Theorem 3.37 in Baby Rudin: Any Counter-Examples In The Other Case?
$begingroup$
Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:
For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$
I think I fully understand the proof by Rudin.
From Theorem 3.37 in Baby Rudin, we can also conclude that following:
For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.
Am I right?
However, I'm unable to figure out the proof of or come up with any counter-examples to the following:
Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.
What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?
real-analysis sequences-and-series limits analysis limsup-and-liminf
$endgroup$
add a comment |
$begingroup$
Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:
For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$
I think I fully understand the proof by Rudin.
From Theorem 3.37 in Baby Rudin, we can also conclude that following:
For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.
Am I right?
However, I'm unable to figure out the proof of or come up with any counter-examples to the following:
Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.
What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?
real-analysis sequences-and-series limits analysis limsup-and-liminf
$endgroup$
1
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
add a comment |
$begingroup$
Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:
For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$
I think I fully understand the proof by Rudin.
From Theorem 3.37 in Baby Rudin, we can also conclude that following:
For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.
Am I right?
However, I'm unable to figure out the proof of or come up with any counter-examples to the following:
Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.
What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?
real-analysis sequences-and-series limits analysis limsup-and-liminf
$endgroup$
Here is Theorem 3.37 in the book Principles Of Mathematical Analysis by Walter Rudin, 3rd edition:
For any sequence $left{ c_n right}$ of positive numbers,
$$ liminf_{ntoinfty} frac{c_{n+1}}{c_n} leq liminf_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} sqrt[n]{c_n} leq limsup_{ntoinfty} frac{c_{n+1}}{c_n}. $$
I think I fully understand the proof by Rudin.
From Theorem 3.37 in Baby Rudin, we can also conclude that following:
For any sequence $left{ c_n right}$ of positive numbers, if the sequence $left{ frac{c_{n+1}}{c_n} right}$ converges in $mathbb{R}$, then so does the sequence $left{ sqrt[n]{c_n} right}$, and then the two limits are equal.
Am I right?
However, I'm unable to figure out the proof of or come up with any counter-examples to the following:
Suppose that $left{ c_n right}$ is a sequence of positive real numbers such that the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R}$. Then so does the sequence $left{ frac{c_{n+1}}{c_n} right}$.
What if the sequence $left{ sqrt[n]{c_n} right}$ converges in $mathbb{R} cup { pm infty }$? Does the sequence $left{ frac{c_{n+1}}{c_n} right}$ then also converge in $mathbb{R} cup { pm infty }$?
real-analysis sequences-and-series limits analysis limsup-and-liminf
real-analysis sequences-and-series limits analysis limsup-and-liminf
asked Dec 3 '18 at 5:27
Saaqib MahmoodSaaqib Mahmood
7,69242377
7,69242377
1
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
add a comment |
1
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
1
1
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
2
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
$endgroup$
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%2f3023678%2ftheorem-3-37-in-baby-rudin-any-counter-examples-in-the-other-case%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
$begingroup$
Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
$endgroup$
add a comment |
$begingroup$
Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
$endgroup$
add a comment |
$begingroup$
Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
$endgroup$
Concerning the first question: yes, you are right.
On the other hand, consider the sequence$$1,1,frac12,frac12,frac14,frac14,ldots$$In the case, $lim_{ntoinfty}sqrt[n]{c_n}=dfrac1{sqrt2}$, but $lim_{ntoinfty}dfrac{c_{n+1}}{c_n}$ doesn't exist.
answered Dec 3 '18 at 6:57
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
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%2f3023678%2ftheorem-3-37-in-baby-rudin-any-counter-examples-in-the-other-case%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
1
$begingroup$
math.stackexchange.com/a/1708611/43949
$endgroup$
– angryavian
Dec 3 '18 at 5:34
2
$begingroup$
Think as $sqrt[n]{c_n}$ as (almost) the geometric mean of $c_{i+1}/c_i$ for $i=0..n$. Applying the $log$ function, we can think of this additively instead. Then the question is to find a sequence of real numbers not converging such that the Cesàro mean converges…
$endgroup$
– Idéophage
Dec 3 '18 at 5:42