Subsequential lims of $x_n=1,{1over2},{2over2},{3over2},{1over4},{4over4},…,{9over...
$begingroup$
Find all subsequential limits of:
$$
x_n=left{1,{1over2},{2over2},{3over2},{1over4},{4over4},dots,{9over 4},dots,{1over2^n},{2over2^n},dots,{3^nover2^n},dotsright}
$$
I've been recently solving a similar problem. After that I've been considerting the following sequence:
$$
y_n = left{1, {1over 2}, {2over2},{3over2},{1over 4},{2over 4},{3over4},{4over4},{5over 4},dots,{1over 2^n},dots,frac{2^n - 1}{2^n},dots right}
$$
Using a similar technique from the linked answer one may introduce a sequence:
$$
r_n = frac{lfloor r2^nrfloor}{2^n}
$$
And then:
$$
r2^n - 1 < lfloor r2^nrfloor le r2^n
$$
Dividing both parts by $2^n$:
$$
r - {1over 2^n} < r_n le r
$$
Which by squeeze theorem gives that:
$$
lim_{ntoinfty}left(r - {1over 2^n}right) < lim_{ntoinfty}r_n le lim_{ntoinfty} r = r
$$
Now if we take $rin [0, 1]$ then $r_n$ should be somewhere in the original $y_n$. Thus $r$ is a subsequentilal limit. Hence the whole set of reals $[0, 1]$ is the set subsequential limits.
If i understand this correctly $y_n$ is nothing but splitting the range $[0, 1]$ into infinitely many nested intervals. Which eventually end up in a single point belonging to each interval. And we can choose any point within that interval appearing to a subsequential limit.
However I'm not sure how to apply this reasoning to $x_n$. Intuitively it seems like $[0, +infty]$ is the set of subsequential limits, but I'm having a hard time proving this.
I'm kindly asking to help me out with that.
calculus sequences-and-series limits
asked Dec 13 '18 at 17:28
romanroman
2,21421224
$endgroup$
add a comment |
$begingroup$
Find all subsequential limits of:
$$
x_n=left{1,{1over2},{2over2},{3over2},{1over4},{4over4},dots,{9over 4},dots,{1over2^n},{2over2^n},dots,{3^nover2^n},dotsright}
$$
I've been recently solving a similar problem. After that I've been considerting the following sequence:
$$
y_n = left{1, {1over 2}, {2over2},{3over2},{1over 4},{2over 4},{3over4},{4over4},{5over 4},dots,{1over 2^n},dots,frac{2^n - 1}{2^n},dots right}
$$
Using a similar technique from the linked answer one may introduce a sequence:
$$
r_n = frac{lfloor r2^nrfloor}{2^n}
$$
And then:
$$
r2^n - 1 < lfloor r2^nrfloor le r2^n
$$
Dividing both parts by $2^n$:
$$
r - {1over 2^n} < r_n le r
$$
Which by squeeze theorem gives that:
$$
lim_{ntoinfty}left(r - {1over 2^n}right) < lim_{ntoinfty}r_n le lim_{ntoinfty} r = r
$$
Now if we take $rin [0, 1]$ then $r_n$ should be somewhere in the original $y_n$. Thus $r$ is a subsequentilal limit. Hence the whole set of reals $[0, 1]$ is the set subsequential limits.
If i understand this correctly $y_n$ is nothing but splitting the range $[0, 1]$ into infinitely many nested intervals. Which eventually end up in a single point belonging to each interval. And we can choose any point within that interval appearing to a subsequential limit.
However I'm not sure how to apply this reasoning to $x_n$. Intuitively it seems like $[0, +infty]$ is the set of subsequential limits, but I'm having a hard time proving this.
I'm kindly asking to help me out with that.
calculus sequences-and-series limits
asked Dec 13 '18 at 17:28
romanroman
2,21421224
$endgroup$
$begingroup$
If you could prove that your set is dense in $mathbb{R_+}+{infty}$, then your claim follows. Also, following the same approach why not consider $dfrac{lfloor ra^nrfloor}{a^n}$ where $a = frac 32 ?$
$endgroup$
– dezdichado
Dec 13 '18 at 18:24
$begingroup$
For a given $r$ and integer $N$ then there is an integer $M$ such that $|r - frac{M}{2^N}| < frac{1}{2^N}$. If $N$ is large enough then $M < 3^N$. Prove this (start from $0$ and go towards $r$ with steps of size $1/2^N$). Use this to construct a subsequence that converges to $r$.
$endgroup$
– Winther
Dec 13 '18 at 18:30
add a comment |
$begingroup$
Find all subsequential limits of:
$$
x_n=left{1,{1over2},{2over2},{3over2},{1over4},{4over4},dots,{9over 4},dots,{1over2^n},{2over2^n},dots,{3^nover2^n},dotsright}
$$
I've been recently solving a similar problem. After that I've been considerting the following sequence:
$$
y_n = left{1, {1over 2}, {2over2},{3over2},{1over 4},{2over 4},{3over4},{4over4},{5over 4},dots,{1over 2^n},dots,frac{2^n - 1}{2^n},dots right}
$$
Using a similar technique from the linked answer one may introduce a sequence:
$$
r_n = frac{lfloor r2^nrfloor}{2^n}
$$
And then:
$$
r2^n - 1 < lfloor r2^nrfloor le r2^n
$$
Dividing both parts by $2^n$:
$$
r - {1over 2^n} < r_n le r
$$
Which by squeeze theorem gives that:
$$
lim_{ntoinfty}left(r - {1over 2^n}right) < lim_{ntoinfty}r_n le lim_{ntoinfty} r = r
$$
Now if we take $rin [0, 1]$ then $r_n$ should be somewhere in the original $y_n$. Thus $r$ is a subsequentilal limit. Hence the whole set of reals $[0, 1]$ is the set subsequential limits.
If i understand this correctly $y_n$ is nothing but splitting the range $[0, 1]$ into infinitely many nested intervals. Which eventually end up in a single point belonging to each interval. And we can choose any point within that interval appearing to a subsequential limit.
However I'm not sure how to apply this reasoning to $x_n$. Intuitively it seems like $[0, +infty]$ is the set of subsequential limits, but I'm having a hard time proving this.
I'm kindly asking to help me out with that.
calculus sequences-and-series limits
asked Dec 13 '18 at 17:28
romanroman
2,21421224
$endgroup$
Find all subsequential limits of:
$$
x_n=left{1,{1over2},{2over2},{3over2},{1over4},{4over4},dots,{9over 4},dots,{1over2^n},{2over2^n},dots,{3^nover2^n},dotsright}
$$
I've been recently solving a similar problem. After that I've been considerting the following sequence:
$$
y_n = left{1, {1over 2}, {2over2},{3over2},{1over 4},{2over 4},{3over4},{4over4},{5over 4},dots,{1over 2^n},dots,frac{2^n - 1}{2^n},dots right}
$$
Using a similar technique from the linked answer one may introduce a sequence:
$$
r_n = frac{lfloor r2^nrfloor}{2^n}
$$
And then:
$$
r2^n - 1 < lfloor r2^nrfloor le r2^n
$$
Dividing both parts by $2^n$:
$$
r - {1over 2^n} < r_n le r
$$
Which by squeeze theorem gives that:
$$
lim_{ntoinfty}left(r - {1over 2^n}right) < lim_{ntoinfty}r_n le lim_{ntoinfty} r = r
$$
Now if we take $rin [0, 1]$ then $r_n$ should be somewhere in the original $y_n$. Thus $r$ is a subsequentilal limit. Hence the whole set of reals $[0, 1]$ is the set subsequential limits.
If i understand this correctly $y_n$ is nothing but splitting the range $[0, 1]$ into infinitely many nested intervals. Which eventually end up in a single point belonging to each interval. And we can choose any point within that interval appearing to a subsequential limit.
However I'm not sure how to apply this reasoning to $x_n$. Intuitively it seems like $[0, +infty]$ is the set of subsequential limits, but I'm having a hard time proving this.
I'm kindly asking to help me out with that.
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Dec 13 '18 at 17:28
romanroman
2,21421224
asked Dec 13 '18 at 17:28
romanroman
2,21421224
asked Dec 13 '18 at 17:28
romanroman
2,21421224
asked Dec 13 '18 at 17:28
romanroman
2,21421224
asked Dec 13 '18 at 17:28
romanroman
2,21421224
2,21421224
$begingroup$
If you could prove that your set is dense in $mathbb{R_+}+{infty}$, then your claim follows. Also, following the same approach why not consider $dfrac{lfloor ra^nrfloor}{a^n}$ where $a = frac 32 ?$
$endgroup$
– dezdichado
Dec 13 '18 at 18:24
$begingroup$
For a given $r$ and integer $N$ then there is an integer $M$ such that $|r - frac{M}{2^N}| < frac{1}{2^N}$. If $N$ is large enough then $M < 3^N$. Prove this (start from $0$ and go towards $r$ with steps of size $1/2^N$). Use this to construct a subsequence that converges to $r$.
$endgroup$
– Winther
Dec 13 '18 at 18:30
add a comment |
$begingroup$
If you could prove that your set is dense in $mathbb{R_+}+{infty}$, then your claim follows. Also, following the same approach why not consider $dfrac{lfloor ra^nrfloor}{a^n}$ where $a = frac 32 ?$
$endgroup$
– dezdichado
Dec 13 '18 at 18:24
$begingroup$
For a given $r$ and integer $N$ then there is an integer $M$ such that $|r - frac{M}{2^N}| < frac{1}{2^N}$. If $N$ is large enough then $M < 3^N$. Prove this (start from $0$ and go towards $r$ with steps of size $1/2^N$). Use this to construct a subsequence that converges to $r$.
$endgroup$
– Winther
Dec 13 '18 at 18:30
$begingroup$
If you could prove that your set is dense in $mathbb{R_+}+{infty}$, then your claim follows. Also, following the same approach why not consider $dfrac{lfloor ra^nrfloor}{a^n}$ where $a = frac 32 ?$
$endgroup$
– dezdichado
Dec 13 '18 at 18:24
$begingroup$
If you could prove that your set is dense in $mathbb{R_+}+{infty}$, then your claim follows. Also, following the same approach why not consider $dfrac{lfloor ra^nrfloor}{a^n}$ where $a = frac 32 ?$
$endgroup$
– dezdichado
Dec 13 '18 at 18:24
$begingroup$
For a given $r$ and integer $N$ then there is an integer $M$ such that $|r - frac{M}{2^N}| < frac{1}{2^N}$. If $N$ is large enough then $M < 3^N$. Prove this (start from $0$ and go towards $r$ with steps of size $1/2^N$). Use this to construct a subsequence that converges to $r$.
$endgroup$
– Winther
Dec 13 '18 at 18:30
$begingroup$
For a given $r$ and integer $N$ then there is an integer $M$ such that $|r - frac{M}{2^N}| < frac{1}{2^N}$. If $N$ is large enough then $M < 3^N$. Prove this (start from $0$ and go towards $r$ with steps of size $1/2^N$). Use this to construct a subsequence that converges to $r$.
$endgroup$
– Winther
Dec 13 '18 at 18:30
add a comment |
0
active
oldest
votes
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%2f3038340%2fsubsequential-lims-of-x-n-1-1-over2-2-over2-3-over2-1-over4-4-over4%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3038340%2fsubsequential-lims-of-x-n-1-1-over2-2-over2-3-over2-1-over4-4-over4%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 you could prove that your set is dense in $mathbb{R_+}+{infty}$, then your claim follows. Also, following the same approach why not consider $dfrac{lfloor ra^nrfloor}{a^n}$ where $a = frac 32 ?$
$endgroup$
– dezdichado
Dec 13 '18 at 18:24
$begingroup$
For a given $r$ and integer $N$ then there is an integer $M$ such that $|r - frac{M}{2^N}| < frac{1}{2^N}$. If $N$ is large enough then $M < 3^N$. Prove this (start from $0$ and go towards $r$ with steps of size $1/2^N$). Use this to construct a subsequence that converges to $r$.
$endgroup$
– Winther
Dec 13 '18 at 18:30