Mixing
Limit of sequence with relation $x_{n+1}=x_n-x_n^2$ [duplicate]
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This question already has an answer here:
$nin N, a_{n+1} = a_n (1 - a_n)$ , $0 < a_0 < 1$. Prove that $limlimits_{ntoinfty} a_ncdot n = 1$.
1 answer
Let the sequence of real numbers is defined as follows: $x_1=frac{1}{2}$ and $x_{n+1}=x_n-x_n^2$. Show that $lim_{nto infty}nx_n=1$.
I've shown that the limit of $x_n$ is zero since this sequence is bounded and monotone. How to show that $nx_nto 1$ as $nto infty$?
I have no ideas how to handle this problem.
It would be interesting to see approach.
real-analysis limits
asked Jan 26 at 22:26
ZFRZFR
5,28331540
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marked as duplicate by rtybase, max_zorn, Lord Shark the Unknown, Eric Wofsey real-analysis
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Jan 27 at 6:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
$nin N, a_{n+1} = a_n (1 - a_n)$ , $0 < a_0 < 1$. Prove that $limlimits_{ntoinfty} a_ncdot n = 1$.
1 answer
Let the sequence of real numbers is defined as follows: $x_1=frac{1}{2}$ and $x_{n+1}=x_n-x_n^2$. Show that $lim_{nto infty}nx_n=1$.
I've shown that the limit of $x_n$ is zero since this sequence is bounded and monotone. How to show that $nx_nto 1$ as $nto infty$?
I have no ideas how to handle this problem.
It would be interesting to see approach.
real-analysis limits
asked Jan 26 at 22:26
ZFRZFR
5,28331540
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marked as duplicate by rtybase, max_zorn, Lord Shark the Unknown, Eric Wofsey real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Jan 27 at 6:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Let $y_n = n x_n$. Then find the recurrence for $y_n$. The initial condition is known. Can you apply the same method to find the limit?
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– Sasha
Jan 26 at 22:28
2
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I'd start by observing that $$frac1{x_{n+1}}=frac{1}{x_n}+1+O(x_n).$$
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– Lord Shark the Unknown
Jan 26 at 22:30
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@Sasha, i cannot answer you remark from the top of my head but i have done this approach before posting a question and if i am not mistaken this approach was not useful
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– ZFR
Jan 26 at 22:56
add a comment |
$begingroup$
This question already has an answer here:
$nin N, a_{n+1} = a_n (1 - a_n)$ , $0 < a_0 < 1$. Prove that $limlimits_{ntoinfty} a_ncdot n = 1$.
1 answer
Let the sequence of real numbers is defined as follows: $x_1=frac{1}{2}$ and $x_{n+1}=x_n-x_n^2$. Show that $lim_{nto infty}nx_n=1$.
I've shown that the limit of $x_n$ is zero since this sequence is bounded and monotone. How to show that $nx_nto 1$ as $nto infty$?
I have no ideas how to handle this problem.
It would be interesting to see approach.
real-analysis limits
asked Jan 26 at 22:26
ZFRZFR
5,28331540
$endgroup$
This question already has an answer here:
$nin N, a_{n+1} = a_n (1 - a_n)$ , $0 < a_0 < 1$. Prove that $limlimits_{ntoinfty} a_ncdot n = 1$.
1 answer
Let the sequence of real numbers is defined as follows: $x_1=frac{1}{2}$ and $x_{n+1}=x_n-x_n^2$. Show that $lim_{nto infty}nx_n=1$.
I've shown that the limit of $x_n$ is zero since this sequence is bounded and monotone. How to show that $nx_nto 1$ as $nto infty$?
I have no ideas how to handle this problem.
It would be interesting to see approach.
This question already has an answer here:
$nin N, a_{n+1} = a_n (1 - a_n)$ , $0 < a_0 < 1$. Prove that $limlimits_{ntoinfty} a_ncdot n = 1$.
1 answer
real-analysis limits
real-analysis limits
asked Jan 26 at 22:26
ZFRZFR
5,28331540
asked Jan 26 at 22:26
ZFRZFR
5,28331540
asked Jan 26 at 22:26
ZFRZFR
5,28331540
asked Jan 26 at 22:26
ZFRZFR
5,28331540
asked Jan 26 at 22:26
ZFRZFR
5,28331540
5,28331540
marked as duplicate by rtybase, max_zorn, Lord Shark the Unknown, Eric Wofsey real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Jan 27 at 6:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by rtybase, max_zorn, Lord Shark the Unknown, Eric Wofsey real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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Jan 27 at 6:58
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Let $y_n = n x_n$. Then find the recurrence for $y_n$. The initial condition is known. Can you apply the same method to find the limit?
$endgroup$
– Sasha
Jan 26 at 22:28
2
$begingroup$
I'd start by observing that $$frac1{x_{n+1}}=frac{1}{x_n}+1+O(x_n).$$
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:30
$begingroup$
@Sasha, i cannot answer you remark from the top of my head but i have done this approach before posting a question and if i am not mistaken this approach was not useful
$endgroup$
– ZFR
Jan 26 at 22:56
add a comment |
$begingroup$
Let $y_n = n x_n$. Then find the recurrence for $y_n$. The initial condition is known. Can you apply the same method to find the limit?
$endgroup$
– Sasha
Jan 26 at 22:28
2
$begingroup$
I'd start by observing that $$frac1{x_{n+1}}=frac{1}{x_n}+1+O(x_n).$$
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:30
$begingroup$
@Sasha, i cannot answer you remark from the top of my head but i have done this approach before posting a question and if i am not mistaken this approach was not useful
$endgroup$
– ZFR
Jan 26 at 22:56
$begingroup$
Let $y_n = n x_n$. Then find the recurrence for $y_n$. The initial condition is known. Can you apply the same method to find the limit?
$endgroup$
– Sasha
Jan 26 at 22:28
$begingroup$
Let $y_n = n x_n$. Then find the recurrence for $y_n$. The initial condition is known. Can you apply the same method to find the limit?
$endgroup$
– Sasha
Jan 26 at 22:28
2
2
$begingroup$
I'd start by observing that $$frac1{x_{n+1}}=frac{1}{x_n}+1+O(x_n).$$
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:30
$begingroup$
I'd start by observing that $$frac1{x_{n+1}}=frac{1}{x_n}+1+O(x_n).$$
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:30
$begingroup$
@Sasha, i cannot answer you remark from the top of my head but i have done this approach before posting a question and if i am not mistaken this approach was not useful
$endgroup$
– ZFR
Jan 26 at 22:56
$begingroup$
@Sasha, i cannot answer you remark from the top of my head but i have done this approach before posting a question and if i am not mistaken this approach was not useful
$endgroup$
– ZFR
Jan 26 at 22:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We have
$$frac1{x_{n+1}}=frac1{x_n(1-x_n)}=frac1{x_n}+1+x_n+x_n^2+cdots$$
(a geometric series). Thus
$$frac1{x_{n+1}}>frac1{x_n}+1$$
and so
$$frac1{x_n}ge n-1+frac1{x_1}=n+1.$$
Therefore $x_n=O(1/n)$. Then
$$frac1{x_{n+1}}=frac1{x_n}+1+O(1/n)$$
and so
$$frac1{x_n}=n+O(ln n).$$
That's enough.
answered Jan 26 at 22:46
Lord Shark the UnknownLord Shark the Unknown
107k1161133
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$begingroup$
Quite nice solution? But is it possible to donit without logarithm?
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– ZFR
Jan 27 at 1:19
$begingroup$
Yes, see the other solution. Or prove that $sum_{k=1}^n1/k=o(n)$. @ZFR
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– Lord Shark the Unknown
Jan 27 at 10:44
add a comment |
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By Stolz-Cesaro's Lemma we get that
$lim_{ntoinfty}nx_n=lim_{ntoinfty}frac{n}{frac{1}{x_n}}=lim_{ntoinfty}frac{1}{frac{1}{x_{n+1}}-frac{1}{x_n}}=lim_{ntoinfty}frac{x_nx_{n+1}}{x_n-x_{n+1}}=lim_{ntoinfty}frac{x_n^2-x_n^3}{x_n^2}=lim_{ntoinfty}(1-x_n)=1$
answered Jan 26 at 23:06
AlexdanutAlexdanut
1538
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have
$$frac1{x_{n+1}}=frac1{x_n(1-x_n)}=frac1{x_n}+1+x_n+x_n^2+cdots$$
(a geometric series). Thus
$$frac1{x_{n+1}}>frac1{x_n}+1$$
and so
$$frac1{x_n}ge n-1+frac1{x_1}=n+1.$$
Therefore $x_n=O(1/n)$. Then
$$frac1{x_{n+1}}=frac1{x_n}+1+O(1/n)$$
and so
$$frac1{x_n}=n+O(ln n).$$
That's enough.
answered Jan 26 at 22:46
Lord Shark the UnknownLord Shark the Unknown
107k1161133
$endgroup$
$begingroup$
Quite nice solution? But is it possible to donit without logarithm?
$endgroup$
– ZFR
Jan 27 at 1:19
$begingroup$
Yes, see the other solution. Or prove that $sum_{k=1}^n1/k=o(n)$. @ZFR
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:44
add a comment |
$begingroup$
We have
$$frac1{x_{n+1}}=frac1{x_n(1-x_n)}=frac1{x_n}+1+x_n+x_n^2+cdots$$
(a geometric series). Thus
$$frac1{x_{n+1}}>frac1{x_n}+1$$
and so
$$frac1{x_n}ge n-1+frac1{x_1}=n+1.$$
Therefore $x_n=O(1/n)$. Then
$$frac1{x_{n+1}}=frac1{x_n}+1+O(1/n)$$
and so
$$frac1{x_n}=n+O(ln n).$$
That's enough.
answered Jan 26 at 22:46
Lord Shark the UnknownLord Shark the Unknown
107k1161133
$endgroup$
$begingroup$
Quite nice solution? But is it possible to donit without logarithm?
$endgroup$
– ZFR
Jan 27 at 1:19
$begingroup$
Yes, see the other solution. Or prove that $sum_{k=1}^n1/k=o(n)$. @ZFR
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:44
add a comment |
$begingroup$
We have
$$frac1{x_{n+1}}=frac1{x_n(1-x_n)}=frac1{x_n}+1+x_n+x_n^2+cdots$$
(a geometric series). Thus
$$frac1{x_{n+1}}>frac1{x_n}+1$$
and so
$$frac1{x_n}ge n-1+frac1{x_1}=n+1.$$
Therefore $x_n=O(1/n)$. Then
$$frac1{x_{n+1}}=frac1{x_n}+1+O(1/n)$$
and so
$$frac1{x_n}=n+O(ln n).$$
That's enough.
answered Jan 26 at 22:46
Lord Shark the UnknownLord Shark the Unknown
107k1161133
$endgroup$
We have
$$frac1{x_{n+1}}=frac1{x_n(1-x_n)}=frac1{x_n}+1+x_n+x_n^2+cdots$$
(a geometric series). Thus
$$frac1{x_{n+1}}>frac1{x_n}+1$$
and so
$$frac1{x_n}ge n-1+frac1{x_1}=n+1.$$
Therefore $x_n=O(1/n)$. Then
$$frac1{x_{n+1}}=frac1{x_n}+1+O(1/n)$$
and so
$$frac1{x_n}=n+O(ln n).$$
That's enough.
answered Jan 26 at 22:46
Lord Shark the UnknownLord Shark the Unknown
107k1161133
answered Jan 26 at 22:46
Lord Shark the UnknownLord Shark the Unknown
107k1161133
answered Jan 26 at 22:46
Lord Shark the UnknownLord Shark the Unknown
107k1161133
answered Jan 26 at 22:46
Lord Shark the UnknownLord Shark the Unknown
107k1161133
107k1161133
$begingroup$
Quite nice solution? But is it possible to donit without logarithm?
$endgroup$
– ZFR
Jan 27 at 1:19
$begingroup$
Yes, see the other solution. Or prove that $sum_{k=1}^n1/k=o(n)$. @ZFR
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:44
add a comment |
$begingroup$
Quite nice solution? But is it possible to donit without logarithm?
$endgroup$
– ZFR
Jan 27 at 1:19
$begingroup$
Yes, see the other solution. Or prove that $sum_{k=1}^n1/k=o(n)$. @ZFR
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:44
$begingroup$
Quite nice solution? But is it possible to donit without logarithm?
$endgroup$
– ZFR
Jan 27 at 1:19
$begingroup$
Quite nice solution? But is it possible to donit without logarithm?
$endgroup$
– ZFR
Jan 27 at 1:19
$begingroup$
Yes, see the other solution. Or prove that $sum_{k=1}^n1/k=o(n)$. @ZFR
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:44
$begingroup$
Yes, see the other solution. Or prove that $sum_{k=1}^n1/k=o(n)$. @ZFR
$endgroup$
– Lord Shark the Unknown
Jan 27 at 10:44
add a comment |
$begingroup$
By Stolz-Cesaro's Lemma we get that
$lim_{ntoinfty}nx_n=lim_{ntoinfty}frac{n}{frac{1}{x_n}}=lim_{ntoinfty}frac{1}{frac{1}{x_{n+1}}-frac{1}{x_n}}=lim_{ntoinfty}frac{x_nx_{n+1}}{x_n-x_{n+1}}=lim_{ntoinfty}frac{x_n^2-x_n^3}{x_n^2}=lim_{ntoinfty}(1-x_n)=1$
answered Jan 26 at 23:06
AlexdanutAlexdanut
1538
$endgroup$
add a comment |
$begingroup$
By Stolz-Cesaro's Lemma we get that
$lim_{ntoinfty}nx_n=lim_{ntoinfty}frac{n}{frac{1}{x_n}}=lim_{ntoinfty}frac{1}{frac{1}{x_{n+1}}-frac{1}{x_n}}=lim_{ntoinfty}frac{x_nx_{n+1}}{x_n-x_{n+1}}=lim_{ntoinfty}frac{x_n^2-x_n^3}{x_n^2}=lim_{ntoinfty}(1-x_n)=1$
answered Jan 26 at 23:06
AlexdanutAlexdanut
1538
$endgroup$
add a comment |
$begingroup$
By Stolz-Cesaro's Lemma we get that
$lim_{ntoinfty}nx_n=lim_{ntoinfty}frac{n}{frac{1}{x_n}}=lim_{ntoinfty}frac{1}{frac{1}{x_{n+1}}-frac{1}{x_n}}=lim_{ntoinfty}frac{x_nx_{n+1}}{x_n-x_{n+1}}=lim_{ntoinfty}frac{x_n^2-x_n^3}{x_n^2}=lim_{ntoinfty}(1-x_n)=1$
answered Jan 26 at 23:06
AlexdanutAlexdanut
1538
$endgroup$
By Stolz-Cesaro's Lemma we get that
$lim_{ntoinfty}nx_n=lim_{ntoinfty}frac{n}{frac{1}{x_n}}=lim_{ntoinfty}frac{1}{frac{1}{x_{n+1}}-frac{1}{x_n}}=lim_{ntoinfty}frac{x_nx_{n+1}}{x_n-x_{n+1}}=lim_{ntoinfty}frac{x_n^2-x_n^3}{x_n^2}=lim_{ntoinfty}(1-x_n)=1$
answered Jan 26 at 23:06
AlexdanutAlexdanut
1538
answered Jan 26 at 23:06
AlexdanutAlexdanut
1538
answered Jan 26 at 23:06
AlexdanutAlexdanut
1538
answered Jan 26 at 23:06
AlexdanutAlexdanut
1538
1538
add a comment |
add a comment |
$begingroup$
Let $y_n = n x_n$. Then find the recurrence for $y_n$. The initial condition is known. Can you apply the same method to find the limit?
$endgroup$
– Sasha
Jan 26 at 22:28
2
$begingroup$
I'd start by observing that $$frac1{x_{n+1}}=frac{1}{x_n}+1+O(x_n).$$
$endgroup$
– Lord Shark the Unknown
Jan 26 at 22:30
$begingroup$
@Sasha, i cannot answer you remark from the top of my head but i have done this approach before posting a question and if i am not mistaken this approach was not useful
$endgroup$
– ZFR
Jan 26 at 22:56