Mixing
If $a_{n+1}=frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is...
$begingroup$
Question:
If $a_{n+1}=frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $limlimits_{ntoinfty}left(frac{4}{3}right)^n(3-a_n)$ ?
My approach:
I am able to prove separately that the sequence $a_n$ is convergent and $left(frac{4}{3}right)^n$ is divergent.
But I somehow cannot find out how will their multiplication behave..
But as I cant conclude about the convergence of the entire sequence within the required limit, I cannot find the limit.
Please help
Thank you.
real-analysis sequences-and-series limits convergence self-learning
edited Jan 26 at 16:39
rtybase
11.5k31534
$endgroup$
|
show 9 more comments
$begingroup$
Question:
If $a_{n+1}=frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $limlimits_{ntoinfty}left(frac{4}{3}right)^n(3-a_n)$ ?
My approach:
I am able to prove separately that the sequence $a_n$ is convergent and $left(frac{4}{3}right)^n$ is divergent.
But I somehow cannot find out how will their multiplication behave..
But as I cant conclude about the convergence of the entire sequence within the required limit, I cannot find the limit.
Please help
Thank you.
real-analysis sequences-and-series limits convergence self-learning
edited Jan 26 at 16:39
rtybase
11.5k31534
$endgroup$
1
$begingroup$
You cannot write $$ lim_n (4/3)^n (3-a_n) = lim_n (4/3)^n lim_n (3-a_n),$$ because at least one term at the RHS does not exist.
$endgroup$
– xbh
Jan 26 at 9:42
1
$begingroup$
Simple example: $(a_n = n)_1^infty$ diverges, $(b_n = 1/n)_1^infty$ converges, but $a_n b_n = 1$ could converge.
$endgroup$
– xbh
Jan 26 at 9:44
$begingroup$
Yes, I get you...Even I doubted this method..Then how can I solve this sum ? Can you help me?
$endgroup$
– user515608
Jan 26 at 10:03
1
$begingroup$
"t[h]e sequence has no upper bound as it is decreasing" ?? Every decreasing sequence has a finite upper bound.
$endgroup$
– Did
Jan 26 at 19:28
1
$begingroup$
"Am I correct or my thinking direction is wrong?" Again, establishing the convergence is easy but identifying the limit is not. // Say, where did you find this question? I am asking because the gap between the level of the question and the level of your knowledge seems huge...
$endgroup$
– Did
Jan 26 at 19:29
|
show 9 more comments
$begingroup$
Question:
If $a_{n+1}=frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $limlimits_{ntoinfty}left(frac{4}{3}right)^n(3-a_n)$ ?
My approach:
I am able to prove separately that the sequence $a_n$ is convergent and $left(frac{4}{3}right)^n$ is divergent.
But I somehow cannot find out how will their multiplication behave..
But as I cant conclude about the convergence of the entire sequence within the required limit, I cannot find the limit.
Please help
Thank you.
real-analysis sequences-and-series limits convergence self-learning
edited Jan 26 at 16:39
rtybase
11.5k31534
$endgroup$
Question:
If $a_{n+1}=frac{3+a_n^2}{a_n+1}$ and $a_1=1$, then what is $limlimits_{ntoinfty}left(frac{4}{3}right)^n(3-a_n)$ ?
My approach:
I am able to prove separately that the sequence $a_n$ is convergent and $left(frac{4}{3}right)^n$ is divergent.
But I somehow cannot find out how will their multiplication behave..
But as I cant conclude about the convergence of the entire sequence within the required limit, I cannot find the limit.
Please help
Thank you.
real-analysis sequences-and-series limits convergence self-learning
real-analysis sequences-and-series limits convergence self-learning
edited Jan 26 at 16:39
rtybase
11.5k31534
edited Jan 26 at 16:39
rtybase
11.5k31534
edited Jan 26 at 16:39
rtybase
11.5k31534
edited Jan 26 at 16:39
rtybase
11.5k31534
edited Jan 26 at 16:39
rtybase
11.5k31534
11.5k31534
asked Jan 26 at 9:29
user515608
1
$begingroup$
You cannot write $$ lim_n (4/3)^n (3-a_n) = lim_n (4/3)^n lim_n (3-a_n),$$ because at least one term at the RHS does not exist.
$endgroup$
– xbh
Jan 26 at 9:42
1
$begingroup$
Simple example: $(a_n = n)_1^infty$ diverges, $(b_n = 1/n)_1^infty$ converges, but $a_n b_n = 1$ could converge.
$endgroup$
– xbh
Jan 26 at 9:44
$begingroup$
Yes, I get you...Even I doubted this method..Then how can I solve this sum ? Can you help me?
$endgroup$
– user515608
Jan 26 at 10:03
1
$begingroup$
"t[h]e sequence has no upper bound as it is decreasing" ?? Every decreasing sequence has a finite upper bound.
$endgroup$
– Did
Jan 26 at 19:28
1
$begingroup$
"Am I correct or my thinking direction is wrong?" Again, establishing the convergence is easy but identifying the limit is not. // Say, where did you find this question? I am asking because the gap between the level of the question and the level of your knowledge seems huge...
$endgroup$
– Did
Jan 26 at 19:29
|
show 9 more comments
1
$begingroup$
You cannot write $$ lim_n (4/3)^n (3-a_n) = lim_n (4/3)^n lim_n (3-a_n),$$ because at least one term at the RHS does not exist.
$endgroup$
– xbh
Jan 26 at 9:42
1
$begingroup$
Simple example: $(a_n = n)_1^infty$ diverges, $(b_n = 1/n)_1^infty$ converges, but $a_n b_n = 1$ could converge.
$endgroup$
– xbh
Jan 26 at 9:44
$begingroup$
Yes, I get you...Even I doubted this method..Then how can I solve this sum ? Can you help me?
$endgroup$
– user515608
Jan 26 at 10:03
1
$begingroup$
"t[h]e sequence has no upper bound as it is decreasing" ?? Every decreasing sequence has a finite upper bound.
$endgroup$
– Did
Jan 26 at 19:28
1
$begingroup$
"Am I correct or my thinking direction is wrong?" Again, establishing the convergence is easy but identifying the limit is not. // Say, where did you find this question? I am asking because the gap between the level of the question and the level of your knowledge seems huge...
$endgroup$
– Did
Jan 26 at 19:29
1
1
$begingroup$
You cannot write $$ lim_n (4/3)^n (3-a_n) = lim_n (4/3)^n lim_n (3-a_n),$$ because at least one term at the RHS does not exist.
$endgroup$
– xbh
Jan 26 at 9:42
$begingroup$
You cannot write $$ lim_n (4/3)^n (3-a_n) = lim_n (4/3)^n lim_n (3-a_n),$$ because at least one term at the RHS does not exist.
$endgroup$
– xbh
Jan 26 at 9:42
1
1
$begingroup$
Simple example: $(a_n = n)_1^infty$ diverges, $(b_n = 1/n)_1^infty$ converges, but $a_n b_n = 1$ could converge.
$endgroup$
– xbh
Jan 26 at 9:44
$begingroup$
Simple example: $(a_n = n)_1^infty$ diverges, $(b_n = 1/n)_1^infty$ converges, but $a_n b_n = 1$ could converge.
$endgroup$
– xbh
Jan 26 at 9:44
$begingroup$
Yes, I get you...Even I doubted this method..Then how can I solve this sum ? Can you help me?
$endgroup$
– user515608
Jan 26 at 10:03
$begingroup$
Yes, I get you...Even I doubted this method..Then how can I solve this sum ? Can you help me?
$endgroup$
– user515608
Jan 26 at 10:03
1
1
$begingroup$
"t[h]e sequence has no upper bound as it is decreasing" ?? Every decreasing sequence has a finite upper bound.
$endgroup$
– Did
Jan 26 at 19:28
$begingroup$
"t[h]e sequence has no upper bound as it is decreasing" ?? Every decreasing sequence has a finite upper bound.
$endgroup$
– Did
Jan 26 at 19:28
1
1
$begingroup$
"Am I correct or my thinking direction is wrong?" Again, establishing the convergence is easy but identifying the limit is not. // Say, where did you find this question? I am asking because the gap between the level of the question and the level of your knowledge seems huge...
$endgroup$
– Did
Jan 26 at 19:29
$begingroup$
"Am I correct or my thinking direction is wrong?" Again, establishing the convergence is easy but identifying the limit is not. // Say, where did you find this question? I am asking because the gap between the level of the question and the level of your knowledge seems huge...
$endgroup$
– Did
Jan 26 at 19:29
|
show 9 more comments
1 Answer
1
active
oldest
votes
$begingroup$
EDIT: I misunderstood the question. This answer only proves convergence, but doesn't find the limit.
Let's try looking at rations again:
$$
begin{align}
R_n = frac{(frac{4}{3})^{n+1}(3-a_{n+1})}{(frac{4}{3})^{n}(3-a_{n})} &=
frac{4}{3} times frac{3-frac{3+a_n^2}{a_n+1}}{3-a_{n}} =
frac{4}{3} times frac{frac{3a_n + 3 - 3 - a_n^2}{a_n+1}}{3-a_{n}} = \
&= frac{4}{3} times frac{a_n(3-a_n)}{(a_n+1)(3-a_{n})} =
frac{4}{3} times frac{a_n}{a_n+1}
end{align}
$$
We know that $a_n < 3$. Since $frac{x}{x+1}$ is increasing, $frac{a_n}{a_n+1} < frac{3}{4}$. Therefore, $R_n < 1$.
Note that we're not doing ratio test (ratio test is for series, not for sequences), and we don't need to take the limit of $R_n$.
Instead, knowing that $R_1<1$, we can conclude that the sequence is decreasing. It's also positive, so it must converge.
answered Jan 26 at 10:02
Todor MarkovTodor Markov
2,420412
$endgroup$
$begingroup$
Thank you..But now that we know, the sequence is convergent, how can I find the given limit, i.e, lim[(4/3)^n](3-a_n) ?
$endgroup$
– user515608
Jan 26 at 10:32
1
$begingroup$
Ah, sorry, I don't know how to get that. You should probably unaccept this answer though, as when it's accepted people will this the question resolved.
$endgroup$
– Todor Markov
Jan 26 at 11:01
$begingroup$
Thank you for your answer anyway :)
$endgroup$
– user515608
Jan 26 at 13:17
$begingroup$
I obtained the limit as a convergent infinite series, but they again involve $a_n$'s, so could not the explicit value of it.
$endgroup$
– i707107
Jan 26 at 17:21
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
EDIT: I misunderstood the question. This answer only proves convergence, but doesn't find the limit.
Let's try looking at rations again:
$$
begin{align}
R_n = frac{(frac{4}{3})^{n+1}(3-a_{n+1})}{(frac{4}{3})^{n}(3-a_{n})} &=
frac{4}{3} times frac{3-frac{3+a_n^2}{a_n+1}}{3-a_{n}} =
frac{4}{3} times frac{frac{3a_n + 3 - 3 - a_n^2}{a_n+1}}{3-a_{n}} = \
&= frac{4}{3} times frac{a_n(3-a_n)}{(a_n+1)(3-a_{n})} =
frac{4}{3} times frac{a_n}{a_n+1}
end{align}
$$
We know that $a_n < 3$. Since $frac{x}{x+1}$ is increasing, $frac{a_n}{a_n+1} < frac{3}{4}$. Therefore, $R_n < 1$.
Note that we're not doing ratio test (ratio test is for series, not for sequences), and we don't need to take the limit of $R_n$.
Instead, knowing that $R_1<1$, we can conclude that the sequence is decreasing. It's also positive, so it must converge.
answered Jan 26 at 10:02
Todor MarkovTodor Markov
2,420412
$endgroup$
$begingroup$
Thank you..But now that we know, the sequence is convergent, how can I find the given limit, i.e, lim[(4/3)^n](3-a_n) ?
$endgroup$
– user515608
Jan 26 at 10:32
1
$begingroup$
Ah, sorry, I don't know how to get that. You should probably unaccept this answer though, as when it's accepted people will this the question resolved.
$endgroup$
– Todor Markov
Jan 26 at 11:01
$begingroup$
Thank you for your answer anyway :)
$endgroup$
– user515608
Jan 26 at 13:17
$begingroup$
I obtained the limit as a convergent infinite series, but they again involve $a_n$'s, so could not the explicit value of it.
$endgroup$
– i707107
Jan 26 at 17:21
add a comment |
$begingroup$
EDIT: I misunderstood the question. This answer only proves convergence, but doesn't find the limit.
Let's try looking at rations again:
$$
begin{align}
R_n = frac{(frac{4}{3})^{n+1}(3-a_{n+1})}{(frac{4}{3})^{n}(3-a_{n})} &=
frac{4}{3} times frac{3-frac{3+a_n^2}{a_n+1}}{3-a_{n}} =
frac{4}{3} times frac{frac{3a_n + 3 - 3 - a_n^2}{a_n+1}}{3-a_{n}} = \
&= frac{4}{3} times frac{a_n(3-a_n)}{(a_n+1)(3-a_{n})} =
frac{4}{3} times frac{a_n}{a_n+1}
end{align}
$$
We know that $a_n < 3$. Since $frac{x}{x+1}$ is increasing, $frac{a_n}{a_n+1} < frac{3}{4}$. Therefore, $R_n < 1$.
Note that we're not doing ratio test (ratio test is for series, not for sequences), and we don't need to take the limit of $R_n$.
Instead, knowing that $R_1<1$, we can conclude that the sequence is decreasing. It's also positive, so it must converge.
answered Jan 26 at 10:02
Todor MarkovTodor Markov
2,420412
$endgroup$
$begingroup$
Thank you..But now that we know, the sequence is convergent, how can I find the given limit, i.e, lim[(4/3)^n](3-a_n) ?
$endgroup$
– user515608
Jan 26 at 10:32
1
$begingroup$
Ah, sorry, I don't know how to get that. You should probably unaccept this answer though, as when it's accepted people will this the question resolved.
$endgroup$
– Todor Markov
Jan 26 at 11:01
$begingroup$
Thank you for your answer anyway :)
$endgroup$
– user515608
Jan 26 at 13:17
$begingroup$
I obtained the limit as a convergent infinite series, but they again involve $a_n$'s, so could not the explicit value of it.
$endgroup$
– i707107
Jan 26 at 17:21
add a comment |
$begingroup$
EDIT: I misunderstood the question. This answer only proves convergence, but doesn't find the limit.
Let's try looking at rations again:
$$
begin{align}
R_n = frac{(frac{4}{3})^{n+1}(3-a_{n+1})}{(frac{4}{3})^{n}(3-a_{n})} &=
frac{4}{3} times frac{3-frac{3+a_n^2}{a_n+1}}{3-a_{n}} =
frac{4}{3} times frac{frac{3a_n + 3 - 3 - a_n^2}{a_n+1}}{3-a_{n}} = \
&= frac{4}{3} times frac{a_n(3-a_n)}{(a_n+1)(3-a_{n})} =
frac{4}{3} times frac{a_n}{a_n+1}
end{align}
$$
We know that $a_n < 3$. Since $frac{x}{x+1}$ is increasing, $frac{a_n}{a_n+1} < frac{3}{4}$. Therefore, $R_n < 1$.
Note that we're not doing ratio test (ratio test is for series, not for sequences), and we don't need to take the limit of $R_n$.
Instead, knowing that $R_1<1$, we can conclude that the sequence is decreasing. It's also positive, so it must converge.
answered Jan 26 at 10:02
Todor MarkovTodor Markov
2,420412
$endgroup$
EDIT: I misunderstood the question. This answer only proves convergence, but doesn't find the limit.
Let's try looking at rations again:
$$
begin{align}
R_n = frac{(frac{4}{3})^{n+1}(3-a_{n+1})}{(frac{4}{3})^{n}(3-a_{n})} &=
frac{4}{3} times frac{3-frac{3+a_n^2}{a_n+1}}{3-a_{n}} =
frac{4}{3} times frac{frac{3a_n + 3 - 3 - a_n^2}{a_n+1}}{3-a_{n}} = \
&= frac{4}{3} times frac{a_n(3-a_n)}{(a_n+1)(3-a_{n})} =
frac{4}{3} times frac{a_n}{a_n+1}
end{align}
$$
We know that $a_n < 3$. Since $frac{x}{x+1}$ is increasing, $frac{a_n}{a_n+1} < frac{3}{4}$. Therefore, $R_n < 1$.
Note that we're not doing ratio test (ratio test is for series, not for sequences), and we don't need to take the limit of $R_n$.
Instead, knowing that $R_1<1$, we can conclude that the sequence is decreasing. It's also positive, so it must converge.
answered Jan 26 at 10:02
Todor MarkovTodor Markov
2,420412
edited Jan 26 at 13:54
answered Jan 26 at 10:02
Todor MarkovTodor Markov
2,420412
answered Jan 26 at 10:02
Todor MarkovTodor Markov
2,420412
answered Jan 26 at 10:02
Todor MarkovTodor Markov
2,420412
2,420412
$begingroup$
Thank you..But now that we know, the sequence is convergent, how can I find the given limit, i.e, lim[(4/3)^n](3-a_n) ?
$endgroup$
– user515608
Jan 26 at 10:32
1
$begingroup$
Ah, sorry, I don't know how to get that. You should probably unaccept this answer though, as when it's accepted people will this the question resolved.
$endgroup$
– Todor Markov
Jan 26 at 11:01
$begingroup$
Thank you for your answer anyway :)
$endgroup$
– user515608
Jan 26 at 13:17
$begingroup$
I obtained the limit as a convergent infinite series, but they again involve $a_n$'s, so could not the explicit value of it.
$endgroup$
– i707107
Jan 26 at 17:21
add a comment |
$begingroup$
Thank you..But now that we know, the sequence is convergent, how can I find the given limit, i.e, lim[(4/3)^n](3-a_n) ?
$endgroup$
– user515608
Jan 26 at 10:32
1
$begingroup$
Ah, sorry, I don't know how to get that. You should probably unaccept this answer though, as when it's accepted people will this the question resolved.
$endgroup$
– Todor Markov
Jan 26 at 11:01
$begingroup$
Thank you for your answer anyway :)
$endgroup$
– user515608
Jan 26 at 13:17
$begingroup$
I obtained the limit as a convergent infinite series, but they again involve $a_n$'s, so could not the explicit value of it.
$endgroup$
– i707107
Jan 26 at 17:21
$begingroup$
Thank you..But now that we know, the sequence is convergent, how can I find the given limit, i.e, lim[(4/3)^n](3-a_n) ?
$endgroup$
– user515608
Jan 26 at 10:32
$begingroup$
Thank you..But now that we know, the sequence is convergent, how can I find the given limit, i.e, lim[(4/3)^n](3-a_n) ?
$endgroup$
– user515608
Jan 26 at 10:32
1
1
$begingroup$
Ah, sorry, I don't know how to get that. You should probably unaccept this answer though, as when it's accepted people will this the question resolved.
$endgroup$
– Todor Markov
Jan 26 at 11:01
$begingroup$
Ah, sorry, I don't know how to get that. You should probably unaccept this answer though, as when it's accepted people will this the question resolved.
$endgroup$
– Todor Markov
Jan 26 at 11:01
$begingroup$
Thank you for your answer anyway :)
$endgroup$
– user515608
Jan 26 at 13:17
$begingroup$
Thank you for your answer anyway :)
$endgroup$
– user515608
Jan 26 at 13:17
$begingroup$
I obtained the limit as a convergent infinite series, but they again involve $a_n$'s, so could not the explicit value of it.
$endgroup$
– i707107
Jan 26 at 17:21
$begingroup$
I obtained the limit as a convergent infinite series, but they again involve $a_n$'s, so could not the explicit value of it.
$endgroup$
– i707107
Jan 26 at 17:21
add a comment |
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1
$begingroup$
You cannot write $$ lim_n (4/3)^n (3-a_n) = lim_n (4/3)^n lim_n (3-a_n),$$ because at least one term at the RHS does not exist.
$endgroup$
– xbh
Jan 26 at 9:42
1
$begingroup$
Simple example: $(a_n = n)_1^infty$ diverges, $(b_n = 1/n)_1^infty$ converges, but $a_n b_n = 1$ could converge.
$endgroup$
– xbh
Jan 26 at 9:44
$begingroup$
Yes, I get you...Even I doubted this method..Then how can I solve this sum ? Can you help me?
$endgroup$
– user515608
Jan 26 at 10:03
1
$begingroup$
"t[h]e sequence has no upper bound as it is decreasing" ?? Every decreasing sequence has a finite upper bound.
$endgroup$
– Did
Jan 26 at 19:28
1
$begingroup$
"Am I correct or my thinking direction is wrong?" Again, establishing the convergence is easy but identifying the limit is not. // Say, where did you find this question? I am asking because the gap between the level of the question and the level of your knowledge seems huge...
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– Did
Jan 26 at 19:29