Prove this sequence converges or diverges to $-infty$
$begingroup$
Let $a_n$ be a sequence such that for every $n$: $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$.
Prove that $a_n$ either converges to a real number $L$ or diverges to $-infty$ $(Lin[-infty,infty))$.
I tried assuming it didn't diverge to $-infty$ in order to show that in that case it must converge to a real number. I tried showing so by cauchy's convergence definition but I failed.
Any ideas?
calculus sequences-and-series cauchy-sequences
asked Jan 1 at 19:50
BelkanBelkan
687
$endgroup$
|
show 1 more comment
$begingroup$
Let $a_n$ be a sequence such that for every $n$: $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$.
Prove that $a_n$ either converges to a real number $L$ or diverges to $-infty$ $(Lin[-infty,infty))$.
I tried assuming it didn't diverge to $-infty$ in order to show that in that case it must converge to a real number. I tried showing so by cauchy's convergence definition but I failed.
Any ideas?
calculus sequences-and-series cauchy-sequences
asked Jan 1 at 19:50
BelkanBelkan
687
$endgroup$
3
$begingroup$
You say you tried showing that it converged, but your attempt failed. Can you upload some screenshot of your attempt to solve the problem? This would help us understand where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 20:08
$begingroup$
I honestly didn't get to anything worth uploading, I'm mostly stuck and trying different things. All I managed is to show is that it doesn't diverge to $+infty$. I also tried splitting into cases for $a_0$ (positive or negative).
$endgroup$
– Belkan
Jan 1 at 20:21
$begingroup$
Perhaps this would help math.stackexchange.com/questions/1842784/….
$endgroup$
– Song
Jan 1 at 20:45
$begingroup$
Here is a suggestion. As in the wrong answer below $a_n$ is bounded from above, say by a constant $a>0$. If $a_n$ does not go to $-infty$ then there exists $M>0$ such that the interval $[-M,a]$ contains infinitely many terms of the sequence. But since $[-M,a]$ is compact it follows that $a_n$ has a convergent subsequent with limit say $lin [-M,a]$. I am trying to figure out how to use this to prove that $l$ is the limit of $a_n$
$endgroup$
– mouthetics
Jan 1 at 21:00
$begingroup$
I tried going that way, but my problem was that $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$ no longer meant the same thing (the indexes of the subsequent aren't necessarily sequential). However it seems like a good direction.
$endgroup$
– Belkan
Jan 1 at 21:05
|
show 1 more comment
$begingroup$
Let $a_n$ be a sequence such that for every $n$: $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$.
Prove that $a_n$ either converges to a real number $L$ or diverges to $-infty$ $(Lin[-infty,infty))$.
I tried assuming it didn't diverge to $-infty$ in order to show that in that case it must converge to a real number. I tried showing so by cauchy's convergence definition but I failed.
Any ideas?
calculus sequences-and-series cauchy-sequences
asked Jan 1 at 19:50
BelkanBelkan
687
$endgroup$
Let $a_n$ be a sequence such that for every $n$: $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$.
Prove that $a_n$ either converges to a real number $L$ or diverges to $-infty$ $(Lin[-infty,infty))$.
I tried assuming it didn't diverge to $-infty$ in order to show that in that case it must converge to a real number. I tried showing so by cauchy's convergence definition but I failed.
Any ideas?
calculus sequences-and-series cauchy-sequences
calculus sequences-and-series cauchy-sequences
asked Jan 1 at 19:50
BelkanBelkan
687
asked Jan 1 at 19:50
BelkanBelkan
687
asked Jan 1 at 19:50
BelkanBelkan
687
asked Jan 1 at 19:50
BelkanBelkan
687
asked Jan 1 at 19:50
BelkanBelkan
687
687
3
$begingroup$
You say you tried showing that it converged, but your attempt failed. Can you upload some screenshot of your attempt to solve the problem? This would help us understand where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 20:08
$begingroup$
I honestly didn't get to anything worth uploading, I'm mostly stuck and trying different things. All I managed is to show is that it doesn't diverge to $+infty$. I also tried splitting into cases for $a_0$ (positive or negative).
$endgroup$
– Belkan
Jan 1 at 20:21
$begingroup$
Perhaps this would help math.stackexchange.com/questions/1842784/….
$endgroup$
– Song
Jan 1 at 20:45
$begingroup$
Here is a suggestion. As in the wrong answer below $a_n$ is bounded from above, say by a constant $a>0$. If $a_n$ does not go to $-infty$ then there exists $M>0$ such that the interval $[-M,a]$ contains infinitely many terms of the sequence. But since $[-M,a]$ is compact it follows that $a_n$ has a convergent subsequent with limit say $lin [-M,a]$. I am trying to figure out how to use this to prove that $l$ is the limit of $a_n$
$endgroup$
– mouthetics
Jan 1 at 21:00
$begingroup$
I tried going that way, but my problem was that $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$ no longer meant the same thing (the indexes of the subsequent aren't necessarily sequential). However it seems like a good direction.
$endgroup$
– Belkan
Jan 1 at 21:05
|
show 1 more comment
3
$begingroup$
You say you tried showing that it converged, but your attempt failed. Can you upload some screenshot of your attempt to solve the problem? This would help us understand where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 20:08
$begingroup$
I honestly didn't get to anything worth uploading, I'm mostly stuck and trying different things. All I managed is to show is that it doesn't diverge to $+infty$. I also tried splitting into cases for $a_0$ (positive or negative).
$endgroup$
– Belkan
Jan 1 at 20:21
$begingroup$
Perhaps this would help math.stackexchange.com/questions/1842784/….
$endgroup$
– Song
Jan 1 at 20:45
$begingroup$
Here is a suggestion. As in the wrong answer below $a_n$ is bounded from above, say by a constant $a>0$. If $a_n$ does not go to $-infty$ then there exists $M>0$ such that the interval $[-M,a]$ contains infinitely many terms of the sequence. But since $[-M,a]$ is compact it follows that $a_n$ has a convergent subsequent with limit say $lin [-M,a]$. I am trying to figure out how to use this to prove that $l$ is the limit of $a_n$
$endgroup$
– mouthetics
Jan 1 at 21:00
$begingroup$
I tried going that way, but my problem was that $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$ no longer meant the same thing (the indexes of the subsequent aren't necessarily sequential). However it seems like a good direction.
$endgroup$
– Belkan
Jan 1 at 21:05
3
3
$begingroup$
You say you tried showing that it converged, but your attempt failed. Can you upload some screenshot of your attempt to solve the problem? This would help us understand where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 20:08
$begingroup$
You say you tried showing that it converged, but your attempt failed. Can you upload some screenshot of your attempt to solve the problem? This would help us understand where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 20:08
$begingroup$
I honestly didn't get to anything worth uploading, I'm mostly stuck and trying different things. All I managed is to show is that it doesn't diverge to $+infty$. I also tried splitting into cases for $a_0$ (positive or negative).
$endgroup$
– Belkan
Jan 1 at 20:21
$begingroup$
I honestly didn't get to anything worth uploading, I'm mostly stuck and trying different things. All I managed is to show is that it doesn't diverge to $+infty$. I also tried splitting into cases for $a_0$ (positive or negative).
$endgroup$
– Belkan
Jan 1 at 20:21
$begingroup$
Perhaps this would help math.stackexchange.com/questions/1842784/….
$endgroup$
– Song
Jan 1 at 20:45
$begingroup$
Perhaps this would help math.stackexchange.com/questions/1842784/….
$endgroup$
– Song
Jan 1 at 20:45
$begingroup$
Here is a suggestion. As in the wrong answer below $a_n$ is bounded from above, say by a constant $a>0$. If $a_n$ does not go to $-infty$ then there exists $M>0$ such that the interval $[-M,a]$ contains infinitely many terms of the sequence. But since $[-M,a]$ is compact it follows that $a_n$ has a convergent subsequent with limit say $lin [-M,a]$. I am trying to figure out how to use this to prove that $l$ is the limit of $a_n$
$endgroup$
– mouthetics
Jan 1 at 21:00
$begingroup$
Here is a suggestion. As in the wrong answer below $a_n$ is bounded from above, say by a constant $a>0$. If $a_n$ does not go to $-infty$ then there exists $M>0$ such that the interval $[-M,a]$ contains infinitely many terms of the sequence. But since $[-M,a]$ is compact it follows that $a_n$ has a convergent subsequent with limit say $lin [-M,a]$. I am trying to figure out how to use this to prove that $l$ is the limit of $a_n$
$endgroup$
– mouthetics
Jan 1 at 21:00
$begingroup$
I tried going that way, but my problem was that $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$ no longer meant the same thing (the indexes of the subsequent aren't necessarily sequential). However it seems like a good direction.
$endgroup$
– Belkan
Jan 1 at 21:05
$begingroup$
I tried going that way, but my problem was that $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$ no longer meant the same thing (the indexes of the subsequent aren't necessarily sequential). However it seems like a good direction.
$endgroup$
– Belkan
Jan 1 at 21:05
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
So we have that
$$a_3leq frac{1}{2}(a_{2}+a_{1})$$
$$a_4leq frac{1}{2}(a_{3}+a_{2})leq 3a_2/4+a_1/4$$
$$a_5leq frac{1}{2}(a_4+a_3)leq 7a_2/8+5a_1/8$$
$$cdots$$
$$a_{n}leq frac{2n-5}{2^{n-2}}a_2+frac{2n-7}{2^{n-2}}a_1.$$
If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$ However
$$a_nleq frac{4n-12}{2^{n-2}}max{a_1,a_2}$$
and so $a_nto -infty.$
answered Jan 1 at 20:13
model_checkermodel_checker
4,33121931
$endgroup$
5
$begingroup$
"If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$" I don't understand this sentence. Isn't it possible that the sequence just oscillates a lot and that the limit doesn't exist at all?
$endgroup$
– Noble Mushtak
Jan 1 at 20:16
$begingroup$
@NobleMushtak: I guess the subtle part is to show that such a sequence cannot oscillate too wildly.
$endgroup$
– Jack D'Aurizio
Jan 2 at 0:37
add a comment |
$begingroup$
Let $delta_n = a_n-a_{n+1}$. From
$$ a_{n+1} leq frac{1}{2}a_n + frac{1}{2}a_{n-1} $$
we get
$$ -delta_n = a_{n+1}-a_n leq frac{1}{2}delta_{n-1} $$
so our sequence is free to decrease as fast (or as slow) as it likes, but if $a_{n}>a_{n-1}$ (i.e. $delta_{n-1}<0$) then $a_{n+1}$ has to lie on the left of the midpoint of $[a_{n-1};a_n]$ and the subsequence of the terms which are greater than the previous one, say ${a_{n_k}}_{kgeq 1}$, is decreasing. We are temporary assuming this actually is a subsequence, i.e. that there are infinite $n$s such that $a_{n+1}>a_n$. Now we may start a dichotomy. If $a_{n_k}$ converges to $-infty$, so does the original sequence. If $a_{n_k}to L$, for any $k$ large enough we have $a_{n_k}in(L-varepsilon,L+varepsilon)$ and
$$ a_{n_k}geq a_{n_k+1} geq a_{n_k+2} geq ldots geq a_{n_{k+1}-2} geq frac{a_{n_{k+1}-1}+a_{n_{k+1}-2}}{2}geqmax(a_{n_{k+1}},a_{n_{k+1}-1}) $$
so for any $min[n_{k},n_{k+1}]$ we have $a_min(L-3varepsilon,L+3varepsilon)$ and ${a_n}_{ngeq 1}$ and ${a_{n_k}}_{kgeq 1}$ share the same limit. At last we have to deal with the case in which $delta_n<0$ holds for a finite number of $n$, i.e. ${a_n}_{ngeq 1}$ is a weakly decreasing sequence from some point on. But again, a weakly decreasing sequence may only diverge to $-infty$ or converge to a finite limit.
answered Jan 2 at 0:35
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
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%2f3058813%2fprove-this-sequence-converges-or-diverges-to-infty%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
$begingroup$
So we have that
$$a_3leq frac{1}{2}(a_{2}+a_{1})$$
$$a_4leq frac{1}{2}(a_{3}+a_{2})leq 3a_2/4+a_1/4$$
$$a_5leq frac{1}{2}(a_4+a_3)leq 7a_2/8+5a_1/8$$
$$cdots$$
$$a_{n}leq frac{2n-5}{2^{n-2}}a_2+frac{2n-7}{2^{n-2}}a_1.$$
If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$ However
$$a_nleq frac{4n-12}{2^{n-2}}max{a_1,a_2}$$
and so $a_nto -infty.$
answered Jan 1 at 20:13
model_checkermodel_checker
4,33121931
$endgroup$
5
$begingroup$
"If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$" I don't understand this sentence. Isn't it possible that the sequence just oscillates a lot and that the limit doesn't exist at all?
$endgroup$
– Noble Mushtak
Jan 1 at 20:16
$begingroup$
@NobleMushtak: I guess the subtle part is to show that such a sequence cannot oscillate too wildly.
$endgroup$
– Jack D'Aurizio
Jan 2 at 0:37
add a comment |
$begingroup$
So we have that
$$a_3leq frac{1}{2}(a_{2}+a_{1})$$
$$a_4leq frac{1}{2}(a_{3}+a_{2})leq 3a_2/4+a_1/4$$
$$a_5leq frac{1}{2}(a_4+a_3)leq 7a_2/8+5a_1/8$$
$$cdots$$
$$a_{n}leq frac{2n-5}{2^{n-2}}a_2+frac{2n-7}{2^{n-2}}a_1.$$
If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$ However
$$a_nleq frac{4n-12}{2^{n-2}}max{a_1,a_2}$$
and so $a_nto -infty.$
answered Jan 1 at 20:13
model_checkermodel_checker
4,33121931
$endgroup$
5
$begingroup$
"If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$" I don't understand this sentence. Isn't it possible that the sequence just oscillates a lot and that the limit doesn't exist at all?
$endgroup$
– Noble Mushtak
Jan 1 at 20:16
$begingroup$
@NobleMushtak: I guess the subtle part is to show that such a sequence cannot oscillate too wildly.
$endgroup$
– Jack D'Aurizio
Jan 2 at 0:37
add a comment |
$begingroup$
So we have that
$$a_3leq frac{1}{2}(a_{2}+a_{1})$$
$$a_4leq frac{1}{2}(a_{3}+a_{2})leq 3a_2/4+a_1/4$$
$$a_5leq frac{1}{2}(a_4+a_3)leq 7a_2/8+5a_1/8$$
$$cdots$$
$$a_{n}leq frac{2n-5}{2^{n-2}}a_2+frac{2n-7}{2^{n-2}}a_1.$$
If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$ However
$$a_nleq frac{4n-12}{2^{n-2}}max{a_1,a_2}$$
and so $a_nto -infty.$
answered Jan 1 at 20:13
model_checkermodel_checker
4,33121931
$endgroup$
So we have that
$$a_3leq frac{1}{2}(a_{2}+a_{1})$$
$$a_4leq frac{1}{2}(a_{3}+a_{2})leq 3a_2/4+a_1/4$$
$$a_5leq frac{1}{2}(a_4+a_3)leq 7a_2/8+5a_1/8$$
$$cdots$$
$$a_{n}leq frac{2n-5}{2^{n-2}}a_2+frac{2n-7}{2^{n-2}}a_1.$$
If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$ However
$$a_nleq frac{4n-12}{2^{n-2}}max{a_1,a_2}$$
and so $a_nto -infty.$
answered Jan 1 at 20:13
model_checkermodel_checker
4,33121931
answered Jan 1 at 20:13
model_checkermodel_checker
4,33121931
answered Jan 1 at 20:13
model_checkermodel_checker
4,33121931
answered Jan 1 at 20:13
model_checkermodel_checker
4,33121931
4,33121931
5
$begingroup$
"If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$" I don't understand this sentence. Isn't it possible that the sequence just oscillates a lot and that the limit doesn't exist at all?
$endgroup$
– Noble Mushtak
Jan 1 at 20:16
$begingroup$
@NobleMushtak: I guess the subtle part is to show that such a sequence cannot oscillate too wildly.
$endgroup$
– Jack D'Aurizio
Jan 2 at 0:37
add a comment |
5
$begingroup$
"If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$" I don't understand this sentence. Isn't it possible that the sequence just oscillates a lot and that the limit doesn't exist at all?
$endgroup$
– Noble Mushtak
Jan 1 at 20:16
$begingroup$
@NobleMushtak: I guess the subtle part is to show that such a sequence cannot oscillate too wildly.
$endgroup$
– Jack D'Aurizio
Jan 2 at 0:37
5
5
$begingroup$
"If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$" I don't understand this sentence. Isn't it possible that the sequence just oscillates a lot and that the limit doesn't exist at all?
$endgroup$
– Noble Mushtak
Jan 1 at 20:16
$begingroup$
"If the sequence is not convergent to a real number $L$ then $a_nto infty$ or $a_nto-infty.$" I don't understand this sentence. Isn't it possible that the sequence just oscillates a lot and that the limit doesn't exist at all?
$endgroup$
– Noble Mushtak
Jan 1 at 20:16
$begingroup$
@NobleMushtak: I guess the subtle part is to show that such a sequence cannot oscillate too wildly.
$endgroup$
– Jack D'Aurizio
Jan 2 at 0:37
$begingroup$
@NobleMushtak: I guess the subtle part is to show that such a sequence cannot oscillate too wildly.
$endgroup$
– Jack D'Aurizio
Jan 2 at 0:37
add a comment |
$begingroup$
Let $delta_n = a_n-a_{n+1}$. From
$$ a_{n+1} leq frac{1}{2}a_n + frac{1}{2}a_{n-1} $$
we get
$$ -delta_n = a_{n+1}-a_n leq frac{1}{2}delta_{n-1} $$
so our sequence is free to decrease as fast (or as slow) as it likes, but if $a_{n}>a_{n-1}$ (i.e. $delta_{n-1}<0$) then $a_{n+1}$ has to lie on the left of the midpoint of $[a_{n-1};a_n]$ and the subsequence of the terms which are greater than the previous one, say ${a_{n_k}}_{kgeq 1}$, is decreasing. We are temporary assuming this actually is a subsequence, i.e. that there are infinite $n$s such that $a_{n+1}>a_n$. Now we may start a dichotomy. If $a_{n_k}$ converges to $-infty$, so does the original sequence. If $a_{n_k}to L$, for any $k$ large enough we have $a_{n_k}in(L-varepsilon,L+varepsilon)$ and
$$ a_{n_k}geq a_{n_k+1} geq a_{n_k+2} geq ldots geq a_{n_{k+1}-2} geq frac{a_{n_{k+1}-1}+a_{n_{k+1}-2}}{2}geqmax(a_{n_{k+1}},a_{n_{k+1}-1}) $$
so for any $min[n_{k},n_{k+1}]$ we have $a_min(L-3varepsilon,L+3varepsilon)$ and ${a_n}_{ngeq 1}$ and ${a_{n_k}}_{kgeq 1}$ share the same limit. At last we have to deal with the case in which $delta_n<0$ holds for a finite number of $n$, i.e. ${a_n}_{ngeq 1}$ is a weakly decreasing sequence from some point on. But again, a weakly decreasing sequence may only diverge to $-infty$ or converge to a finite limit.
answered Jan 2 at 0:35
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
Let $delta_n = a_n-a_{n+1}$. From
$$ a_{n+1} leq frac{1}{2}a_n + frac{1}{2}a_{n-1} $$
we get
$$ -delta_n = a_{n+1}-a_n leq frac{1}{2}delta_{n-1} $$
so our sequence is free to decrease as fast (or as slow) as it likes, but if $a_{n}>a_{n-1}$ (i.e. $delta_{n-1}<0$) then $a_{n+1}$ has to lie on the left of the midpoint of $[a_{n-1};a_n]$ and the subsequence of the terms which are greater than the previous one, say ${a_{n_k}}_{kgeq 1}$, is decreasing. We are temporary assuming this actually is a subsequence, i.e. that there are infinite $n$s such that $a_{n+1}>a_n$. Now we may start a dichotomy. If $a_{n_k}$ converges to $-infty$, so does the original sequence. If $a_{n_k}to L$, for any $k$ large enough we have $a_{n_k}in(L-varepsilon,L+varepsilon)$ and
$$ a_{n_k}geq a_{n_k+1} geq a_{n_k+2} geq ldots geq a_{n_{k+1}-2} geq frac{a_{n_{k+1}-1}+a_{n_{k+1}-2}}{2}geqmax(a_{n_{k+1}},a_{n_{k+1}-1}) $$
so for any $min[n_{k},n_{k+1}]$ we have $a_min(L-3varepsilon,L+3varepsilon)$ and ${a_n}_{ngeq 1}$ and ${a_{n_k}}_{kgeq 1}$ share the same limit. At last we have to deal with the case in which $delta_n<0$ holds for a finite number of $n$, i.e. ${a_n}_{ngeq 1}$ is a weakly decreasing sequence from some point on. But again, a weakly decreasing sequence may only diverge to $-infty$ or converge to a finite limit.
answered Jan 2 at 0:35
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
Let $delta_n = a_n-a_{n+1}$. From
$$ a_{n+1} leq frac{1}{2}a_n + frac{1}{2}a_{n-1} $$
we get
$$ -delta_n = a_{n+1}-a_n leq frac{1}{2}delta_{n-1} $$
so our sequence is free to decrease as fast (or as slow) as it likes, but if $a_{n}>a_{n-1}$ (i.e. $delta_{n-1}<0$) then $a_{n+1}$ has to lie on the left of the midpoint of $[a_{n-1};a_n]$ and the subsequence of the terms which are greater than the previous one, say ${a_{n_k}}_{kgeq 1}$, is decreasing. We are temporary assuming this actually is a subsequence, i.e. that there are infinite $n$s such that $a_{n+1}>a_n$. Now we may start a dichotomy. If $a_{n_k}$ converges to $-infty$, so does the original sequence. If $a_{n_k}to L$, for any $k$ large enough we have $a_{n_k}in(L-varepsilon,L+varepsilon)$ and
$$ a_{n_k}geq a_{n_k+1} geq a_{n_k+2} geq ldots geq a_{n_{k+1}-2} geq frac{a_{n_{k+1}-1}+a_{n_{k+1}-2}}{2}geqmax(a_{n_{k+1}},a_{n_{k+1}-1}) $$
so for any $min[n_{k},n_{k+1}]$ we have $a_min(L-3varepsilon,L+3varepsilon)$ and ${a_n}_{ngeq 1}$ and ${a_{n_k}}_{kgeq 1}$ share the same limit. At last we have to deal with the case in which $delta_n<0$ holds for a finite number of $n$, i.e. ${a_n}_{ngeq 1}$ is a weakly decreasing sequence from some point on. But again, a weakly decreasing sequence may only diverge to $-infty$ or converge to a finite limit.
answered Jan 2 at 0:35
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
Let $delta_n = a_n-a_{n+1}$. From
$$ a_{n+1} leq frac{1}{2}a_n + frac{1}{2}a_{n-1} $$
we get
$$ -delta_n = a_{n+1}-a_n leq frac{1}{2}delta_{n-1} $$
so our sequence is free to decrease as fast (or as slow) as it likes, but if $a_{n}>a_{n-1}$ (i.e. $delta_{n-1}<0$) then $a_{n+1}$ has to lie on the left of the midpoint of $[a_{n-1};a_n]$ and the subsequence of the terms which are greater than the previous one, say ${a_{n_k}}_{kgeq 1}$, is decreasing. We are temporary assuming this actually is a subsequence, i.e. that there are infinite $n$s such that $a_{n+1}>a_n$. Now we may start a dichotomy. If $a_{n_k}$ converges to $-infty$, so does the original sequence. If $a_{n_k}to L$, for any $k$ large enough we have $a_{n_k}in(L-varepsilon,L+varepsilon)$ and
$$ a_{n_k}geq a_{n_k+1} geq a_{n_k+2} geq ldots geq a_{n_{k+1}-2} geq frac{a_{n_{k+1}-1}+a_{n_{k+1}-2}}{2}geqmax(a_{n_{k+1}},a_{n_{k+1}-1}) $$
so for any $min[n_{k},n_{k+1}]$ we have $a_min(L-3varepsilon,L+3varepsilon)$ and ${a_n}_{ngeq 1}$ and ${a_{n_k}}_{kgeq 1}$ share the same limit. At last we have to deal with the case in which $delta_n<0$ holds for a finite number of $n$, i.e. ${a_n}_{ngeq 1}$ is a weakly decreasing sequence from some point on. But again, a weakly decreasing sequence may only diverge to $-infty$ or converge to a finite limit.
answered Jan 2 at 0:35
Jack D'AurizioJack D'Aurizio
292k33284672
answered Jan 2 at 0:35
Jack D'AurizioJack D'Aurizio
292k33284672
answered Jan 2 at 0:35
Jack D'AurizioJack D'Aurizio
292k33284672
answered Jan 2 at 0:35
Jack D'AurizioJack D'Aurizio
292k33284672
292k33284672
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%2f3058813%2fprove-this-sequence-converges-or-diverges-to-infty%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
3
$begingroup$
You say you tried showing that it converged, but your attempt failed. Can you upload some screenshot of your attempt to solve the problem? This would help us understand where you went wrong.
$endgroup$
– Noble Mushtak
Jan 1 at 20:08
$begingroup$
I honestly didn't get to anything worth uploading, I'm mostly stuck and trying different things. All I managed is to show is that it doesn't diverge to $+infty$. I also tried splitting into cases for $a_0$ (positive or negative).
$endgroup$
– Belkan
Jan 1 at 20:21
$begingroup$
Perhaps this would help math.stackexchange.com/questions/1842784/….
$endgroup$
– Song
Jan 1 at 20:45
$begingroup$
Here is a suggestion. As in the wrong answer below $a_n$ is bounded from above, say by a constant $a>0$. If $a_n$ does not go to $-infty$ then there exists $M>0$ such that the interval $[-M,a]$ contains infinitely many terms of the sequence. But since $[-M,a]$ is compact it follows that $a_n$ has a convergent subsequent with limit say $lin [-M,a]$. I am trying to figure out how to use this to prove that $l$ is the limit of $a_n$
$endgroup$
– mouthetics
Jan 1 at 21:00
$begingroup$
I tried going that way, but my problem was that $a_nlefrac{1}{2}(a_{n-1}+a_{n-2})$ no longer meant the same thing (the indexes of the subsequent aren't necessarily sequential). However it seems like a good direction.
$endgroup$
– Belkan
Jan 1 at 21:05