Mixing
Help with filling in the details to show that $limlimits_{ntoinfty}...
$begingroup$
So we have,
$$begin{align}
lim_{ntoinfty} sum_{k=1}^{n}left(frac{k}{n}right)^n &= lim_{ntoinfty} sum_{j=0}^{n-1}bigg(frac{n-j}{n}bigg)^n \
&= lim_{ntoinfty} bigg(1+bigg(1-frac{1}{n}bigg)^n+...+bigg(1-frac{n-1}{n}bigg)^nbigg) \
&= 1+e^{-1}+e^{-2}+... \
&= frac{e}{e-1}
end{align},$$
but my problem is going from the second to the third line. As the limit involves both the summand and the sum and I am not sure how this line is "legal", that is what is it that allows is to apply the limit first to each term then to the sum, viz why is that $limlimits_{ntoinfty} sumlimits_{k=0}^{n-1}bigg(1+frac{-k}{n}bigg)^n=sumlimits_{k=0}^inftylimlimits_{ntoinfty}bigg(1+frac{-k}{n}bigg)^n$? Is there a double limit? But can you have a double limit involving the same variable? Is this some special case of Fubini's Theorem, if so I am really struggling to see how? Or maybe it is a Riemann sum? If so so I'm not sure how to show t manipulate it to show that. Any help will be greatly appreciated.
Thanks in advance.
real-analysis limits
asked Jan 8 at 7:49
user152874user152874
21719
$endgroup$
|
show 6 more comments
$begingroup$
So we have,
$$begin{align}
lim_{ntoinfty} sum_{k=1}^{n}left(frac{k}{n}right)^n &= lim_{ntoinfty} sum_{j=0}^{n-1}bigg(frac{n-j}{n}bigg)^n \
&= lim_{ntoinfty} bigg(1+bigg(1-frac{1}{n}bigg)^n+...+bigg(1-frac{n-1}{n}bigg)^nbigg) \
&= 1+e^{-1}+e^{-2}+... \
&= frac{e}{e-1}
end{align},$$
but my problem is going from the second to the third line. As the limit involves both the summand and the sum and I am not sure how this line is "legal", that is what is it that allows is to apply the limit first to each term then to the sum, viz why is that $limlimits_{ntoinfty} sumlimits_{k=0}^{n-1}bigg(1+frac{-k}{n}bigg)^n=sumlimits_{k=0}^inftylimlimits_{ntoinfty}bigg(1+frac{-k}{n}bigg)^n$? Is there a double limit? But can you have a double limit involving the same variable? Is this some special case of Fubini's Theorem, if so I am really struggling to see how? Or maybe it is a Riemann sum? If so so I'm not sure how to show t manipulate it to show that. Any help will be greatly appreciated.
Thanks in advance.
real-analysis limits
asked Jan 8 at 7:49
user152874user152874
21719
$endgroup$
$begingroup$
You are right to be skeptical of this move. It is definitely something which has to be carefully justified.
$endgroup$
– Arthur
Jan 8 at 7:51
5
$begingroup$
You can justify it by monotone convergence theorem. See this earlier post math.stackexchange.com/questions/3003092/to-evaluate-the-limit/…
$endgroup$
– Song
Jan 8 at 8:00
$begingroup$
Thanks, this clears it up but if you use the dominated convergence theorem, what is the dominant function?
$endgroup$
– user152874
Jan 8 at 8:16
1
$begingroup$
@user152874 If you don't add @, I can't get an alarm ... Well, in this case the given measure is a counting measure, usually denoted by $c$. So if $a :mathbb{N}to mathbb{R}$ is a sequence, $int_mathbb{N} a_k dc(k) = sum_{k=1}^infty a_k$.
$endgroup$
– Song
Jan 10 at 14:19
2
$begingroup$
Possible duplicate of How should I calculate $lim_{nrightarrow infty} frac{1^n+2^n+3^n+...+n^n}{n^n}$
$endgroup$
– StubbornAtom
Jan 23 at 14:01
|
show 6 more comments
$begingroup$
So we have,
$$begin{align}
lim_{ntoinfty} sum_{k=1}^{n}left(frac{k}{n}right)^n &= lim_{ntoinfty} sum_{j=0}^{n-1}bigg(frac{n-j}{n}bigg)^n \
&= lim_{ntoinfty} bigg(1+bigg(1-frac{1}{n}bigg)^n+...+bigg(1-frac{n-1}{n}bigg)^nbigg) \
&= 1+e^{-1}+e^{-2}+... \
&= frac{e}{e-1}
end{align},$$
but my problem is going from the second to the third line. As the limit involves both the summand and the sum and I am not sure how this line is "legal", that is what is it that allows is to apply the limit first to each term then to the sum, viz why is that $limlimits_{ntoinfty} sumlimits_{k=0}^{n-1}bigg(1+frac{-k}{n}bigg)^n=sumlimits_{k=0}^inftylimlimits_{ntoinfty}bigg(1+frac{-k}{n}bigg)^n$? Is there a double limit? But can you have a double limit involving the same variable? Is this some special case of Fubini's Theorem, if so I am really struggling to see how? Or maybe it is a Riemann sum? If so so I'm not sure how to show t manipulate it to show that. Any help will be greatly appreciated.
Thanks in advance.
real-analysis limits
asked Jan 8 at 7:49
user152874user152874
21719
$endgroup$
So we have,
$$begin{align}
lim_{ntoinfty} sum_{k=1}^{n}left(frac{k}{n}right)^n &= lim_{ntoinfty} sum_{j=0}^{n-1}bigg(frac{n-j}{n}bigg)^n \
&= lim_{ntoinfty} bigg(1+bigg(1-frac{1}{n}bigg)^n+...+bigg(1-frac{n-1}{n}bigg)^nbigg) \
&= 1+e^{-1}+e^{-2}+... \
&= frac{e}{e-1}
end{align},$$
but my problem is going from the second to the third line. As the limit involves both the summand and the sum and I am not sure how this line is "legal", that is what is it that allows is to apply the limit first to each term then to the sum, viz why is that $limlimits_{ntoinfty} sumlimits_{k=0}^{n-1}bigg(1+frac{-k}{n}bigg)^n=sumlimits_{k=0}^inftylimlimits_{ntoinfty}bigg(1+frac{-k}{n}bigg)^n$? Is there a double limit? But can you have a double limit involving the same variable? Is this some special case of Fubini's Theorem, if so I am really struggling to see how? Or maybe it is a Riemann sum? If so so I'm not sure how to show t manipulate it to show that. Any help will be greatly appreciated.
Thanks in advance.
real-analysis limits
real-analysis limits
asked Jan 8 at 7:49
user152874user152874
21719
asked Jan 8 at 7:49
user152874user152874
21719
edited Jan 8 at 8:02
user152874
asked Jan 8 at 7:49
user152874user152874
21719
asked Jan 8 at 7:49
user152874user152874
21719
asked Jan 8 at 7:49
user152874user152874
21719
21719
$begingroup$
You are right to be skeptical of this move. It is definitely something which has to be carefully justified.
$endgroup$
– Arthur
Jan 8 at 7:51
5
$begingroup$
You can justify it by monotone convergence theorem. See this earlier post math.stackexchange.com/questions/3003092/to-evaluate-the-limit/…
$endgroup$
– Song
Jan 8 at 8:00
$begingroup$
Thanks, this clears it up but if you use the dominated convergence theorem, what is the dominant function?
$endgroup$
– user152874
Jan 8 at 8:16
1
$begingroup$
@user152874 If you don't add @, I can't get an alarm ... Well, in this case the given measure is a counting measure, usually denoted by $c$. So if $a :mathbb{N}to mathbb{R}$ is a sequence, $int_mathbb{N} a_k dc(k) = sum_{k=1}^infty a_k$.
$endgroup$
– Song
Jan 10 at 14:19
2
$begingroup$
Possible duplicate of How should I calculate $lim_{nrightarrow infty} frac{1^n+2^n+3^n+...+n^n}{n^n}$
$endgroup$
– StubbornAtom
Jan 23 at 14:01
|
show 6 more comments
$begingroup$
You are right to be skeptical of this move. It is definitely something which has to be carefully justified.
$endgroup$
– Arthur
Jan 8 at 7:51
5
$begingroup$
You can justify it by monotone convergence theorem. See this earlier post math.stackexchange.com/questions/3003092/to-evaluate-the-limit/…
$endgroup$
– Song
Jan 8 at 8:00
$begingroup$
Thanks, this clears it up but if you use the dominated convergence theorem, what is the dominant function?
$endgroup$
– user152874
Jan 8 at 8:16
1
$begingroup$
@user152874 If you don't add @, I can't get an alarm ... Well, in this case the given measure is a counting measure, usually denoted by $c$. So if $a :mathbb{N}to mathbb{R}$ is a sequence, $int_mathbb{N} a_k dc(k) = sum_{k=1}^infty a_k$.
$endgroup$
– Song
Jan 10 at 14:19
2
$begingroup$
Possible duplicate of How should I calculate $lim_{nrightarrow infty} frac{1^n+2^n+3^n+...+n^n}{n^n}$
$endgroup$
– StubbornAtom
Jan 23 at 14:01
$begingroup$
You are right to be skeptical of this move. It is definitely something which has to be carefully justified.
$endgroup$
– Arthur
Jan 8 at 7:51
$begingroup$
You are right to be skeptical of this move. It is definitely something which has to be carefully justified.
$endgroup$
– Arthur
Jan 8 at 7:51
5
5
$begingroup$
You can justify it by monotone convergence theorem. See this earlier post math.stackexchange.com/questions/3003092/to-evaluate-the-limit/…
$endgroup$
– Song
Jan 8 at 8:00
$begingroup$
You can justify it by monotone convergence theorem. See this earlier post math.stackexchange.com/questions/3003092/to-evaluate-the-limit/…
$endgroup$
– Song
Jan 8 at 8:00
$begingroup$
Thanks, this clears it up but if you use the dominated convergence theorem, what is the dominant function?
$endgroup$
– user152874
Jan 8 at 8:16
$begingroup$
Thanks, this clears it up but if you use the dominated convergence theorem, what is the dominant function?
$endgroup$
– user152874
Jan 8 at 8:16
1
1
$begingroup$
@user152874 If you don't add @, I can't get an alarm ... Well, in this case the given measure is a counting measure, usually denoted by $c$. So if $a :mathbb{N}to mathbb{R}$ is a sequence, $int_mathbb{N} a_k dc(k) = sum_{k=1}^infty a_k$.
$endgroup$
– Song
Jan 10 at 14:19
$begingroup$
@user152874 If you don't add @, I can't get an alarm ... Well, in this case the given measure is a counting measure, usually denoted by $c$. So if $a :mathbb{N}to mathbb{R}$ is a sequence, $int_mathbb{N} a_k dc(k) = sum_{k=1}^infty a_k$.
$endgroup$
– Song
Jan 10 at 14:19
2
2
$begingroup$
Possible duplicate of How should I calculate $lim_{nrightarrow infty} frac{1^n+2^n+3^n+...+n^n}{n^n}$
$endgroup$
– StubbornAtom
Jan 23 at 14:01
$begingroup$
Possible duplicate of How should I calculate $lim_{nrightarrow infty} frac{1^n+2^n+3^n+...+n^n}{n^n}$
$endgroup$
– StubbornAtom
Jan 23 at 14:01
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This requires justification.
First use the first $m$ terms with $m$ fixed to see that $sumlimits_{k=1}^{n} (frac k n)^{n} geq 1+(1-frac 1n)^{n}+cdots+(1-frac m n)^{n}$ for $n >m$ which gives $liminfsumlimits_{k=1}^{n} (frac k n)^{n} geq 1+e^{-1}+e^{-2}+cdots+e^{-m}$ for each $m$. Now let $m to infty$.
Next use the inequality $1-x leq e^{-x}, x geq 0$ to prove that for every $n$ we have $ sumlimits_{k=1}^{n} (frac k n)^{n} leq $ RHS. [$(1-frac j n)^{n} leq (e^{-j/n})^{n}=e^{-j}$ for each $j$].
answered Jan 8 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065902%2fhelp-with-filling-in-the-details-to-show-that-lim-limits-n-to-infty-sum-lim%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This requires justification.
First use the first $m$ terms with $m$ fixed to see that $sumlimits_{k=1}^{n} (frac k n)^{n} geq 1+(1-frac 1n)^{n}+cdots+(1-frac m n)^{n}$ for $n >m$ which gives $liminfsumlimits_{k=1}^{n} (frac k n)^{n} geq 1+e^{-1}+e^{-2}+cdots+e^{-m}$ for each $m$. Now let $m to infty$.
Next use the inequality $1-x leq e^{-x}, x geq 0$ to prove that for every $n$ we have $ sumlimits_{k=1}^{n} (frac k n)^{n} leq $ RHS. [$(1-frac j n)^{n} leq (e^{-j/n})^{n}=e^{-j}$ for each $j$].
answered Jan 8 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
add a comment |
$begingroup$
This requires justification.
First use the first $m$ terms with $m$ fixed to see that $sumlimits_{k=1}^{n} (frac k n)^{n} geq 1+(1-frac 1n)^{n}+cdots+(1-frac m n)^{n}$ for $n >m$ which gives $liminfsumlimits_{k=1}^{n} (frac k n)^{n} geq 1+e^{-1}+e^{-2}+cdots+e^{-m}$ for each $m$. Now let $m to infty$.
Next use the inequality $1-x leq e^{-x}, x geq 0$ to prove that for every $n$ we have $ sumlimits_{k=1}^{n} (frac k n)^{n} leq $ RHS. [$(1-frac j n)^{n} leq (e^{-j/n})^{n}=e^{-j}$ for each $j$].
answered Jan 8 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
add a comment |
$begingroup$
This requires justification.
First use the first $m$ terms with $m$ fixed to see that $sumlimits_{k=1}^{n} (frac k n)^{n} geq 1+(1-frac 1n)^{n}+cdots+(1-frac m n)^{n}$ for $n >m$ which gives $liminfsumlimits_{k=1}^{n} (frac k n)^{n} geq 1+e^{-1}+e^{-2}+cdots+e^{-m}$ for each $m$. Now let $m to infty$.
Next use the inequality $1-x leq e^{-x}, x geq 0$ to prove that for every $n$ we have $ sumlimits_{k=1}^{n} (frac k n)^{n} leq $ RHS. [$(1-frac j n)^{n} leq (e^{-j/n})^{n}=e^{-j}$ for each $j$].
answered Jan 8 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
This requires justification.
First use the first $m$ terms with $m$ fixed to see that $sumlimits_{k=1}^{n} (frac k n)^{n} geq 1+(1-frac 1n)^{n}+cdots+(1-frac m n)^{n}$ for $n >m$ which gives $liminfsumlimits_{k=1}^{n} (frac k n)^{n} geq 1+e^{-1}+e^{-2}+cdots+e^{-m}$ for each $m$. Now let $m to infty$.
Next use the inequality $1-x leq e^{-x}, x geq 0$ to prove that for every $n$ we have $ sumlimits_{k=1}^{n} (frac k n)^{n} leq $ RHS. [$(1-frac j n)^{n} leq (e^{-j/n})^{n}=e^{-j}$ for each $j$].
answered Jan 8 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
edited Jan 9 at 23:16
answered Jan 8 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 8 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 8 at 8:01
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
56.6k42159
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065902%2fhelp-with-filling-in-the-details-to-show-that-lim-limits-n-to-infty-sum-lim%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
You are right to be skeptical of this move. It is definitely something which has to be carefully justified.
$endgroup$
– Arthur
Jan 8 at 7:51
5
$begingroup$
You can justify it by monotone convergence theorem. See this earlier post math.stackexchange.com/questions/3003092/to-evaluate-the-limit/…
$endgroup$
– Song
Jan 8 at 8:00
$begingroup$
Thanks, this clears it up but if you use the dominated convergence theorem, what is the dominant function?
$endgroup$
– user152874
Jan 8 at 8:16
1
$begingroup$
@user152874 If you don't add @, I can't get an alarm ... Well, in this case the given measure is a counting measure, usually denoted by $c$. So if $a :mathbb{N}to mathbb{R}$ is a sequence, $int_mathbb{N} a_k dc(k) = sum_{k=1}^infty a_k$.
$endgroup$
– Song
Jan 10 at 14:19
2
$begingroup$
Possible duplicate of How should I calculate $lim_{nrightarrow infty} frac{1^n+2^n+3^n+...+n^n}{n^n}$
$endgroup$
– StubbornAtom
Jan 23 at 14:01