Mixing
If the radius of convergence of $sum^{infty}_{n=1}a_n x^n$ is $R$, what is the radius of convergence of...
$begingroup$
If the radius of convergence of $sum^{infty}_{n=1}a_n x^n$ is $R$, what is the radius of convergence of $sum^{infty}_{n=1}a_n x^{n!}$.
My trial
begin{align} limsuplimits_{nto infty} sqrt[n]{|a_n x^{n!}|}&=limsuplimits_{nto infty} sqrt[n]{|a_n |}cdot limlimits_{nto infty} sqrt[n]{|x|^{n!}}\&=dfrac{1}{R}cdot limlimits_{nto infty} sqrt[n]{|x|^{n!}}< 1end{align}
This implies that
begin{align} limlimits_{nto infty} sqrt[n]{|x|^{n!}}< R&implies ;{|x|^{(n-1)!}}< R,;;forall;ngeq N,;;text{for some};N,\&implies ;{|x|}< R^{1/(n-1)!},;;forall;ngeq N,;;text{for some};N,\&implies ;{|x|}leq 1,;;text{as};nto infty.end{align}
I'm I correct? If yes, can I say that the radius of convergence $1?$. If no, can anyone help out or provide a hint?
real-analysis sequences-and-series analysis
asked Jan 25 at 18:54
Omojola MichealOmojola Micheal
1,986324
$endgroup$
add a comment |
$begingroup$
If the radius of convergence of $sum^{infty}_{n=1}a_n x^n$ is $R$, what is the radius of convergence of $sum^{infty}_{n=1}a_n x^{n!}$.
My trial
begin{align} limsuplimits_{nto infty} sqrt[n]{|a_n x^{n!}|}&=limsuplimits_{nto infty} sqrt[n]{|a_n |}cdot limlimits_{nto infty} sqrt[n]{|x|^{n!}}\&=dfrac{1}{R}cdot limlimits_{nto infty} sqrt[n]{|x|^{n!}}< 1end{align}
This implies that
begin{align} limlimits_{nto infty} sqrt[n]{|x|^{n!}}< R&implies ;{|x|^{(n-1)!}}< R,;;forall;ngeq N,;;text{for some};N,\&implies ;{|x|}< R^{1/(n-1)!},;;forall;ngeq N,;;text{for some};N,\&implies ;{|x|}leq 1,;;text{as};nto infty.end{align}
I'm I correct? If yes, can I say that the radius of convergence $1?$. If no, can anyone help out or provide a hint?
real-analysis sequences-and-series analysis
asked Jan 25 at 18:54
Omojola MichealOmojola Micheal
1,986324
$endgroup$
$begingroup$
If your ultimate conclusion is $1$, then you are correct, but the approach is shaky. Radius of convergence is a constant that does not depend on $n$.
$endgroup$
– RRL
Jan 25 at 22:11
add a comment |
$begingroup$
If the radius of convergence of $sum^{infty}_{n=1}a_n x^n$ is $R$, what is the radius of convergence of $sum^{infty}_{n=1}a_n x^{n!}$.
My trial
begin{align} limsuplimits_{nto infty} sqrt[n]{|a_n x^{n!}|}&=limsuplimits_{nto infty} sqrt[n]{|a_n |}cdot limlimits_{nto infty} sqrt[n]{|x|^{n!}}\&=dfrac{1}{R}cdot limlimits_{nto infty} sqrt[n]{|x|^{n!}}< 1end{align}
This implies that
begin{align} limlimits_{nto infty} sqrt[n]{|x|^{n!}}< R&implies ;{|x|^{(n-1)!}}< R,;;forall;ngeq N,;;text{for some};N,\&implies ;{|x|}< R^{1/(n-1)!},;;forall;ngeq N,;;text{for some};N,\&implies ;{|x|}leq 1,;;text{as};nto infty.end{align}
I'm I correct? If yes, can I say that the radius of convergence $1?$. If no, can anyone help out or provide a hint?
real-analysis sequences-and-series analysis
asked Jan 25 at 18:54
Omojola MichealOmojola Micheal
1,986324
$endgroup$
If the radius of convergence of $sum^{infty}_{n=1}a_n x^n$ is $R$, what is the radius of convergence of $sum^{infty}_{n=1}a_n x^{n!}$.
My trial
begin{align} limsuplimits_{nto infty} sqrt[n]{|a_n x^{n!}|}&=limsuplimits_{nto infty} sqrt[n]{|a_n |}cdot limlimits_{nto infty} sqrt[n]{|x|^{n!}}\&=dfrac{1}{R}cdot limlimits_{nto infty} sqrt[n]{|x|^{n!}}< 1end{align}
This implies that
begin{align} limlimits_{nto infty} sqrt[n]{|x|^{n!}}< R&implies ;{|x|^{(n-1)!}}< R,;;forall;ngeq N,;;text{for some};N,\&implies ;{|x|}< R^{1/(n-1)!},;;forall;ngeq N,;;text{for some};N,\&implies ;{|x|}leq 1,;;text{as};nto infty.end{align}
I'm I correct? If yes, can I say that the radius of convergence $1?$. If no, can anyone help out or provide a hint?
real-analysis sequences-and-series analysis
real-analysis sequences-and-series analysis
asked Jan 25 at 18:54
Omojola MichealOmojola Micheal
1,986324
asked Jan 25 at 18:54
Omojola MichealOmojola Micheal
1,986324
edited Jan 26 at 0:09
Omojola Micheal
asked Jan 25 at 18:54
Omojola MichealOmojola Micheal
1,986324
asked Jan 25 at 18:54
Omojola MichealOmojola Micheal
1,986324
asked Jan 25 at 18:54
Omojola MichealOmojola Micheal
1,986324
1,986324
$begingroup$
If your ultimate conclusion is $1$, then you are correct, but the approach is shaky. Radius of convergence is a constant that does not depend on $n$.
$endgroup$
– RRL
Jan 25 at 22:11
add a comment |
$begingroup$
If your ultimate conclusion is $1$, then you are correct, but the approach is shaky. Radius of convergence is a constant that does not depend on $n$.
$endgroup$
– RRL
Jan 25 at 22:11
$begingroup$
If your ultimate conclusion is $1$, then you are correct, but the approach is shaky. Radius of convergence is a constant that does not depend on $n$.
$endgroup$
– RRL
Jan 25 at 22:11
$begingroup$
If your ultimate conclusion is $1$, then you are correct, but the approach is shaky. Radius of convergence is a constant that does not depend on $n$.
$endgroup$
– RRL
Jan 25 at 22:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $displaystylesum_{n=1}^infty a_n x^{n!} = sum_{k=1}^infty b_k x^k$ where $displaystyle b_k = begin{cases} a_n, & k = n!\ 0, & k neq n! end{cases}$
Now you need to consider
$$limsup_{k to infty}, |b_kx^k|^{1/k} = limsup_{n to infty} |a_n|^{1/n!} |x| < 1,$$
which implies that the radius of convergence is $left(limsup_{n to infty} |a_n|^{1/n!}right)^{-1}$.
If $|a_n|^{1/n} to R^{-1}$ then to what does $|a_n|^{1/n!} = (|a_n|^{1/n})^{1/(n-1)!}$ converge ?
answered Jan 25 at 19:23
RRLRRL
52.8k42573
$endgroup$
$begingroup$
Yes, that true! (+1)
$endgroup$
– Omojola Micheal
Jan 26 at 0:07
$begingroup$
@RRL: So, what happens when $R=0,infty$?
$endgroup$
– Micheal
Jan 30 at 7:39
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%2f3087486%2fif-the-radius-of-convergence-of-sum-infty-n-1a-n-xn-is-r-what-is-the%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
$begingroup$
Note that $displaystylesum_{n=1}^infty a_n x^{n!} = sum_{k=1}^infty b_k x^k$ where $displaystyle b_k = begin{cases} a_n, & k = n!\ 0, & k neq n! end{cases}$
Now you need to consider
$$limsup_{k to infty}, |b_kx^k|^{1/k} = limsup_{n to infty} |a_n|^{1/n!} |x| < 1,$$
which implies that the radius of convergence is $left(limsup_{n to infty} |a_n|^{1/n!}right)^{-1}$.
If $|a_n|^{1/n} to R^{-1}$ then to what does $|a_n|^{1/n!} = (|a_n|^{1/n})^{1/(n-1)!}$ converge ?
answered Jan 25 at 19:23
RRLRRL
52.8k42573
$endgroup$
$begingroup$
Yes, that true! (+1)
$endgroup$
– Omojola Micheal
Jan 26 at 0:07
$begingroup$
@RRL: So, what happens when $R=0,infty$?
$endgroup$
– Micheal
Jan 30 at 7:39
add a comment |
$begingroup$
Note that $displaystylesum_{n=1}^infty a_n x^{n!} = sum_{k=1}^infty b_k x^k$ where $displaystyle b_k = begin{cases} a_n, & k = n!\ 0, & k neq n! end{cases}$
Now you need to consider
$$limsup_{k to infty}, |b_kx^k|^{1/k} = limsup_{n to infty} |a_n|^{1/n!} |x| < 1,$$
which implies that the radius of convergence is $left(limsup_{n to infty} |a_n|^{1/n!}right)^{-1}$.
If $|a_n|^{1/n} to R^{-1}$ then to what does $|a_n|^{1/n!} = (|a_n|^{1/n})^{1/(n-1)!}$ converge ?
answered Jan 25 at 19:23
RRLRRL
52.8k42573
$endgroup$
$begingroup$
Yes, that true! (+1)
$endgroup$
– Omojola Micheal
Jan 26 at 0:07
$begingroup$
@RRL: So, what happens when $R=0,infty$?
$endgroup$
– Micheal
Jan 30 at 7:39
add a comment |
$begingroup$
Note that $displaystylesum_{n=1}^infty a_n x^{n!} = sum_{k=1}^infty b_k x^k$ where $displaystyle b_k = begin{cases} a_n, & k = n!\ 0, & k neq n! end{cases}$
Now you need to consider
$$limsup_{k to infty}, |b_kx^k|^{1/k} = limsup_{n to infty} |a_n|^{1/n!} |x| < 1,$$
which implies that the radius of convergence is $left(limsup_{n to infty} |a_n|^{1/n!}right)^{-1}$.
If $|a_n|^{1/n} to R^{-1}$ then to what does $|a_n|^{1/n!} = (|a_n|^{1/n})^{1/(n-1)!}$ converge ?
answered Jan 25 at 19:23
RRLRRL
52.8k42573
$endgroup$
Note that $displaystylesum_{n=1}^infty a_n x^{n!} = sum_{k=1}^infty b_k x^k$ where $displaystyle b_k = begin{cases} a_n, & k = n!\ 0, & k neq n! end{cases}$
Now you need to consider
$$limsup_{k to infty}, |b_kx^k|^{1/k} = limsup_{n to infty} |a_n|^{1/n!} |x| < 1,$$
which implies that the radius of convergence is $left(limsup_{n to infty} |a_n|^{1/n!}right)^{-1}$.
If $|a_n|^{1/n} to R^{-1}$ then to what does $|a_n|^{1/n!} = (|a_n|^{1/n})^{1/(n-1)!}$ converge ?
answered Jan 25 at 19:23
RRLRRL
52.8k42573
edited Jan 25 at 22:08
answered Jan 25 at 19:23
RRLRRL
52.8k42573
answered Jan 25 at 19:23
RRLRRL
52.8k42573
answered Jan 25 at 19:23
RRLRRL
52.8k42573
52.8k42573
$begingroup$
Yes, that true! (+1)
$endgroup$
– Omojola Micheal
Jan 26 at 0:07
$begingroup$
@RRL: So, what happens when $R=0,infty$?
$endgroup$
– Micheal
Jan 30 at 7:39
add a comment |
$begingroup$
Yes, that true! (+1)
$endgroup$
– Omojola Micheal
Jan 26 at 0:07
$begingroup$
@RRL: So, what happens when $R=0,infty$?
$endgroup$
– Micheal
Jan 30 at 7:39
$begingroup$
Yes, that true! (+1)
$endgroup$
– Omojola Micheal
Jan 26 at 0:07
$begingroup$
Yes, that true! (+1)
$endgroup$
– Omojola Micheal
Jan 26 at 0:07
$begingroup$
@RRL: So, what happens when $R=0,infty$?
$endgroup$
– Micheal
Jan 30 at 7:39
$begingroup$
@RRL: So, what happens when $R=0,infty$?
$endgroup$
– Micheal
Jan 30 at 7:39
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%2f3087486%2fif-the-radius-of-convergence-of-sum-infty-n-1a-n-xn-is-r-what-is-the%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
$begingroup$
If your ultimate conclusion is $1$, then you are correct, but the approach is shaky. Radius of convergence is a constant that does not depend on $n$.
$endgroup$
– RRL
Jan 25 at 22:11