Mixing
Relation between root test and radius of convergence
$begingroup$
We are able to use calculus to compute the radius of convergence of a power series. Now I want to look at a analytical way to that procedure. Say, we have a power series $sum a_nx^n$. What if we should know that $frac{a_{n+1}}{a_n} to z$. What does that mean for the radius of convergence of $sum a_nx^n$. Of course, the radius of convergence should be $frac{1}{z}$. But how can I proof that when $frac{a_{n+1}}{a_n} to z$, the radius of convergence has to be $frac{1}{z}$.
We could use the inequality $ limsup a_n^{1/n} leq limsup frac{a_{n+1}}{a_n}$ to find $limsup a_n^{1/n}$ and compute the radius of convergence. But how can I show that $frac{a_{n+1}}{a_n} to z Rightarrow limsup a_n^{1/n} = z $?
real-analysis convergence
asked Jan 29 at 11:47
PaulPaul
12
$endgroup$
add a comment |
$begingroup$
We are able to use calculus to compute the radius of convergence of a power series. Now I want to look at a analytical way to that procedure. Say, we have a power series $sum a_nx^n$. What if we should know that $frac{a_{n+1}}{a_n} to z$. What does that mean for the radius of convergence of $sum a_nx^n$. Of course, the radius of convergence should be $frac{1}{z}$. But how can I proof that when $frac{a_{n+1}}{a_n} to z$, the radius of convergence has to be $frac{1}{z}$.
We could use the inequality $ limsup a_n^{1/n} leq limsup frac{a_{n+1}}{a_n}$ to find $limsup a_n^{1/n}$ and compute the radius of convergence. But how can I show that $frac{a_{n+1}}{a_n} to z Rightarrow limsup a_n^{1/n} = z $?
real-analysis convergence
asked Jan 29 at 11:47
PaulPaul
12
$endgroup$
add a comment |
$begingroup$
We are able to use calculus to compute the radius of convergence of a power series. Now I want to look at a analytical way to that procedure. Say, we have a power series $sum a_nx^n$. What if we should know that $frac{a_{n+1}}{a_n} to z$. What does that mean for the radius of convergence of $sum a_nx^n$. Of course, the radius of convergence should be $frac{1}{z}$. But how can I proof that when $frac{a_{n+1}}{a_n} to z$, the radius of convergence has to be $frac{1}{z}$.
We could use the inequality $ limsup a_n^{1/n} leq limsup frac{a_{n+1}}{a_n}$ to find $limsup a_n^{1/n}$ and compute the radius of convergence. But how can I show that $frac{a_{n+1}}{a_n} to z Rightarrow limsup a_n^{1/n} = z $?
real-analysis convergence
asked Jan 29 at 11:47
PaulPaul
12
$endgroup$
We are able to use calculus to compute the radius of convergence of a power series. Now I want to look at a analytical way to that procedure. Say, we have a power series $sum a_nx^n$. What if we should know that $frac{a_{n+1}}{a_n} to z$. What does that mean for the radius of convergence of $sum a_nx^n$. Of course, the radius of convergence should be $frac{1}{z}$. But how can I proof that when $frac{a_{n+1}}{a_n} to z$, the radius of convergence has to be $frac{1}{z}$.
We could use the inequality $ limsup a_n^{1/n} leq limsup frac{a_{n+1}}{a_n}$ to find $limsup a_n^{1/n}$ and compute the radius of convergence. But how can I show that $frac{a_{n+1}}{a_n} to z Rightarrow limsup a_n^{1/n} = z $?
real-analysis convergence
real-analysis convergence
asked Jan 29 at 11:47
PaulPaul
12
asked Jan 29 at 11:47
PaulPaul
12
edited Jan 29 at 11:54
Paul
asked Jan 29 at 11:47
PaulPaul
12
asked Jan 29 at 11:47
PaulPaul
12
asked Jan 29 at 11:47
PaulPaul
12
12
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This answer is for an earlier version of the question.
$lim sup a_n^{1/n}=z$ does not imply that $frac {a_{n+1}} {a_n} to 1$. For example, if $a_n=2n$ for $n$ even and $n$ for $n$ odd then $lim sup a_n^{1/n}=1$ but $frac {a_{n+1}} {a_n}$ does not have a limit. Note that ratio test only gives a sufficient condition for convergence and this condition is not necessary.
answered Jan 29 at 11:53
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
$begingroup$
You are right, I will change the last sentence. Excuse me
$endgroup$
– Paul
Jan 29 at 11:53
add a comment |
$begingroup$
We suppose that in $sum a_nx^n$ all $a_n$ are $ ne 0$ and that $frac{a_{n+1}}{a_n} to z$.
For $x ne 0$ let $b_n(x):=a_nx^n$. Then we have
$frac{|b_{n+1}(x)|}{|b_n(x)|}=|x| |frac{a_{n+1}}{a_n}| to |x||z|.$
Case 1: $z=0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}=0<1$ and the ratio-test says: $sum a_nx^n$ converges absolutely for all $x$.
Case 2: $z ne 0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}<1 iff |x|<frac{1}{|z|}$.
The ratio-test gives: $sum a_nx^n$ converges absolutely for all $x$ with $|x|<frac{1}{|z|}$ and $sum a_nx^n$ is divergent for all $x$ with $|x|>frac{1}{|z|}$ .
Conclusion: the radius of convergence is $frac{1}{|z|}$ with $"frac{1}{0}= infty".$
answered Jan 29 at 12:47
FredFred
48.9k11849
$endgroup$
add a comment |
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%2f3092073%2frelation-between-root-test-and-radius-of-convergence%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This answer is for an earlier version of the question.
$lim sup a_n^{1/n}=z$ does not imply that $frac {a_{n+1}} {a_n} to 1$. For example, if $a_n=2n$ for $n$ even and $n$ for $n$ odd then $lim sup a_n^{1/n}=1$ but $frac {a_{n+1}} {a_n}$ does not have a limit. Note that ratio test only gives a sufficient condition for convergence and this condition is not necessary.
answered Jan 29 at 11:53
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
$begingroup$
You are right, I will change the last sentence. Excuse me
$endgroup$
– Paul
Jan 29 at 11:53
add a comment |
$begingroup$
This answer is for an earlier version of the question.
$lim sup a_n^{1/n}=z$ does not imply that $frac {a_{n+1}} {a_n} to 1$. For example, if $a_n=2n$ for $n$ even and $n$ for $n$ odd then $lim sup a_n^{1/n}=1$ but $frac {a_{n+1}} {a_n}$ does not have a limit. Note that ratio test only gives a sufficient condition for convergence and this condition is not necessary.
answered Jan 29 at 11:53
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
$begingroup$
You are right, I will change the last sentence. Excuse me
$endgroup$
– Paul
Jan 29 at 11:53
add a comment |
$begingroup$
This answer is for an earlier version of the question.
$lim sup a_n^{1/n}=z$ does not imply that $frac {a_{n+1}} {a_n} to 1$. For example, if $a_n=2n$ for $n$ even and $n$ for $n$ odd then $lim sup a_n^{1/n}=1$ but $frac {a_{n+1}} {a_n}$ does not have a limit. Note that ratio test only gives a sufficient condition for convergence and this condition is not necessary.
answered Jan 29 at 11:53
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
$endgroup$
This answer is for an earlier version of the question.
$lim sup a_n^{1/n}=z$ does not imply that $frac {a_{n+1}} {a_n} to 1$. For example, if $a_n=2n$ for $n$ even and $n$ for $n$ odd then $lim sup a_n^{1/n}=1$ but $frac {a_{n+1}} {a_n}$ does not have a limit. Note that ratio test only gives a sufficient condition for convergence and this condition is not necessary.
answered Jan 29 at 11:53
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
edited Jan 29 at 23:10
answered Jan 29 at 11:53
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Jan 29 at 11:53
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
answered Jan 29 at 11:53
Kavi Rama MurthyKavi Rama Murthy
71.4k53170
71.4k53170
$begingroup$
You are right, I will change the last sentence. Excuse me
$endgroup$
– Paul
Jan 29 at 11:53
add a comment |
$begingroup$
You are right, I will change the last sentence. Excuse me
$endgroup$
– Paul
Jan 29 at 11:53
$begingroup$
You are right, I will change the last sentence. Excuse me
$endgroup$
– Paul
Jan 29 at 11:53
$begingroup$
You are right, I will change the last sentence. Excuse me
$endgroup$
– Paul
Jan 29 at 11:53
add a comment |
$begingroup$
We suppose that in $sum a_nx^n$ all $a_n$ are $ ne 0$ and that $frac{a_{n+1}}{a_n} to z$.
For $x ne 0$ let $b_n(x):=a_nx^n$. Then we have
$frac{|b_{n+1}(x)|}{|b_n(x)|}=|x| |frac{a_{n+1}}{a_n}| to |x||z|.$
Case 1: $z=0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}=0<1$ and the ratio-test says: $sum a_nx^n$ converges absolutely for all $x$.
Case 2: $z ne 0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}<1 iff |x|<frac{1}{|z|}$.
The ratio-test gives: $sum a_nx^n$ converges absolutely for all $x$ with $|x|<frac{1}{|z|}$ and $sum a_nx^n$ is divergent for all $x$ with $|x|>frac{1}{|z|}$ .
Conclusion: the radius of convergence is $frac{1}{|z|}$ with $"frac{1}{0}= infty".$
answered Jan 29 at 12:47
FredFred
48.9k11849
$endgroup$
add a comment |
$begingroup$
We suppose that in $sum a_nx^n$ all $a_n$ are $ ne 0$ and that $frac{a_{n+1}}{a_n} to z$.
For $x ne 0$ let $b_n(x):=a_nx^n$. Then we have
$frac{|b_{n+1}(x)|}{|b_n(x)|}=|x| |frac{a_{n+1}}{a_n}| to |x||z|.$
Case 1: $z=0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}=0<1$ and the ratio-test says: $sum a_nx^n$ converges absolutely for all $x$.
Case 2: $z ne 0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}<1 iff |x|<frac{1}{|z|}$.
The ratio-test gives: $sum a_nx^n$ converges absolutely for all $x$ with $|x|<frac{1}{|z|}$ and $sum a_nx^n$ is divergent for all $x$ with $|x|>frac{1}{|z|}$ .
Conclusion: the radius of convergence is $frac{1}{|z|}$ with $"frac{1}{0}= infty".$
answered Jan 29 at 12:47
FredFred
48.9k11849
$endgroup$
add a comment |
$begingroup$
We suppose that in $sum a_nx^n$ all $a_n$ are $ ne 0$ and that $frac{a_{n+1}}{a_n} to z$.
For $x ne 0$ let $b_n(x):=a_nx^n$. Then we have
$frac{|b_{n+1}(x)|}{|b_n(x)|}=|x| |frac{a_{n+1}}{a_n}| to |x||z|.$
Case 1: $z=0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}=0<1$ and the ratio-test says: $sum a_nx^n$ converges absolutely for all $x$.
Case 2: $z ne 0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}<1 iff |x|<frac{1}{|z|}$.
The ratio-test gives: $sum a_nx^n$ converges absolutely for all $x$ with $|x|<frac{1}{|z|}$ and $sum a_nx^n$ is divergent for all $x$ with $|x|>frac{1}{|z|}$ .
Conclusion: the radius of convergence is $frac{1}{|z|}$ with $"frac{1}{0}= infty".$
answered Jan 29 at 12:47
FredFred
48.9k11849
$endgroup$
We suppose that in $sum a_nx^n$ all $a_n$ are $ ne 0$ and that $frac{a_{n+1}}{a_n} to z$.
For $x ne 0$ let $b_n(x):=a_nx^n$. Then we have
$frac{|b_{n+1}(x)|}{|b_n(x)|}=|x| |frac{a_{n+1}}{a_n}| to |x||z|.$
Case 1: $z=0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}=0<1$ and the ratio-test says: $sum a_nx^n$ converges absolutely for all $x$.
Case 2: $z ne 0$. Then $lim_{n to infty}frac{|b_{n+1}(x)|}{|b_n(x)|}<1 iff |x|<frac{1}{|z|}$.
The ratio-test gives: $sum a_nx^n$ converges absolutely for all $x$ with $|x|<frac{1}{|z|}$ and $sum a_nx^n$ is divergent for all $x$ with $|x|>frac{1}{|z|}$ .
Conclusion: the radius of convergence is $frac{1}{|z|}$ with $"frac{1}{0}= infty".$
answered Jan 29 at 12:47
FredFred
48.9k11849
answered Jan 29 at 12:47
FredFred
48.9k11849
answered Jan 29 at 12:47
FredFred
48.9k11849
answered Jan 29 at 12:47
FredFred
48.9k11849
48.9k11849
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%2f3092073%2frelation-between-root-test-and-radius-of-convergence%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
