Sep 10 at 10:27
 |Â
show 3 more comments
up vote
1
down vote
accepted
It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.
My textbook says that the extreme value of the function (maximum) is $frac 2019$.
– Cesare
Sep 10 at 10:16
@Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
– José Carlos Santos
Sep 10 at 10:19
@Winther I saw that. Why do you think otherwise?
– José Carlos Santos
Sep 10 at 10:20
@Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
– José Carlos Santos
Sep 10 at 10:25
Therefore $a_4$ should be less than $frac 209$?
– Cesare
Sep 10 at 10:27
 |Â
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.
It follows from your definition that $a_n+1>a_n$ is $n=1$ or $n=2$ and that $a_n+1<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $frac209$, since $a_3=frac209$.
edited Sep 10 at 10:24
answered Sep 10 at 10:13


José Carlos Santos
124k17101186
124k17101186
My textbook says that the extreme value of the function (maximum) is $frac 2019$.
– Cesare
Sep 10 at 10:16
@Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
– José Carlos Santos
Sep 10 at 10:19
@Winther I saw that. Why do you think otherwise?
– José Carlos Santos
Sep 10 at 10:20
@Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
– José Carlos Santos
Sep 10 at 10:25
Therefore $a_4$ should be less than $frac 209$?
– Cesare
Sep 10 at 10:27
 |Â
show 3 more comments
My textbook says that the extreme value of the function (maximum) is $frac 2019$.
– Cesare
Sep 10 at 10:16
@Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
– José Carlos Santos
Sep 10 at 10:19
@Winther I saw that. Why do you think otherwise?
– José Carlos Santos
Sep 10 at 10:20
@Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
– José Carlos Santos
Sep 10 at 10:25
Therefore $a_4$ should be less than $frac 209$?
– Cesare
Sep 10 at 10:27
My textbook says that the extreme value of the function (maximum) is $frac 2019$.
– Cesare
Sep 10 at 10:16
My textbook says that the extreme value of the function (maximum) is $frac 2019$.
– Cesare
Sep 10 at 10:16
@Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
– José Carlos Santos
Sep 10 at 10:19
@Cesare Since$$a_2=frac1^2+1+42times1^2+1=frac63=2>frac2019,$$that's trivially false.
– José Carlos Santos
Sep 10 at 10:19
@Winther I saw that. Why do you think otherwise?
– José Carlos Santos
Sep 10 at 10:20
@Winther I saw that. Why do you think otherwise?
– José Carlos Santos
Sep 10 at 10:20
@Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
– José Carlos Santos
Sep 10 at 10:25
@Winther I used the correct sequence, but I made a computational mistake. You are right: the maximum is $frac209$. I've edited my answer. Thank you.
– José Carlos Santos
Sep 10 at 10:25
Therefore $a_4$ should be less than $frac 209$?
– Cesare
Sep 10 at 10:27
Therefore $a_4$ should be less than $frac 209$?
– Cesare
Sep 10 at 10:27
 |Â
show 3 more comments
up vote
1
down vote
In general for a recursion of the form:
$$
a_n+1 = f(n)a_n, a_1 > 0.
$$
if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.
in this case:
$f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...
add a comment |Â
up vote
1
down vote
In general for a recursion of the form:
$$
a_n+1 = f(n)a_n, a_1 > 0.
$$
if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.
in this case:
$f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In general for a recursion of the form:
$$
a_n+1 = f(n)a_n, a_1 > 0.
$$
if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.
in this case:
$f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...
In general for a recursion of the form:
$$
a_n+1 = f(n)a_n, a_1 > 0.
$$
if $f(n) - 1$ has only one zero $> 0$ and $lim_ntoinftyf(n) < 1$ then you know $a_n$ will attain its maximum for the largest $n$ for which you have $f(n)geq 1$.
in this case:
$f(n)= fracn^2+n+42n^2+1 > 1Rightarrow - n^2 +n +3 > 0 Rightarrow n < frac1+sqrt132 approx 2.303$. Thus your maximum is $a_3$ ...
answered Sep 10 at 10:25
denklo
3335
3335
add a comment |Â
add a comment |Â
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1
There's something wrong here. The limit is obviously $frac12$. Are you sure that you got the definition right?
– José Carlos Santos
Sep 10 at 9:59
@JoséCarlosSantos yes, what we did was finding $lim fraca_n+1a_n$ this gives $1/2$ which for the rule of division (not sure what's it called in English) gives that since $1/2 < 1$ the limit $rightarrow 0$.
– Cesare
Sep 10 at 10:01
1
So, you are sure that the definition of $a_n+1$ does not use $a_n$.
– José Carlos Santos
Sep 10 at 10:02
It must be something wrong. Who cares about $a_1$ if $a_n+1$ is just a function of $n$ ?
– Claude Leibovici
Sep 10 at 10:03
@JoséCarlosSantos sorry, edited the question. I forgot to put the $a_n$.
– Cesare
Sep 10 at 10:05