Does calculating the limit of a given real-valued (recursive) sequence already imply its convergence?
$begingroup$
Assuming we are given a real-valued recursive sequence $(a_n)_{n in mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = varphi(a_n)$
EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:
$$a_1 = sqrt{2}, a_{n+1} = sqrt{2a_n}$$
In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.
However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$lim_{n to infty} a_n$$
without even proving whether or not $(a_n)_n$ is convergent at all.
Ive been asked that if we basically could find an $a$ for which $$ lim_{n to infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.
I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.
Can you possibly help me with an answer to that question? Can i simply compute $lim_{n to infty} a_n = a$ prior to proving the existence of such an $a$ at all?
This is all restricted to $mathbb{R}$ but any general answer is highly appreciated as well.
Thank you very much for any help.
real-analysis calculus sequences-and-series convergence
asked Dec 6 '18 at 22:18
ZestZest
26513
$endgroup$
|
show 5 more comments
$begingroup$
Assuming we are given a real-valued recursive sequence $(a_n)_{n in mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = varphi(a_n)$
EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:
$$a_1 = sqrt{2}, a_{n+1} = sqrt{2a_n}$$
In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.
However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$lim_{n to infty} a_n$$
without even proving whether or not $(a_n)_n$ is convergent at all.
Ive been asked that if we basically could find an $a$ for which $$ lim_{n to infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.
I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.
Can you possibly help me with an answer to that question? Can i simply compute $lim_{n to infty} a_n = a$ prior to proving the existence of such an $a$ at all?
This is all restricted to $mathbb{R}$ but any general answer is highly appreciated as well.
Thank you very much for any help.
real-analysis calculus sequences-and-series convergence
asked Dec 6 '18 at 22:18
ZestZest
26513
$endgroup$
$begingroup$
I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
$endgroup$
– gammatester
Dec 6 '18 at 22:32
$begingroup$
I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
$endgroup$
– saulspatz
Dec 6 '18 at 22:40
$begingroup$
If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
$endgroup$
– Henry
Dec 6 '18 at 22:41
$begingroup$
Sorry for any confusion i might have caused. I will edit my question to provide a sequence
$endgroup$
– Zest
Dec 6 '18 at 22:41
$begingroup$
It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
$endgroup$
– Steven Stadnicki
Dec 6 '18 at 22:49
|
show 5 more comments
$begingroup$
Assuming we are given a real-valued recursive sequence $(a_n)_{n in mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = varphi(a_n)$
EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:
$$a_1 = sqrt{2}, a_{n+1} = sqrt{2a_n}$$
In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.
However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$lim_{n to infty} a_n$$
without even proving whether or not $(a_n)_n$ is convergent at all.
Ive been asked that if we basically could find an $a$ for which $$ lim_{n to infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.
I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.
Can you possibly help me with an answer to that question? Can i simply compute $lim_{n to infty} a_n = a$ prior to proving the existence of such an $a$ at all?
This is all restricted to $mathbb{R}$ but any general answer is highly appreciated as well.
Thank you very much for any help.
real-analysis calculus sequences-and-series convergence
asked Dec 6 '18 at 22:18
ZestZest
26513
$endgroup$
Assuming we are given a real-valued recursive sequence $(a_n)_{n in mathbb{N}}$ by its starting point $a_1$ and its recursive function $a_{n+1} = varphi(a_n)$
EDIT: Since this question caused some confusion or i failed to be clear i'd like to add the recursive sequence:
$$a_1 = sqrt{2}, a_{n+1} = sqrt{2a_n}$$
In order to prove the convergence of $(a_n)_{n}$ i would try to prove boundedness and monotonicity (if that's the case). Then i would try to compute its limit.
However, a few days ago i've been asked whether it would not be sufficient enough to simply compute $$lim_{n to infty} a_n$$
without even proving whether or not $(a_n)_n$ is convergent at all.
Ive been asked that if we basically could find an $a$ for which $$ lim_{n to infty} a_n = a$$ holds
we could immediately conclude that the given sequence was convergent in the first place.
I was quite sceptical but i failed to provide a reasonable answer why it would certainly not sufficient enough. My first reaction was that it might be possible that $a$ might not be the limit but only an accumulation point. But i'm not sure if that's either the case or if i'm completely mistaken.
Can you possibly help me with an answer to that question? Can i simply compute $lim_{n to infty} a_n = a$ prior to proving the existence of such an $a$ at all?
This is all restricted to $mathbb{R}$ but any general answer is highly appreciated as well.
Thank you very much for any help.
real-analysis calculus sequences-and-series convergence
real-analysis calculus sequences-and-series convergence
asked Dec 6 '18 at 22:18
ZestZest
26513
asked Dec 6 '18 at 22:18
ZestZest
26513
edited Dec 6 '18 at 22:43
Zest
asked Dec 6 '18 at 22:18
ZestZest
26513
asked Dec 6 '18 at 22:18
ZestZest
26513
asked Dec 6 '18 at 22:18
ZestZest
26513
26513
$begingroup$
I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
$endgroup$
– gammatester
Dec 6 '18 at 22:32
$begingroup$
I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
$endgroup$
– saulspatz
Dec 6 '18 at 22:40
$begingroup$
If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
$endgroup$
– Henry
Dec 6 '18 at 22:41
$begingroup$
Sorry for any confusion i might have caused. I will edit my question to provide a sequence
$endgroup$
– Zest
Dec 6 '18 at 22:41
$begingroup$
It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
$endgroup$
– Steven Stadnicki
Dec 6 '18 at 22:49
|
show 5 more comments
$begingroup$
I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
$endgroup$
– gammatester
Dec 6 '18 at 22:32
$begingroup$
I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
$endgroup$
– saulspatz
Dec 6 '18 at 22:40
$begingroup$
If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
$endgroup$
– Henry
Dec 6 '18 at 22:41
$begingroup$
Sorry for any confusion i might have caused. I will edit my question to provide a sequence
$endgroup$
– Zest
Dec 6 '18 at 22:41
$begingroup$
It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
$endgroup$
– Steven Stadnicki
Dec 6 '18 at 22:49
$begingroup$
I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
$endgroup$
– gammatester
Dec 6 '18 at 22:32
$begingroup$
I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
$endgroup$
– gammatester
Dec 6 '18 at 22:32
$begingroup$
I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
$endgroup$
– saulspatz
Dec 6 '18 at 22:40
$begingroup$
I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
$endgroup$
– saulspatz
Dec 6 '18 at 22:40
$begingroup$
If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
$endgroup$
– Henry
Dec 6 '18 at 22:41
$begingroup$
If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
$endgroup$
– Henry
Dec 6 '18 at 22:41
$begingroup$
Sorry for any confusion i might have caused. I will edit my question to provide a sequence
$endgroup$
– Zest
Dec 6 '18 at 22:41
$begingroup$
Sorry for any confusion i might have caused. I will edit my question to provide a sequence
$endgroup$
– Zest
Dec 6 '18 at 22:41
$begingroup$
It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
$endgroup$
– Steven Stadnicki
Dec 6 '18 at 22:49
$begingroup$
It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
$endgroup$
– Steven Stadnicki
Dec 6 '18 at 22:49
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?
answered Dec 6 '18 at 22:51
William SunWilliam Sun
473211
$endgroup$
$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54
1
$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55
$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06
1
$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08
1
$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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oldest
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oldest
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active
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$begingroup$
Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?
answered Dec 6 '18 at 22:51
William SunWilliam Sun
473211
$endgroup$
$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54
1
$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55
$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06
1
$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08
1
$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19
|
show 2 more comments
$begingroup$
Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?
answered Dec 6 '18 at 22:51
William SunWilliam Sun
473211
$endgroup$
$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54
1
$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55
$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06
1
$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08
1
$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19
|
show 2 more comments
$begingroup$
Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?
answered Dec 6 '18 at 22:51
William SunWilliam Sun
473211
$endgroup$
Take a sequence $(a_n)$ defined by $a_0=0$ and $a_n=2a_{n-1}+1$. If you assume that the limit $L=lim_{ntoinfty}(a_n)$ exists then by taking the limit of both sides you get $L=-1$. Does this answer your question?
answered Dec 6 '18 at 22:51
William SunWilliam Sun
473211
edited Dec 6 '18 at 22:58
answered Dec 6 '18 at 22:51
William SunWilliam Sun
473211
answered Dec 6 '18 at 22:51
William SunWilliam Sun
473211
answered Dec 6 '18 at 22:51
William SunWilliam Sun
473211
473211
$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54
1
$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55
$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06
1
$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08
1
$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19
|
show 2 more comments
$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54
1
$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55
$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06
1
$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08
1
$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19
$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54
$begingroup$
this is exactly what i was looking for. i really appreciate every comment and your answer. thank you very much.
$endgroup$
– Zest
Dec 6 '18 at 22:54
1
1
$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55
$begingroup$
@Zest You are welcome! I’m glad to help.
$endgroup$
– William Sun
Dec 6 '18 at 22:55
$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06
$begingroup$
one last question: the given sequence isn't convergent since it is not bounded right?
$endgroup$
– Zest
Dec 6 '18 at 23:06
1
1
$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08
$begingroup$
@Zest The answer is Yes assuming you are working in $mathbb{R}$, in which case all convergent sequences are bounded.
$endgroup$
– William Sun
Dec 6 '18 at 23:08
1
1
$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19
$begingroup$
@Zest $L=-1$ is not a limit for the recursively defined sequence in my example. The reasoning is “If the limit of the sequence exists, then it must be $-1$.” But this does not provide with any clue whether the limit exists or not.
$endgroup$
– William Sun
Dec 6 '18 at 23:19
|
show 2 more comments
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$begingroup$
I do not know what the exact question is. With $varphi(x) = x^2$ your hypotetical limit will satisfy $a=a^2$ which would give $a=1$ or $a=0$. But of course the sequence diverges for $|a_1| >1$
$endgroup$
– gammatester
Dec 6 '18 at 22:32
$begingroup$
I don't know what you are asking. A sequence is convergent if and only if the limit exists. What distinction are you making?
$endgroup$
– saulspatz
Dec 6 '18 at 22:40
$begingroup$
If $limlimits_{n to infty} a_n$ exists then the sequence converges, but it difficult to see how you know it exists just by calculating successive terms which appear to approach a particular value without reaching it
$endgroup$
– Henry
Dec 6 '18 at 22:41
$begingroup$
Sorry for any confusion i might have caused. I will edit my question to provide a sequence
$endgroup$
– Zest
Dec 6 '18 at 22:41
$begingroup$
It's still not clear what you're asking here; I think explaining how you compute the limit would help matters here. Note that saying e.g. 'the limit is the unique value $L$ satisfying $L=sqrt{2L}$' isn't actually computation in any relevant sense...
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– Steven Stadnicki
Dec 6 '18 at 22:49