Confusion in applying Cauchy's Theorem of limits
$begingroup$
Cauchy's theorem of limits states that if $ lim_{n to infty} a_n=L ,$ then $$ lim_{n to infty} frac{a_1+a_2+cdots+a_n}{n}=L $$
If I apply this in the series $$S = lim_{ntoinfty} dfrac{1}{n}[e^{frac{1}{n}} + e^{frac{2}{n}} + e^{frac{3}{n}} + e^{frac{4}{n}} + e^{frac{5}{n}} + e^{frac{6}{n}} + ... + e^{frac{n}{n}}]$$
Here $a_n=e^{n/n},$ Hence, $lim_{n to infty} a_n=e $, so $S=e$.
I know this is incorrect because the integration gives the answer $e-1$. It seems to me that the reason for this is related to the terms of the sequence being dependent on $n$. I can't be sure of this because the same theorem gives the correct answer for the limit of this series(below) which is given in my book as a solved example of the above theorem $$ lim_{n to infty} left( frac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}} right)=1 $$
real-analysis sequences-and-series limits
edited Sep 27 '18 at 17:45
dmtri
1,4492521
asked Sep 27 '18 at 17:36
jirenjiren
766
$endgroup$
|
show 4 more comments
$begingroup$
Cauchy's theorem of limits states that if $ lim_{n to infty} a_n=L ,$ then $$ lim_{n to infty} frac{a_1+a_2+cdots+a_n}{n}=L $$
If I apply this in the series $$S = lim_{ntoinfty} dfrac{1}{n}[e^{frac{1}{n}} + e^{frac{2}{n}} + e^{frac{3}{n}} + e^{frac{4}{n}} + e^{frac{5}{n}} + e^{frac{6}{n}} + ... + e^{frac{n}{n}}]$$
Here $a_n=e^{n/n},$ Hence, $lim_{n to infty} a_n=e $, so $S=e$.
I know this is incorrect because the integration gives the answer $e-1$. It seems to me that the reason for this is related to the terms of the sequence being dependent on $n$. I can't be sure of this because the same theorem gives the correct answer for the limit of this series(below) which is given in my book as a solved example of the above theorem $$ lim_{n to infty} left( frac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}} right)=1 $$
real-analysis sequences-and-series limits
edited Sep 27 '18 at 17:45
dmtri
1,4492521
asked Sep 27 '18 at 17:36
jirenjiren
766
$endgroup$
4
$begingroup$
The sequence you're calling $(a_n)$ is changing as a function of $n$. In effect you have a sequence of two variables, $a_{m,n} = e^{m/n}$ for $1 leq m leq n$.
$endgroup$
– Bungo
Sep 27 '18 at 17:39
$begingroup$
@Bungo Then what's the difference between $(a_n)$ and the series given at the end.
$endgroup$
– jiren
Sep 27 '18 at 17:42
1
$begingroup$
The series given at the end is not a direct application of the theorem you state, on the face of it. Your book must be doing more.
$endgroup$
– MPW
Sep 27 '18 at 17:59
1
$begingroup$
If $a_n=e^{n/n}$ then $a_1+cdots+a_n$ is $e^{1/1} + e^{2/2} + cdots + e^{n/n}$, not $e^{1/n} + e^{2/n}+ cdots + e^{n/n}$ as you wrote in your second displayed equation.
$endgroup$
– Matthew Towers
Sep 27 '18 at 18:02
$begingroup$
@MPW In my book the general term of the series is given as $a_k= frac{n}{sqrt{n^2+k}}, k=1,2cdots n$ and after that, without giving any explanation, $lim a_n= lim frac{1}{sqrt{1+1/n}}=1$, so using cauchy's theorem, limit of the series is also 1.
$endgroup$
– jiren
Sep 27 '18 at 18:14
|
show 4 more comments
$begingroup$
Cauchy's theorem of limits states that if $ lim_{n to infty} a_n=L ,$ then $$ lim_{n to infty} frac{a_1+a_2+cdots+a_n}{n}=L $$
If I apply this in the series $$S = lim_{ntoinfty} dfrac{1}{n}[e^{frac{1}{n}} + e^{frac{2}{n}} + e^{frac{3}{n}} + e^{frac{4}{n}} + e^{frac{5}{n}} + e^{frac{6}{n}} + ... + e^{frac{n}{n}}]$$
Here $a_n=e^{n/n},$ Hence, $lim_{n to infty} a_n=e $, so $S=e$.
I know this is incorrect because the integration gives the answer $e-1$. It seems to me that the reason for this is related to the terms of the sequence being dependent on $n$. I can't be sure of this because the same theorem gives the correct answer for the limit of this series(below) which is given in my book as a solved example of the above theorem $$ lim_{n to infty} left( frac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}} right)=1 $$
real-analysis sequences-and-series limits
edited Sep 27 '18 at 17:45
dmtri
1,4492521
asked Sep 27 '18 at 17:36
jirenjiren
766
$endgroup$
Cauchy's theorem of limits states that if $ lim_{n to infty} a_n=L ,$ then $$ lim_{n to infty} frac{a_1+a_2+cdots+a_n}{n}=L $$
If I apply this in the series $$S = lim_{ntoinfty} dfrac{1}{n}[e^{frac{1}{n}} + e^{frac{2}{n}} + e^{frac{3}{n}} + e^{frac{4}{n}} + e^{frac{5}{n}} + e^{frac{6}{n}} + ... + e^{frac{n}{n}}]$$
Here $a_n=e^{n/n},$ Hence, $lim_{n to infty} a_n=e $, so $S=e$.
I know this is incorrect because the integration gives the answer $e-1$. It seems to me that the reason for this is related to the terms of the sequence being dependent on $n$. I can't be sure of this because the same theorem gives the correct answer for the limit of this series(below) which is given in my book as a solved example of the above theorem $$ lim_{n to infty} left( frac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}} right)=1 $$
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited Sep 27 '18 at 17:45
dmtri
1,4492521
asked Sep 27 '18 at 17:36
jirenjiren
766
edited Sep 27 '18 at 17:45
dmtri
1,4492521
asked Sep 27 '18 at 17:36
jirenjiren
766
edited Sep 27 '18 at 17:45
dmtri
1,4492521
edited Sep 27 '18 at 17:45
dmtri
1,4492521
edited Sep 27 '18 at 17:45
dmtri
1,4492521
1,4492521
asked Sep 27 '18 at 17:36
jirenjiren
766
asked Sep 27 '18 at 17:36
jirenjiren
766
asked Sep 27 '18 at 17:36
jirenjiren
766
766
4
$begingroup$
The sequence you're calling $(a_n)$ is changing as a function of $n$. In effect you have a sequence of two variables, $a_{m,n} = e^{m/n}$ for $1 leq m leq n$.
$endgroup$
– Bungo
Sep 27 '18 at 17:39
$begingroup$
@Bungo Then what's the difference between $(a_n)$ and the series given at the end.
$endgroup$
– jiren
Sep 27 '18 at 17:42
1
$begingroup$
The series given at the end is not a direct application of the theorem you state, on the face of it. Your book must be doing more.
$endgroup$
– MPW
Sep 27 '18 at 17:59
1
$begingroup$
If $a_n=e^{n/n}$ then $a_1+cdots+a_n$ is $e^{1/1} + e^{2/2} + cdots + e^{n/n}$, not $e^{1/n} + e^{2/n}+ cdots + e^{n/n}$ as you wrote in your second displayed equation.
$endgroup$
– Matthew Towers
Sep 27 '18 at 18:02
$begingroup$
@MPW In my book the general term of the series is given as $a_k= frac{n}{sqrt{n^2+k}}, k=1,2cdots n$ and after that, without giving any explanation, $lim a_n= lim frac{1}{sqrt{1+1/n}}=1$, so using cauchy's theorem, limit of the series is also 1.
$endgroup$
– jiren
Sep 27 '18 at 18:14
|
show 4 more comments
4
$begingroup$
The sequence you're calling $(a_n)$ is changing as a function of $n$. In effect you have a sequence of two variables, $a_{m,n} = e^{m/n}$ for $1 leq m leq n$.
$endgroup$
– Bungo
Sep 27 '18 at 17:39
$begingroup$
@Bungo Then what's the difference between $(a_n)$ and the series given at the end.
$endgroup$
– jiren
Sep 27 '18 at 17:42
1
$begingroup$
The series given at the end is not a direct application of the theorem you state, on the face of it. Your book must be doing more.
$endgroup$
– MPW
Sep 27 '18 at 17:59
1
$begingroup$
If $a_n=e^{n/n}$ then $a_1+cdots+a_n$ is $e^{1/1} + e^{2/2} + cdots + e^{n/n}$, not $e^{1/n} + e^{2/n}+ cdots + e^{n/n}$ as you wrote in your second displayed equation.
$endgroup$
– Matthew Towers
Sep 27 '18 at 18:02
$begingroup$
@MPW In my book the general term of the series is given as $a_k= frac{n}{sqrt{n^2+k}}, k=1,2cdots n$ and after that, without giving any explanation, $lim a_n= lim frac{1}{sqrt{1+1/n}}=1$, so using cauchy's theorem, limit of the series is also 1.
$endgroup$
– jiren
Sep 27 '18 at 18:14
4
4
$begingroup$
The sequence you're calling $(a_n)$ is changing as a function of $n$. In effect you have a sequence of two variables, $a_{m,n} = e^{m/n}$ for $1 leq m leq n$.
$endgroup$
– Bungo
Sep 27 '18 at 17:39
$begingroup$
The sequence you're calling $(a_n)$ is changing as a function of $n$. In effect you have a sequence of two variables, $a_{m,n} = e^{m/n}$ for $1 leq m leq n$.
$endgroup$
– Bungo
Sep 27 '18 at 17:39
$begingroup$
@Bungo Then what's the difference between $(a_n)$ and the series given at the end.
$endgroup$
– jiren
Sep 27 '18 at 17:42
$begingroup$
@Bungo Then what's the difference between $(a_n)$ and the series given at the end.
$endgroup$
– jiren
Sep 27 '18 at 17:42
1
1
$begingroup$
The series given at the end is not a direct application of the theorem you state, on the face of it. Your book must be doing more.
$endgroup$
– MPW
Sep 27 '18 at 17:59
$begingroup$
The series given at the end is not a direct application of the theorem you state, on the face of it. Your book must be doing more.
$endgroup$
– MPW
Sep 27 '18 at 17:59
1
1
$begingroup$
If $a_n=e^{n/n}$ then $a_1+cdots+a_n$ is $e^{1/1} + e^{2/2} + cdots + e^{n/n}$, not $e^{1/n} + e^{2/n}+ cdots + e^{n/n}$ as you wrote in your second displayed equation.
$endgroup$
– Matthew Towers
Sep 27 '18 at 18:02
$begingroup$
If $a_n=e^{n/n}$ then $a_1+cdots+a_n$ is $e^{1/1} + e^{2/2} + cdots + e^{n/n}$, not $e^{1/n} + e^{2/n}+ cdots + e^{n/n}$ as you wrote in your second displayed equation.
$endgroup$
– Matthew Towers
Sep 27 '18 at 18:02
$begingroup$
@MPW In my book the general term of the series is given as $a_k= frac{n}{sqrt{n^2+k}}, k=1,2cdots n$ and after that, without giving any explanation, $lim a_n= lim frac{1}{sqrt{1+1/n}}=1$, so using cauchy's theorem, limit of the series is also 1.
$endgroup$
– jiren
Sep 27 '18 at 18:14
$begingroup$
@MPW In my book the general term of the series is given as $a_k= frac{n}{sqrt{n^2+k}}, k=1,2cdots n$ and after that, without giving any explanation, $lim a_n= lim frac{1}{sqrt{1+1/n}}=1$, so using cauchy's theorem, limit of the series is also 1.
$endgroup$
– jiren
Sep 27 '18 at 18:14
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
You can't apply the theorem in the first case since $$a_n=e^{nover n}=e$$which doesn't make sense. Another reason is that $a_k=e^{kover n}$ that says $a_k$ is dependent to $n$ (!!) which is quite obviously wrong . Instead $$lim_{nto infty}{1over n}sum_{i=1}^{n} e^{iover n}=int_{0}^{1}e^xdx=e-1$$also for the second one, I'm yet suspicious because of a same argument as the former one. In this case we can easily write that$$ {nover sqrt{n^2+n}}lefrac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}}le{nover sqrt{n^2+1}}$$and use squeeze theorem to attain the result.
answered Dec 1 '18 at 9:11
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
The main issue is that your series UP TO $n$ are $a_i=e^{frac{i}{n}}$ but this defines the sequence only UP TO some $n$ which is fixed and thus taking this $n$ fixed tells you nothing about the later terms.
answered Sep 27 '18 at 17:42
asdfasdf
3,691519
$endgroup$
add a comment |
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2 Answers
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$begingroup$
You can't apply the theorem in the first case since $$a_n=e^{nover n}=e$$which doesn't make sense. Another reason is that $a_k=e^{kover n}$ that says $a_k$ is dependent to $n$ (!!) which is quite obviously wrong . Instead $$lim_{nto infty}{1over n}sum_{i=1}^{n} e^{iover n}=int_{0}^{1}e^xdx=e-1$$also for the second one, I'm yet suspicious because of a same argument as the former one. In this case we can easily write that$$ {nover sqrt{n^2+n}}lefrac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}}le{nover sqrt{n^2+1}}$$and use squeeze theorem to attain the result.
answered Dec 1 '18 at 9:11
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
You can't apply the theorem in the first case since $$a_n=e^{nover n}=e$$which doesn't make sense. Another reason is that $a_k=e^{kover n}$ that says $a_k$ is dependent to $n$ (!!) which is quite obviously wrong . Instead $$lim_{nto infty}{1over n}sum_{i=1}^{n} e^{iover n}=int_{0}^{1}e^xdx=e-1$$also for the second one, I'm yet suspicious because of a same argument as the former one. In this case we can easily write that$$ {nover sqrt{n^2+n}}lefrac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}}le{nover sqrt{n^2+1}}$$and use squeeze theorem to attain the result.
answered Dec 1 '18 at 9:11
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
add a comment |
$begingroup$
You can't apply the theorem in the first case since $$a_n=e^{nover n}=e$$which doesn't make sense. Another reason is that $a_k=e^{kover n}$ that says $a_k$ is dependent to $n$ (!!) which is quite obviously wrong . Instead $$lim_{nto infty}{1over n}sum_{i=1}^{n} e^{iover n}=int_{0}^{1}e^xdx=e-1$$also for the second one, I'm yet suspicious because of a same argument as the former one. In this case we can easily write that$$ {nover sqrt{n^2+n}}lefrac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}}le{nover sqrt{n^2+1}}$$and use squeeze theorem to attain the result.
answered Dec 1 '18 at 9:11
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
You can't apply the theorem in the first case since $$a_n=e^{nover n}=e$$which doesn't make sense. Another reason is that $a_k=e^{kover n}$ that says $a_k$ is dependent to $n$ (!!) which is quite obviously wrong . Instead $$lim_{nto infty}{1over n}sum_{i=1}^{n} e^{iover n}=int_{0}^{1}e^xdx=e-1$$also for the second one, I'm yet suspicious because of a same argument as the former one. In this case we can easily write that$$ {nover sqrt{n^2+n}}lefrac{1}{sqrt{n^2+1}}+frac{1}{sqrt{n^2+2}}+cdots+frac{1}{sqrt{n^2+n}}le{nover sqrt{n^2+1}}$$and use squeeze theorem to attain the result.
answered Dec 1 '18 at 9:11
Mostafa AyazMostafa Ayaz
15.3k3939
answered Dec 1 '18 at 9:11
Mostafa AyazMostafa Ayaz
15.3k3939
answered Dec 1 '18 at 9:11
Mostafa AyazMostafa Ayaz
15.3k3939
answered Dec 1 '18 at 9:11
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
add a comment |
add a comment |
$begingroup$
The main issue is that your series UP TO $n$ are $a_i=e^{frac{i}{n}}$ but this defines the sequence only UP TO some $n$ which is fixed and thus taking this $n$ fixed tells you nothing about the later terms.
answered Sep 27 '18 at 17:42
asdfasdf
3,691519
$endgroup$
add a comment |
$begingroup$
The main issue is that your series UP TO $n$ are $a_i=e^{frac{i}{n}}$ but this defines the sequence only UP TO some $n$ which is fixed and thus taking this $n$ fixed tells you nothing about the later terms.
answered Sep 27 '18 at 17:42
asdfasdf
3,691519
$endgroup$
add a comment |
$begingroup$
The main issue is that your series UP TO $n$ are $a_i=e^{frac{i}{n}}$ but this defines the sequence only UP TO some $n$ which is fixed and thus taking this $n$ fixed tells you nothing about the later terms.
answered Sep 27 '18 at 17:42
asdfasdf
3,691519
$endgroup$
The main issue is that your series UP TO $n$ are $a_i=e^{frac{i}{n}}$ but this defines the sequence only UP TO some $n$ which is fixed and thus taking this $n$ fixed tells you nothing about the later terms.
answered Sep 27 '18 at 17:42
asdfasdf
3,691519
answered Sep 27 '18 at 17:42
asdfasdf
3,691519
answered Sep 27 '18 at 17:42
asdfasdf
3,691519
answered Sep 27 '18 at 17:42
asdfasdf
3,691519
3,691519
add a comment |
add a comment |
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4
$begingroup$
The sequence you're calling $(a_n)$ is changing as a function of $n$. In effect you have a sequence of two variables, $a_{m,n} = e^{m/n}$ for $1 leq m leq n$.
$endgroup$
– Bungo
Sep 27 '18 at 17:39
$begingroup$
@Bungo Then what's the difference between $(a_n)$ and the series given at the end.
$endgroup$
– jiren
Sep 27 '18 at 17:42
1
$begingroup$
The series given at the end is not a direct application of the theorem you state, on the face of it. Your book must be doing more.
$endgroup$
– MPW
Sep 27 '18 at 17:59
1
$begingroup$
If $a_n=e^{n/n}$ then $a_1+cdots+a_n$ is $e^{1/1} + e^{2/2} + cdots + e^{n/n}$, not $e^{1/n} + e^{2/n}+ cdots + e^{n/n}$ as you wrote in your second displayed equation.
$endgroup$
– Matthew Towers
Sep 27 '18 at 18:02
$begingroup$
@MPW In my book the general term of the series is given as $a_k= frac{n}{sqrt{n^2+k}}, k=1,2cdots n$ and after that, without giving any explanation, $lim a_n= lim frac{1}{sqrt{1+1/n}}=1$, so using cauchy's theorem, limit of the series is also 1.
$endgroup$
– jiren
Sep 27 '18 at 18:14