Mixing
Radius of convergence of $sum_{n geq 0}a_n z^{n!}$ given that of $sum_{n geq 0}a_n z^{n}$
$begingroup$
What would be the radius of convergence of $sum_{n geq 0}a_n z^{n!}$ given that the radius of convergence of $sum_{n geq 0}a_n z^{n}$ is $L$.
real-analysis sequences-and-series complex-analysis power-series
$endgroup$
add a comment |
$begingroup$
What would be the radius of convergence of $sum_{n geq 0}a_n z^{n!}$ given that the radius of convergence of $sum_{n geq 0}a_n z^{n}$ is $L$.
real-analysis sequences-and-series complex-analysis power-series
$endgroup$
add a comment |
$begingroup$
What would be the radius of convergence of $sum_{n geq 0}a_n z^{n!}$ given that the radius of convergence of $sum_{n geq 0}a_n z^{n}$ is $L$.
real-analysis sequences-and-series complex-analysis power-series
$endgroup$
What would be the radius of convergence of $sum_{n geq 0}a_n z^{n!}$ given that the radius of convergence of $sum_{n geq 0}a_n z^{n}$ is $L$.
real-analysis sequences-and-series complex-analysis power-series
real-analysis sequences-and-series complex-analysis power-series
edited Jan 29 at 4:55
asked Jan 29 at 4:42
user397197
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: let $b_n=a_k$ if $n=k!$ and $0$ if $n$ is not a factorial. We have to find the radius of convergence of $sum b_nz^{n}$. Use root test. Can you see that the radius of convergence is $1$ provided $0<L<infty$?.
answered Jan 29 at 5:59
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Well, not quite irrespective. Certain possibilities open if $L=0$ or $L=infty$. Short of that, you are right.
$endgroup$
– Ivan Neretin
Jan 29 at 6:14
$begingroup$
@IvanNeretin You are right.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 6:17
$begingroup$
@KaviRamaMurthy what of the cases where $L=0$ or $L=infty$
$endgroup$
– user397197
Jan 29 at 6:41
$begingroup$
@stackuser Noting can be said about the cases $L=0$ and $L=infty$. The radius of convergence of $sum a_n z^{n!}$ can be anything.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 7:27
1
$begingroup$
Try $dfrac1{2^{n!}}$.
$endgroup$
– Ivan Neretin
Jan 29 at 15:23
|
show 6 more comments
$begingroup$
Using the Cauchy-Hadamard theorem:
$r=frac1{limsup_{ntoinfty}sqrt[n!]{a_n}}=frac1{limsup_{ntoinfty}sqrt[(n-1)!]L}=1$, for $Lneq0,infty$.
answered Jan 29 at 6:26
Chris CusterChris Custer
14.2k3827
$endgroup$
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%2f3091751%2fradius-of-convergence-of-sum-n-geq-0a-n-zn-given-that-of-sum-n-geq%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
$begingroup$
Hint: let $b_n=a_k$ if $n=k!$ and $0$ if $n$ is not a factorial. We have to find the radius of convergence of $sum b_nz^{n}$. Use root test. Can you see that the radius of convergence is $1$ provided $0<L<infty$?.
answered Jan 29 at 5:59
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Well, not quite irrespective. Certain possibilities open if $L=0$ or $L=infty$. Short of that, you are right.
$endgroup$
– Ivan Neretin
Jan 29 at 6:14
$begingroup$
@IvanNeretin You are right.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 6:17
$begingroup$
@KaviRamaMurthy what of the cases where $L=0$ or $L=infty$
$endgroup$
– user397197
Jan 29 at 6:41
$begingroup$
@stackuser Noting can be said about the cases $L=0$ and $L=infty$. The radius of convergence of $sum a_n z^{n!}$ can be anything.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 7:27
1
$begingroup$
Try $dfrac1{2^{n!}}$.
$endgroup$
– Ivan Neretin
Jan 29 at 15:23
|
show 6 more comments
$begingroup$
Hint: let $b_n=a_k$ if $n=k!$ and $0$ if $n$ is not a factorial. We have to find the radius of convergence of $sum b_nz^{n}$. Use root test. Can you see that the radius of convergence is $1$ provided $0<L<infty$?.
answered Jan 29 at 5:59
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
$begingroup$
Well, not quite irrespective. Certain possibilities open if $L=0$ or $L=infty$. Short of that, you are right.
$endgroup$
– Ivan Neretin
Jan 29 at 6:14
$begingroup$
@IvanNeretin You are right.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 6:17
$begingroup$
@KaviRamaMurthy what of the cases where $L=0$ or $L=infty$
$endgroup$
– user397197
Jan 29 at 6:41
$begingroup$
@stackuser Noting can be said about the cases $L=0$ and $L=infty$. The radius of convergence of $sum a_n z^{n!}$ can be anything.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 7:27
1
$begingroup$
Try $dfrac1{2^{n!}}$.
$endgroup$
– Ivan Neretin
Jan 29 at 15:23
|
show 6 more comments
$begingroup$
Hint: let $b_n=a_k$ if $n=k!$ and $0$ if $n$ is not a factorial. We have to find the radius of convergence of $sum b_nz^{n}$. Use root test. Can you see that the radius of convergence is $1$ provided $0<L<infty$?.
answered Jan 29 at 5:59
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
$endgroup$
Hint: let $b_n=a_k$ if $n=k!$ and $0$ if $n$ is not a factorial. We have to find the radius of convergence of $sum b_nz^{n}$. Use root test. Can you see that the radius of convergence is $1$ provided $0<L<infty$?.
answered Jan 29 at 5:59
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
edited Jan 29 at 6:16
answered Jan 29 at 5:59
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 29 at 5:59
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
answered Jan 29 at 5:59
Kavi Rama MurthyKavi Rama Murthy
71.2k53170
71.2k53170
$begingroup$
Well, not quite irrespective. Certain possibilities open if $L=0$ or $L=infty$. Short of that, you are right.
$endgroup$
– Ivan Neretin
Jan 29 at 6:14
$begingroup$
@IvanNeretin You are right.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 6:17
$begingroup$
@KaviRamaMurthy what of the cases where $L=0$ or $L=infty$
$endgroup$
– user397197
Jan 29 at 6:41
$begingroup$
@stackuser Noting can be said about the cases $L=0$ and $L=infty$. The radius of convergence of $sum a_n z^{n!}$ can be anything.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 7:27
1
$begingroup$
Try $dfrac1{2^{n!}}$.
$endgroup$
– Ivan Neretin
Jan 29 at 15:23
|
show 6 more comments
$begingroup$
Well, not quite irrespective. Certain possibilities open if $L=0$ or $L=infty$. Short of that, you are right.
$endgroup$
– Ivan Neretin
Jan 29 at 6:14
$begingroup$
@IvanNeretin You are right.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 6:17
$begingroup$
@KaviRamaMurthy what of the cases where $L=0$ or $L=infty$
$endgroup$
– user397197
Jan 29 at 6:41
$begingroup$
@stackuser Noting can be said about the cases $L=0$ and $L=infty$. The radius of convergence of $sum a_n z^{n!}$ can be anything.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 7:27
1
$begingroup$
Try $dfrac1{2^{n!}}$.
$endgroup$
– Ivan Neretin
Jan 29 at 15:23
$begingroup$
Well, not quite irrespective. Certain possibilities open if $L=0$ or $L=infty$. Short of that, you are right.
$endgroup$
– Ivan Neretin
Jan 29 at 6:14
$begingroup$
Well, not quite irrespective. Certain possibilities open if $L=0$ or $L=infty$. Short of that, you are right.
$endgroup$
– Ivan Neretin
Jan 29 at 6:14
$begingroup$
@IvanNeretin You are right.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 6:17
$begingroup$
@IvanNeretin You are right.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 6:17
$begingroup$
@KaviRamaMurthy what of the cases where $L=0$ or $L=infty$
$endgroup$
– user397197
Jan 29 at 6:41
$begingroup$
@KaviRamaMurthy what of the cases where $L=0$ or $L=infty$
$endgroup$
– user397197
Jan 29 at 6:41
$begingroup$
@stackuser Noting can be said about the cases $L=0$ and $L=infty$. The radius of convergence of $sum a_n z^{n!}$ can be anything.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 7:27
$begingroup$
@stackuser Noting can be said about the cases $L=0$ and $L=infty$. The radius of convergence of $sum a_n z^{n!}$ can be anything.
$endgroup$
– Kavi Rama Murthy
Jan 29 at 7:27
1
1
$begingroup$
Try $dfrac1{2^{n!}}$.
$endgroup$
– Ivan Neretin
Jan 29 at 15:23
$begingroup$
Try $dfrac1{2^{n!}}$.
$endgroup$
– Ivan Neretin
Jan 29 at 15:23
|
show 6 more comments
$begingroup$
Using the Cauchy-Hadamard theorem:
$r=frac1{limsup_{ntoinfty}sqrt[n!]{a_n}}=frac1{limsup_{ntoinfty}sqrt[(n-1)!]L}=1$, for $Lneq0,infty$.
answered Jan 29 at 6:26
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Using the Cauchy-Hadamard theorem:
$r=frac1{limsup_{ntoinfty}sqrt[n!]{a_n}}=frac1{limsup_{ntoinfty}sqrt[(n-1)!]L}=1$, for $Lneq0,infty$.
answered Jan 29 at 6:26
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
Using the Cauchy-Hadamard theorem:
$r=frac1{limsup_{ntoinfty}sqrt[n!]{a_n}}=frac1{limsup_{ntoinfty}sqrt[(n-1)!]L}=1$, for $Lneq0,infty$.
answered Jan 29 at 6:26
Chris CusterChris Custer
14.2k3827
$endgroup$
Using the Cauchy-Hadamard theorem:
$r=frac1{limsup_{ntoinfty}sqrt[n!]{a_n}}=frac1{limsup_{ntoinfty}sqrt[(n-1)!]L}=1$, for $Lneq0,infty$.
answered Jan 29 at 6:26
Chris CusterChris Custer
14.2k3827
edited Jan 29 at 6:42
answered Jan 29 at 6:26
Chris CusterChris Custer
14.2k3827
answered Jan 29 at 6:26
Chris CusterChris Custer
14.2k3827
answered Jan 29 at 6:26
Chris CusterChris Custer
14.2k3827
14.2k3827
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%2f3091751%2fradius-of-convergence-of-sum-n-geq-0a-n-zn-given-that-of-sum-n-geq%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
