If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$...
$begingroup$
If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$ converges.
My attempt:
I think it isn't enough to $a_n$ converges.
I know $lim limits_{n to infty }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n} $
But I don't know how how to finish this proof. Could someone give me some hints?
real-analysis sequences-and-series limits-without-lhopital
$endgroup$
add a comment |
$begingroup$
If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$ converges.
My attempt:
I think it isn't enough to $a_n$ converges.
I know $lim limits_{n to infty }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n} $
But I don't know how how to finish this proof. Could someone give me some hints?
real-analysis sequences-and-series limits-without-lhopital
$endgroup$
add a comment |
$begingroup$
If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$ converges.
My attempt:
I think it isn't enough to $a_n$ converges.
I know $lim limits_{n to infty }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n} $
But I don't know how how to finish this proof. Could someone give me some hints?
real-analysis sequences-and-series limits-without-lhopital
$endgroup$
If $lim limits_{n to infty } (a_{n+1}-a_n)=0$ and $(a_{4n})_{nge1}$ converges decide whether or not $(a_n)$ converges.
My attempt:
I think it isn't enough to $a_n$ converges.
I know $lim limits_{n to infty }( a_{4n+1}-a_{4n})=0$, and because $(a_{4n+1}−a_{4n})$ is a subsequence of $(a_{n+1}-a_n)$ it also converges to 0, so $lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n} $
But I don't know how how to finish this proof. Could someone give me some hints?
real-analysis sequences-and-series limits-without-lhopital
real-analysis sequences-and-series limits-without-lhopital
edited Nov 30 '18 at 8:33
Mostafa Ayaz
15.5k3939
15.5k3939
asked Nov 29 '18 at 23:32
matematicccmatematiccc
1275
1275
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,
- Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.
- Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.
Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,
$$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$
Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.
$endgroup$
add a comment |
$begingroup$
Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?
$endgroup$
add a comment |
$begingroup$
By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?
$endgroup$
$begingroup$
So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
$endgroup$
– matematiccc
Nov 29 '18 at 23:45
$begingroup$
Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
$endgroup$
– platty
Nov 29 '18 at 23:46
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%2f3019396%2fif-lim-limits-n-to-infty-a-n1-a-n-0-and-a-4n-n-ge1-conver%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,
- Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.
- Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.
Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,
$$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$
Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.
$endgroup$
add a comment |
$begingroup$
Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,
- Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.
- Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.
Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,
$$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$
Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.
$endgroup$
add a comment |
$begingroup$
Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,
- Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.
- Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.
Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,
$$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$
Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.
$endgroup$
Write $ell = lim_{ntoinfty} a_{4n}$. For each $epsilon > 0$,
- Pick $N_1$ such that $n geq N_1$ implies $|a_{n+1} - a_n| < epsilon/5$.
- Pick $N_2$ such that $4k geq N_2$ implies $|a_{4k} - ell| < epsilon/5$.
Write $N = max{N_1, N_2}$. Then for each $n geq N$, with $n = 4k+r$ with $r in {0,1,2,3}$,
$$ |a_n - ell| leq left( sum_{j=r}^{3} |a_{4k+j+1} - a_{4k+j}| right) + |a_{4(k+1)} - ell| < epsilon. $$
Therefore the $epsilon$-$N$ definition of $a_n to ell$ is established and hence $a_n to ell$.
edited Nov 30 '18 at 17:06
answered Nov 30 '18 at 8:53
Sangchul LeeSangchul Lee
93.3k12167270
93.3k12167270
add a comment |
add a comment |
$begingroup$
Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?
$endgroup$
add a comment |
$begingroup$
Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?
$endgroup$
add a comment |
$begingroup$
Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?
$endgroup$
Hints: let $j,kin mathbb N$ and write $j=4j'+r,k=4k'+s$ with $0leq r,s <4$. Using the fact that $a_{n+1}-a_n to 0$ show that $a_{n+2}-a_n to 0$ and $a_{n+3}-a_n to 0$. Conclude that $a_j-a_{4j'} to 0$ and $a_k-a_{4k'} to 0$. Now can you complete the proof?
edited Nov 29 '18 at 23:46
José Carlos Santos
160k22126232
160k22126232
answered Nov 29 '18 at 23:38
Kavi Rama MurthyKavi Rama Murthy
58.8k42161
58.8k42161
add a comment |
add a comment |
$begingroup$
By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?
$endgroup$
$begingroup$
So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
$endgroup$
– matematiccc
Nov 29 '18 at 23:45
$begingroup$
Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
$endgroup$
– platty
Nov 29 '18 at 23:46
add a comment |
$begingroup$
By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?
$endgroup$
$begingroup$
So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
$endgroup$
– matematiccc
Nov 29 '18 at 23:45
$begingroup$
Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
$endgroup$
– platty
Nov 29 '18 at 23:46
add a comment |
$begingroup$
By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?
$endgroup$
By the same technique you used, $(a_{4n+2} - a_{4n+1})$ is a subsequence of $(a_{n+1} - a_n)$, so it has to converge to $0$ as well. Does that help?
answered Nov 29 '18 at 23:36
plattyplatty
3,370320
3,370320
$begingroup$
So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
$endgroup$
– matematiccc
Nov 29 '18 at 23:45
$begingroup$
Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
$endgroup$
– platty
Nov 29 '18 at 23:46
add a comment |
$begingroup$
So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
$endgroup$
– matematiccc
Nov 29 '18 at 23:45
$begingroup$
Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
$endgroup$
– platty
Nov 29 '18 at 23:46
$begingroup$
So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
$endgroup$
– matematiccc
Nov 29 '18 at 23:45
$begingroup$
So $lim limits_{n to infty } a_{4n}=lim limits_{n to infty } a_{4n+1}=lim limits_{n to infty } a_{4n+2}+lim limits_{n to infty } a_{4n+3}$, and from this, it equals $lim limits_{n to infty } a_{n}$ so $ a_n $ converges, true?
$endgroup$
– matematiccc
Nov 29 '18 at 23:45
$begingroup$
Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
$endgroup$
– platty
Nov 29 '18 at 23:46
$begingroup$
Yep (to be precise, you may want to break it down into $epsilon$ and $N$, but that's the basic idea).
$endgroup$
– platty
Nov 29 '18 at 23:46
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%2f3019396%2fif-lim-limits-n-to-infty-a-n1-a-n-0-and-a-4n-n-ge1-conver%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