Why is it possible here to sum equivalent
$begingroup$
We have a sequence :
$$u_n = frac{e^{-u_{n-1}}}{n+1}$$
It's easy to prove that : $lim_{n to infty} u_n = 0$ and that $lim_{n to infty} nu_n = 1$ so that : $u_n sim frac{1}{n}$.
Then I am asked to find the nature of $ sum (-1)^n u_n$.
In the correction of the exercise they do the following :
$$mid u_n mid = frac{1-u_{n-1} +O(u_{n-1})}{n+1} = frac{1}{n} - frac{1}{n^2} + O(frac{1}{n^2})$$
I don't understand why is it possible to replace $u_{n-1}$ by the equivalent of $u_{n-1}$ which is $frac{1}{n-1}$ since it's a sum. Normally in a sum we need to do some tricks in order to replace a sequence by it's equivalent.
When I compute the limit of $ frac{1-u_{n-1}}{1-1/n}$ it does go to $1$ at infinity. But they are not giving any explanations of why it is possible here ? Is there a rule or something I missed ?
Thank you !
real-analysis sequences-and-series asymptotics
asked Jan 6 at 11:12
hfrjeaklhfuzlhfrjeaklhfuzl
132
$endgroup$
add a comment |
$begingroup$
We have a sequence :
$$u_n = frac{e^{-u_{n-1}}}{n+1}$$
It's easy to prove that : $lim_{n to infty} u_n = 0$ and that $lim_{n to infty} nu_n = 1$ so that : $u_n sim frac{1}{n}$.
Then I am asked to find the nature of $ sum (-1)^n u_n$.
In the correction of the exercise they do the following :
$$mid u_n mid = frac{1-u_{n-1} +O(u_{n-1})}{n+1} = frac{1}{n} - frac{1}{n^2} + O(frac{1}{n^2})$$
I don't understand why is it possible to replace $u_{n-1}$ by the equivalent of $u_{n-1}$ which is $frac{1}{n-1}$ since it's a sum. Normally in a sum we need to do some tricks in order to replace a sequence by it's equivalent.
When I compute the limit of $ frac{1-u_{n-1}}{1-1/n}$ it does go to $1$ at infinity. But they are not giving any explanations of why it is possible here ? Is there a rule or something I missed ?
Thank you !
real-analysis sequences-and-series asymptotics
asked Jan 6 at 11:12
hfrjeaklhfuzlhfrjeaklhfuzl
132
$endgroup$
$begingroup$
I suppose your big-Oh's should be little-oh's?
$endgroup$
– Hagen von Eitzen
Jan 6 at 11:26
add a comment |
$begingroup$
We have a sequence :
$$u_n = frac{e^{-u_{n-1}}}{n+1}$$
It's easy to prove that : $lim_{n to infty} u_n = 0$ and that $lim_{n to infty} nu_n = 1$ so that : $u_n sim frac{1}{n}$.
Then I am asked to find the nature of $ sum (-1)^n u_n$.
In the correction of the exercise they do the following :
$$mid u_n mid = frac{1-u_{n-1} +O(u_{n-1})}{n+1} = frac{1}{n} - frac{1}{n^2} + O(frac{1}{n^2})$$
I don't understand why is it possible to replace $u_{n-1}$ by the equivalent of $u_{n-1}$ which is $frac{1}{n-1}$ since it's a sum. Normally in a sum we need to do some tricks in order to replace a sequence by it's equivalent.
When I compute the limit of $ frac{1-u_{n-1}}{1-1/n}$ it does go to $1$ at infinity. But they are not giving any explanations of why it is possible here ? Is there a rule or something I missed ?
Thank you !
real-analysis sequences-and-series asymptotics
asked Jan 6 at 11:12
hfrjeaklhfuzlhfrjeaklhfuzl
132
$endgroup$
We have a sequence :
$$u_n = frac{e^{-u_{n-1}}}{n+1}$$
It's easy to prove that : $lim_{n to infty} u_n = 0$ and that $lim_{n to infty} nu_n = 1$ so that : $u_n sim frac{1}{n}$.
Then I am asked to find the nature of $ sum (-1)^n u_n$.
In the correction of the exercise they do the following :
$$mid u_n mid = frac{1-u_{n-1} +O(u_{n-1})}{n+1} = frac{1}{n} - frac{1}{n^2} + O(frac{1}{n^2})$$
I don't understand why is it possible to replace $u_{n-1}$ by the equivalent of $u_{n-1}$ which is $frac{1}{n-1}$ since it's a sum. Normally in a sum we need to do some tricks in order to replace a sequence by it's equivalent.
When I compute the limit of $ frac{1-u_{n-1}}{1-1/n}$ it does go to $1$ at infinity. But they are not giving any explanations of why it is possible here ? Is there a rule or something I missed ?
Thank you !
real-analysis sequences-and-series asymptotics
real-analysis sequences-and-series asymptotics
asked Jan 6 at 11:12
hfrjeaklhfuzlhfrjeaklhfuzl
132
asked Jan 6 at 11:12
hfrjeaklhfuzlhfrjeaklhfuzl
132
asked Jan 6 at 11:12
hfrjeaklhfuzlhfrjeaklhfuzl
132
asked Jan 6 at 11:12
hfrjeaklhfuzlhfrjeaklhfuzl
132
asked Jan 6 at 11:12
hfrjeaklhfuzlhfrjeaklhfuzl
132
132
$begingroup$
I suppose your big-Oh's should be little-oh's?
$endgroup$
– Hagen von Eitzen
Jan 6 at 11:26
add a comment |
$begingroup$
I suppose your big-Oh's should be little-oh's?
$endgroup$
– Hagen von Eitzen
Jan 6 at 11:26
$begingroup$
I suppose your big-Oh's should be little-oh's?
$endgroup$
– Hagen von Eitzen
Jan 6 at 11:26
$begingroup$
I suppose your big-Oh's should be little-oh's?
$endgroup$
– Hagen von Eitzen
Jan 6 at 11:26
add a comment |
1 Answer
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$begingroup$
Consider the sequences $a_n=frac{1}{n+1}$, $b_n=u_n-a_n$.
Then $sum_n{(-1)^na_n}$ is well-known to be convergent.
Furthermore, $b_n=frac{1}{n+1}(e^{-u_{n-1}}-1)sim frac{-u_{n-1}}{n} sim -n^{-2}$, thus $sum_n{b_n(-1)^n}$ is absolutely convergent.
answered Jan 6 at 11:18
MindlackMindlack
4,910211
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider the sequences $a_n=frac{1}{n+1}$, $b_n=u_n-a_n$.
Then $sum_n{(-1)^na_n}$ is well-known to be convergent.
Furthermore, $b_n=frac{1}{n+1}(e^{-u_{n-1}}-1)sim frac{-u_{n-1}}{n} sim -n^{-2}$, thus $sum_n{b_n(-1)^n}$ is absolutely convergent.
answered Jan 6 at 11:18
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Consider the sequences $a_n=frac{1}{n+1}$, $b_n=u_n-a_n$.
Then $sum_n{(-1)^na_n}$ is well-known to be convergent.
Furthermore, $b_n=frac{1}{n+1}(e^{-u_{n-1}}-1)sim frac{-u_{n-1}}{n} sim -n^{-2}$, thus $sum_n{b_n(-1)^n}$ is absolutely convergent.
answered Jan 6 at 11:18
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Consider the sequences $a_n=frac{1}{n+1}$, $b_n=u_n-a_n$.
Then $sum_n{(-1)^na_n}$ is well-known to be convergent.
Furthermore, $b_n=frac{1}{n+1}(e^{-u_{n-1}}-1)sim frac{-u_{n-1}}{n} sim -n^{-2}$, thus $sum_n{b_n(-1)^n}$ is absolutely convergent.
answered Jan 6 at 11:18
MindlackMindlack
4,910211
$endgroup$
Consider the sequences $a_n=frac{1}{n+1}$, $b_n=u_n-a_n$.
Then $sum_n{(-1)^na_n}$ is well-known to be convergent.
Furthermore, $b_n=frac{1}{n+1}(e^{-u_{n-1}}-1)sim frac{-u_{n-1}}{n} sim -n^{-2}$, thus $sum_n{b_n(-1)^n}$ is absolutely convergent.
answered Jan 6 at 11:18
MindlackMindlack
4,910211
edited Jan 6 at 11:33
answered Jan 6 at 11:18
MindlackMindlack
4,910211
answered Jan 6 at 11:18
MindlackMindlack
4,910211
answered Jan 6 at 11:18
MindlackMindlack
4,910211
4,910211
add a comment |
add a comment |
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$begingroup$
I suppose your big-Oh's should be little-oh's?
$endgroup$
– Hagen von Eitzen
Jan 6 at 11:26