Mixing
If $limlimits_{ntoinfty}sqrt[n]a_n = 1$ and $a_n$ subsequence converges/diverges do we know that $a_n$ in...
We also assume that $(a_n) gt 0$.
What if $a_n$ has a subsequence divergent to infinity to convergent to 0?
My feeling is that if subsequence is divergent, then whole sequence diverges, since we can take $a_n = n$, but I can't think if any way to prove that. And what about subsequence converginig to 0?
Thank you for your help.
sequences-and-series limits
asked Nov 22 '18 at 9:38
JoséJosé
11
add a comment |
We also assume that $(a_n) gt 0$.
What if $a_n$ has a subsequence divergent to infinity to convergent to 0?
My feeling is that if subsequence is divergent, then whole sequence diverges, since we can take $a_n = n$, but I can't think if any way to prove that. And what about subsequence converginig to 0?
Thank you for your help.
sequences-and-series limits
asked Nov 22 '18 at 9:38
JoséJosé
11
1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 '18 at 9:40
2
Your question is really unclear.
– gimusi
Nov 22 '18 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 '18 at 9:49
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 '18 at 9:49
add a comment |
We also assume that $(a_n) gt 0$.
What if $a_n$ has a subsequence divergent to infinity to convergent to 0?
My feeling is that if subsequence is divergent, then whole sequence diverges, since we can take $a_n = n$, but I can't think if any way to prove that. And what about subsequence converginig to 0?
Thank you for your help.
sequences-and-series limits
asked Nov 22 '18 at 9:38
JoséJosé
11
We also assume that $(a_n) gt 0$.
What if $a_n$ has a subsequence divergent to infinity to convergent to 0?
My feeling is that if subsequence is divergent, then whole sequence diverges, since we can take $a_n = n$, but I can't think if any way to prove that. And what about subsequence converginig to 0?
Thank you for your help.
sequences-and-series limits
sequences-and-series limits
asked Nov 22 '18 at 9:38
JoséJosé
11
asked Nov 22 '18 at 9:38
JoséJosé
11
edited Nov 22 '18 at 9:46
José
asked Nov 22 '18 at 9:38
JoséJosé
11
asked Nov 22 '18 at 9:38
JoséJosé
11
asked Nov 22 '18 at 9:38
JoséJosé
11
11
1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 '18 at 9:40
2
Your question is really unclear.
– gimusi
Nov 22 '18 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 '18 at 9:49
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 '18 at 9:49
add a comment |
1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 '18 at 9:40
2
Your question is really unclear.
– gimusi
Nov 22 '18 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 '18 at 9:49
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 '18 at 9:49
1
1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 '18 at 9:40
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 '18 at 9:40
2
2
Your question is really unclear.
– gimusi
Nov 22 '18 at 9:43
Your question is really unclear.
– gimusi
Nov 22 '18 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 '18 at 9:49
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 '18 at 9:49
1
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 '18 at 9:49
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 '18 at 9:49
add a comment |
1 Answer
1
active
oldest
votes
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 '18 at 9:46
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
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%2f3008917%2fif-lim-limits-n-to-infty-sqrtna-n-1-and-a-n-subsequence-converges-div%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
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 '18 at 9:46
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
add a comment |
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 '18 at 9:46
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
add a comment |
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 '18 at 9:46
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 '18 at 9:46
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 22 '18 at 9:46
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 22 '18 at 9:46
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 22 '18 at 9:46
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
52.1k32055
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3008917%2fif-lim-limits-n-to-infty-sqrtna-n-1-and-a-n-subsequence-converges-div%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
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 '18 at 9:40
2
Your question is really unclear.
– gimusi
Nov 22 '18 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 '18 at 9:49
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 '18 at 9:49