Solution verification: evaluate $limlimits_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda...
$begingroup$
Problem
Evaluate
$$lim_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda n}}{n^{lambda n}}$$
where $lambda>0.$
Solution
Denote
$$S_n:=sum_{k=1}^{n}left(frac{k}{n}right)^{lambda n}=sum_{k=0}^{n-1}left(1-frac{k}{n}right)^{lambda n}.tag1$$
On one hand, we choose some $pin mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n geq sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}.tag2$$
Let $n to infty$ within $(2)$. We obtain
$$liminf_{n to infty}S_ngeq liminf_{n to infty} sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}liminf_{n to infty}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}e^{-lambda k}.tag3$$
Notice that $(3)$ holds for all $p$. We may let $p to infty$ and obtain
$$liminf_{n to infty}S_ngeq sum_{k=0}^{infty}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag4$$
On the other hand, from $1+xleq e^x$ we may derive
$left(1-dfrac{k}{n}right)^{lambda n}leq e^{-lambda k}.$
Thus
$$S_n leq sum_{k=0}^{n-1}e^{-lambda k}.tag5$$
Let $n to infty$ within $(5)$. We obtain
$$limsup_{n to infty}S_n leq limsup_{n to infty}sum_{k=0}^{n-1}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag6$$
See $(4)$ and $(6)$. We may conclude
$$frac{e^{lambda}}{e^{lambda}-1}leq liminf_{n to infty}S_nleq limsup_{n to infty}S_nleq frac{e^{lambda}}{e^{lambda}-1},tag7$$
which implies
$$lim_{n to infty}S_n=frac{e^{lambda}}{e^{lambda}-1}.tag8$$
Please correct me if I'm wrong. Thanks.
limits proof-verification limsup-and-liminf
$endgroup$
add a comment |
$begingroup$
Problem
Evaluate
$$lim_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda n}}{n^{lambda n}}$$
where $lambda>0.$
Solution
Denote
$$S_n:=sum_{k=1}^{n}left(frac{k}{n}right)^{lambda n}=sum_{k=0}^{n-1}left(1-frac{k}{n}right)^{lambda n}.tag1$$
On one hand, we choose some $pin mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n geq sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}.tag2$$
Let $n to infty$ within $(2)$. We obtain
$$liminf_{n to infty}S_ngeq liminf_{n to infty} sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}liminf_{n to infty}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}e^{-lambda k}.tag3$$
Notice that $(3)$ holds for all $p$. We may let $p to infty$ and obtain
$$liminf_{n to infty}S_ngeq sum_{k=0}^{infty}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag4$$
On the other hand, from $1+xleq e^x$ we may derive
$left(1-dfrac{k}{n}right)^{lambda n}leq e^{-lambda k}.$
Thus
$$S_n leq sum_{k=0}^{n-1}e^{-lambda k}.tag5$$
Let $n to infty$ within $(5)$. We obtain
$$limsup_{n to infty}S_n leq limsup_{n to infty}sum_{k=0}^{n-1}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag6$$
See $(4)$ and $(6)$. We may conclude
$$frac{e^{lambda}}{e^{lambda}-1}leq liminf_{n to infty}S_nleq limsup_{n to infty}S_nleq frac{e^{lambda}}{e^{lambda}-1},tag7$$
which implies
$$lim_{n to infty}S_n=frac{e^{lambda}}{e^{lambda}-1}.tag8$$
Please correct me if I'm wrong. Thanks.
limits proof-verification limsup-and-liminf
$endgroup$
$begingroup$
Yes, it's correct.
$endgroup$
– zoidberg
Dec 7 '18 at 4:35
add a comment |
$begingroup$
Problem
Evaluate
$$lim_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda n}}{n^{lambda n}}$$
where $lambda>0.$
Solution
Denote
$$S_n:=sum_{k=1}^{n}left(frac{k}{n}right)^{lambda n}=sum_{k=0}^{n-1}left(1-frac{k}{n}right)^{lambda n}.tag1$$
On one hand, we choose some $pin mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n geq sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}.tag2$$
Let $n to infty$ within $(2)$. We obtain
$$liminf_{n to infty}S_ngeq liminf_{n to infty} sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}liminf_{n to infty}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}e^{-lambda k}.tag3$$
Notice that $(3)$ holds for all $p$. We may let $p to infty$ and obtain
$$liminf_{n to infty}S_ngeq sum_{k=0}^{infty}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag4$$
On the other hand, from $1+xleq e^x$ we may derive
$left(1-dfrac{k}{n}right)^{lambda n}leq e^{-lambda k}.$
Thus
$$S_n leq sum_{k=0}^{n-1}e^{-lambda k}.tag5$$
Let $n to infty$ within $(5)$. We obtain
$$limsup_{n to infty}S_n leq limsup_{n to infty}sum_{k=0}^{n-1}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag6$$
See $(4)$ and $(6)$. We may conclude
$$frac{e^{lambda}}{e^{lambda}-1}leq liminf_{n to infty}S_nleq limsup_{n to infty}S_nleq frac{e^{lambda}}{e^{lambda}-1},tag7$$
which implies
$$lim_{n to infty}S_n=frac{e^{lambda}}{e^{lambda}-1}.tag8$$
Please correct me if I'm wrong. Thanks.
limits proof-verification limsup-and-liminf
$endgroup$
Problem
Evaluate
$$lim_{n to infty}frac{1^{lambda n}+2^{lambda n}+cdots+n^{lambda n}}{n^{lambda n}}$$
where $lambda>0.$
Solution
Denote
$$S_n:=sum_{k=1}^{n}left(frac{k}{n}right)^{lambda n}=sum_{k=0}^{n-1}left(1-frac{k}{n}right)^{lambda n}.tag1$$
On one hand, we choose some $pin mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n geq sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}.tag2$$
Let $n to infty$ within $(2)$. We obtain
$$liminf_{n to infty}S_ngeq liminf_{n to infty} sum_{k=0}^{p}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}liminf_{n to infty}left(1-frac{k}{n}right)^{lambda n}=sum_{k=0}^{p}e^{-lambda k}.tag3$$
Notice that $(3)$ holds for all $p$. We may let $p to infty$ and obtain
$$liminf_{n to infty}S_ngeq sum_{k=0}^{infty}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag4$$
On the other hand, from $1+xleq e^x$ we may derive
$left(1-dfrac{k}{n}right)^{lambda n}leq e^{-lambda k}.$
Thus
$$S_n leq sum_{k=0}^{n-1}e^{-lambda k}.tag5$$
Let $n to infty$ within $(5)$. We obtain
$$limsup_{n to infty}S_n leq limsup_{n to infty}sum_{k=0}^{n-1}e^{-lambda k}=frac{e^{lambda}}{e^{lambda}-1}.tag6$$
See $(4)$ and $(6)$. We may conclude
$$frac{e^{lambda}}{e^{lambda}-1}leq liminf_{n to infty}S_nleq limsup_{n to infty}S_nleq frac{e^{lambda}}{e^{lambda}-1},tag7$$
which implies
$$lim_{n to infty}S_n=frac{e^{lambda}}{e^{lambda}-1}.tag8$$
Please correct me if I'm wrong. Thanks.
limits proof-verification limsup-and-liminf
limits proof-verification limsup-and-liminf
edited Dec 7 '18 at 13:22
mengdie1982
asked Dec 7 '18 at 4:22
mengdie1982mengdie1982
4,912618
4,912618
$begingroup$
Yes, it's correct.
$endgroup$
– zoidberg
Dec 7 '18 at 4:35
add a comment |
$begingroup$
Yes, it's correct.
$endgroup$
– zoidberg
Dec 7 '18 at 4:35
$begingroup$
Yes, it's correct.
$endgroup$
– zoidberg
Dec 7 '18 at 4:35
$begingroup$
Yes, it's correct.
$endgroup$
– zoidberg
Dec 7 '18 at 4:35
add a comment |
0
active
oldest
votes
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%2f3029458%2fsolution-verification-evaluate-lim-limits-n-to-infty-frac1-lambda-n2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3029458%2fsolution-verification-evaluate-lim-limits-n-to-infty-frac1-lambda-n2%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
$begingroup$
Yes, it's correct.
$endgroup$
– zoidberg
Dec 7 '18 at 4:35