Prove $lim_nsupx_n+1over x_n < 1$ implies $lim_n x_n = 0$Prove existence of $X_n_n=1^infty$, $Y_n_n=1^infty subseteq S subseteq mathbbR$, $S neq emptyset$ and $S$ boundedProve that $lim_ntoinfty supx_n=lim_ntoinfty infx_n=L$ IFF the sequence $(x_n)$ converges.Proof of $ suplimits_ninmathbbN (x_n+y_n) lesup x_n+sup y_n$Proof verification that if $y_n = sum_k=1^n x_n$ converges then $lim_ntoinfty x_n = 0$Request for assistance on finishing the proof that $lim_ntoinftyn x_n = 0$ given some initial conditions.Find the limit $lim_ntoinftyfracx_n + x_n^2 + cdots + x_n^k - kx_n - 1$ given $x_n ne 1$ and $lim x_n = 1$Prove $lim_ntoinfty(|x_n + y_n| - |x_n - y_n|) = +infty iff lim_ntoinfty|x_n| =lim_ntoinfty|y_n| =lim_ntoinftyx_ny_n =+infty$$x_n$ is a bounded above sequence such that $x_n+1 - x_n ge a_n$, where $sum a_k$ converges. Prove $x_n$ converges.$x_n$ is an unbounded increasing sequence. Prove that $exists lim_n K_nover n =L iff exists lim_nnover x_n =L$, where $K_n$…Prove that convergence of a sequence implies boundedness of its variation.
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Prove $lim_nsupx_n+1over x_n
Prove existence of $X_n_n=1^infty$, $Y_n_n=1^infty subseteq S subseteq mathbbR$, $S neq emptyset$ and $S$ boundedProve that $lim_ntoinfty supx_n=lim_ntoinfty infx_n=L$ IFF the sequence $(x_n)$ converges.Proof of $ suplimits_ninmathbbN (x_n+y_n) lesup x_n+sup y_n$Proof verification that if $y_n = sum_k=1^n x_n$ converges then $lim_ntoinfty x_n = 0$Request for assistance on finishing the proof that $lim_ntoinftyn x_n = 0$ given some initial conditions.Find the limit $lim_ntoinftyfracx_n + x_n^2 + cdots + x_n^k - kx_n - 1$ given $x_n ne 1$ and $lim x_n = 1$Prove $lim_ntoinfty(|x_n + y_n| - |x_n - y_n|) = +infty iff lim_ntoinfty|x_n| =lim_ntoinfty|y_n| =lim_ntoinftyx_ny_n =+infty$$x_n$ is a bounded above sequence such that $x_n+1 - x_n ge a_n$, where $sum a_k$ converges. Prove $x_n$ converges.$x_n$ is an unbounded increasing sequence. Prove that $exists lim_n K_nover n =L iff exists lim_nnover x_n =L$, where $K_n$…Prove that convergence of a sequence implies boundedness of its variation.
$begingroup$
Let $x_n$ be a sequence $x_n ne 0$ such that:
$$
lim_ntoinftysupx_n+1over x_n < 1
$$
Prove:
$$
lim_ntoinftyx_n = 0
$$
I would like to verify the below.
Let:
$$
y_n = fracx_n+1x_n
$$
If $y_n$ converges, then:
$$
existslim_ntoinftysup y_n = L implies exists lim_ntoinftysup |y_n| = |L|
$$
Since $L < 1$, then $|L| in [0, 1)$. Also:
$$
lim_ntoinftyy_n le lim_ntoinftysup y_n < 1
$$
Thus:
$$
0 le lim_ntoinfty|y_n| le lim_ntoinftysup |y_n| < 1
$$
Therefore:
$$
0 le lim_ntoinftyleft|fracx_n+1x_nright| < 1
$$
Which means $exists NinBbb N$ such that for $n > N$ the sequence of absolute values starts to strictly decrease. Thus:
$$
exists NinBbb N : forall n > N : |x_n+1| < |x_n|
$$
But $|x_n| ge 0$, then by monotone convergence theorem:
$$
lim_ntoinfty left|fracx_n+1x_nright| in [0, 1) implies lim_ntoinfty|x_n| = 0 implies lim_ntoinftyx_n = 0
$$
Could someone please verify the proof above?
sequences-and-series limits proof-verification supremum-and-infimum
asked Mar 29 at 15:16
romanroman
2,43721226
$endgroup$
|
show 2 more comments
$begingroup$
Let $x_n$ be a sequence $x_n ne 0$ such that:
$$
lim_ntoinftysupx_n+1over x_n < 1
$$
Prove:
$$
lim_ntoinftyx_n = 0
$$
I would like to verify the below.
Let:
$$
y_n = fracx_n+1x_n
$$
If $y_n$ converges, then:
$$
existslim_ntoinftysup y_n = L implies exists lim_ntoinftysup |y_n| = |L|
$$
Since $L < 1$, then $|L| in [0, 1)$. Also:
$$
lim_ntoinftyy_n le lim_ntoinftysup y_n < 1
$$
Thus:
$$
0 le lim_ntoinfty|y_n| le lim_ntoinftysup |y_n| < 1
$$
Therefore:
$$
0 le lim_ntoinftyleft|fracx_n+1x_nright| < 1
$$
Which means $exists NinBbb N$ such that for $n > N$ the sequence of absolute values starts to strictly decrease. Thus:
$$
exists NinBbb N : forall n > N : |x_n+1| < |x_n|
$$
But $|x_n| ge 0$, then by monotone convergence theorem:
$$
lim_ntoinfty left|fracx_n+1x_nright| in [0, 1) implies lim_ntoinfty|x_n| = 0 implies lim_ntoinftyx_n = 0
$$
Could someone please verify the proof above?
sequences-and-series limits proof-verification supremum-and-infimum
asked Mar 29 at 15:16
romanroman
2,43721226
$endgroup$
1
$begingroup$
That is wrong, a counter-example is $x_n = (-2)^n$ with $fracx_n+1x_n = -2 < 1$. Or do you assume $x_n > 0$?
$endgroup$
– Martin R
Mar 29 at 15:24
$begingroup$
@MartinR Nope, I did not assume that. By your counterexample, the above is indeed wrong.
$endgroup$
– roman
Mar 29 at 15:30
$begingroup$
Note that $L < 1$ does not imply $|L| in [0, 1)$.
$endgroup$
– Martin R
Mar 29 at 15:32
$begingroup$
Your last edit changes the problem completely so that my answer does not apply anymore. I suggest that you revert to the original version and ask a new question for the updated problem. Alternatively please unaccept my answer so that I can delete it.
$endgroup$
– Martin R
Mar 29 at 17:21
$begingroup$
@MartinR But your answer is the one I was looking for, why do you want to delete it? As it does apply to the problem. Thanks to you and José Carlos Santos I've figured out the problem was incorrectly stated in the book. And that is the only thing reflected in the update to the question along with some thought from your answer. Of course, I can unaccept your answer if you still insist.
$endgroup$
– roman
Mar 29 at 17:27
|
show 2 more comments
$begingroup$
Let $x_n$ be a sequence $x_n ne 0$ such that:
$$
lim_ntoinftysupx_n+1over x_n < 1
$$
Prove:
$$
lim_ntoinftyx_n = 0
$$
I would like to verify the below.
Let:
$$
y_n = fracx_n+1x_n
$$
If $y_n$ converges, then:
$$
existslim_ntoinftysup y_n = L implies exists lim_ntoinftysup |y_n| = |L|
$$
Since $L < 1$, then $|L| in [0, 1)$. Also:
$$
lim_ntoinftyy_n le lim_ntoinftysup y_n < 1
$$
Thus:
$$
0 le lim_ntoinfty|y_n| le lim_ntoinftysup |y_n| < 1
$$
Therefore:
$$
0 le lim_ntoinftyleft|fracx_n+1x_nright| < 1
$$
Which means $exists NinBbb N$ such that for $n > N$ the sequence of absolute values starts to strictly decrease. Thus:
$$
exists NinBbb N : forall n > N : |x_n+1| < |x_n|
$$
But $|x_n| ge 0$, then by monotone convergence theorem:
$$
lim_ntoinfty left|fracx_n+1x_nright| in [0, 1) implies lim_ntoinfty|x_n| = 0 implies lim_ntoinftyx_n = 0
$$
Could someone please verify the proof above?
sequences-and-series limits proof-verification supremum-and-infimum
asked Mar 29 at 15:16
romanroman
2,43721226
$endgroup$
Let $x_n$ be a sequence $x_n ne 0$ such that:
$$
lim_ntoinftysupx_n+1over x_n < 1
$$
Prove:
$$
lim_ntoinftyx_n = 0
$$
I would like to verify the below.
Let:
$$
y_n = fracx_n+1x_n
$$
If $y_n$ converges, then:
$$
existslim_ntoinftysup y_n = L implies exists lim_ntoinftysup |y_n| = |L|
$$
Since $L < 1$, then $|L| in [0, 1)$. Also:
$$
lim_ntoinftyy_n le lim_ntoinftysup y_n < 1
$$
Thus:
$$
0 le lim_ntoinfty|y_n| le lim_ntoinftysup |y_n| < 1
$$
Therefore:
$$
0 le lim_ntoinftyleft|fracx_n+1x_nright| < 1
$$
Which means $exists NinBbb N$ such that for $n > N$ the sequence of absolute values starts to strictly decrease. Thus:
$$
exists NinBbb N : forall n > N : |x_n+1| < |x_n|
$$
But $|x_n| ge 0$, then by monotone convergence theorem:
$$
lim_ntoinfty left|fracx_n+1x_nright| in [0, 1) implies lim_ntoinfty|x_n| = 0 implies lim_ntoinftyx_n = 0
$$
Could someone please verify the proof above?
sequences-and-series limits proof-verification supremum-and-infimum
sequences-and-series limits proof-verification supremum-and-infimum
asked Mar 29 at 15:16
romanroman
2,43721226
asked Mar 29 at 15:16
romanroman
2,43721226
edited Mar 29 at 17:37
roman
asked Mar 29 at 15:16
romanroman
2,43721226
asked Mar 29 at 15:16
romanroman
2,43721226
asked Mar 29 at 15:16
romanroman
2,43721226
2,43721226
1
$begingroup$
That is wrong, a counter-example is $x_n = (-2)^n$ with $fracx_n+1x_n = -2 < 1$. Or do you assume $x_n > 0$?
$endgroup$
– Martin R
Mar 29 at 15:24
$begingroup$
@MartinR Nope, I did not assume that. By your counterexample, the above is indeed wrong.
$endgroup$
– roman
Mar 29 at 15:30
$begingroup$
Note that $L < 1$ does not imply $|L| in [0, 1)$.
$endgroup$
– Martin R
Mar 29 at 15:32
$begingroup$
Your last edit changes the problem completely so that my answer does not apply anymore. I suggest that you revert to the original version and ask a new question for the updated problem. Alternatively please unaccept my answer so that I can delete it.
$endgroup$
– Martin R
Mar 29 at 17:21
$begingroup$
@MartinR But your answer is the one I was looking for, why do you want to delete it? As it does apply to the problem. Thanks to you and José Carlos Santos I've figured out the problem was incorrectly stated in the book. And that is the only thing reflected in the update to the question along with some thought from your answer. Of course, I can unaccept your answer if you still insist.
$endgroup$
– roman
Mar 29 at 17:27
|
show 2 more comments
1
$begingroup$
That is wrong, a counter-example is $x_n = (-2)^n$ with $fracx_n+1x_n = -2 < 1$. Or do you assume $x_n > 0$?
$endgroup$
– Martin R
Mar 29 at 15:24
$begingroup$
@MartinR Nope, I did not assume that. By your counterexample, the above is indeed wrong.
$endgroup$
– roman
Mar 29 at 15:30
$begingroup$
Note that $L < 1$ does not imply $|L| in [0, 1)$.
$endgroup$
– Martin R
Mar 29 at 15:32
$begingroup$
Your last edit changes the problem completely so that my answer does not apply anymore. I suggest that you revert to the original version and ask a new question for the updated problem. Alternatively please unaccept my answer so that I can delete it.
$endgroup$
– Martin R
Mar 29 at 17:21
$begingroup$
@MartinR But your answer is the one I was looking for, why do you want to delete it? As it does apply to the problem. Thanks to you and José Carlos Santos I've figured out the problem was incorrectly stated in the book. And that is the only thing reflected in the update to the question along with some thought from your answer. Of course, I can unaccept your answer if you still insist.
$endgroup$
– roman
Mar 29 at 17:27
1
1
$begingroup$
That is wrong, a counter-example is $x_n = (-2)^n$ with $fracx_n+1x_n = -2 < 1$. Or do you assume $x_n > 0$?
$endgroup$
– Martin R
Mar 29 at 15:24
$begingroup$
That is wrong, a counter-example is $x_n = (-2)^n$ with $fracx_n+1x_n = -2 < 1$. Or do you assume $x_n > 0$?
$endgroup$
– Martin R
Mar 29 at 15:24
$begingroup$
@MartinR Nope, I did not assume that. By your counterexample, the above is indeed wrong.
$endgroup$
– roman
Mar 29 at 15:30
$begingroup$
@MartinR Nope, I did not assume that. By your counterexample, the above is indeed wrong.
$endgroup$
– roman
Mar 29 at 15:30
$begingroup$
Note that $L < 1$ does not imply $|L| in [0, 1)$.
$endgroup$
– Martin R
Mar 29 at 15:32
$begingroup$
Note that $L < 1$ does not imply $|L| in [0, 1)$.
$endgroup$
– Martin R
Mar 29 at 15:32
$begingroup$
Your last edit changes the problem completely so that my answer does not apply anymore. I suggest that you revert to the original version and ask a new question for the updated problem. Alternatively please unaccept my answer so that I can delete it.
$endgroup$
– Martin R
Mar 29 at 17:21
$begingroup$
Your last edit changes the problem completely so that my answer does not apply anymore. I suggest that you revert to the original version and ask a new question for the updated problem. Alternatively please unaccept my answer so that I can delete it.
$endgroup$
– Martin R
Mar 29 at 17:21
$begingroup$
@MartinR But your answer is the one I was looking for, why do you want to delete it? As it does apply to the problem. Thanks to you and José Carlos Santos I've figured out the problem was incorrectly stated in the book. And that is the only thing reflected in the update to the question along with some thought from your answer. Of course, I can unaccept your answer if you still insist.
$endgroup$
– roman
Mar 29 at 17:27
$begingroup$
@MartinR But your answer is the one I was looking for, why do you want to delete it? As it does apply to the problem. Thanks to you and José Carlos Santos I've figured out the problem was incorrectly stated in the book. And that is the only thing reflected in the update to the question along with some thought from your answer. Of course, I can unaccept your answer if you still insist.
$endgroup$
– roman
Mar 29 at 17:27
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The statement is wrong, a counter-example is $x_n = (-2)^n$ with
$limsup_ntoinftyx_n+1over x_n = -2 < 1$.
There are several flaws in the proof:
$y_n = fracx_n+1x_n$ is not necessarily convergent.
$limsup y_n = L$ does not imply $limsup |y_n| = |L|$, a counter-example is $y_n = -2 + (-1)^n$.
$L < 1$ does not imply $|L| in [0, 1)$, a counter-example is $L=2$.- If $|x_n|$ is monotone decreasing then it is convergent, but not necessarily against zero.
You'll have to require that $limsup_ntoinfty left|x_n+1over x_nright| < 1$ instead. Then (as José already pointed out),
$$
0 le |x_n| le c^n-n_0 |x_n_0|
$$
for some $c in [0, 1)$ and $n ge n_0$, and $x_n to 0 $ follows.
answered Mar 29 at 15:59
Martin RMartin R
30.7k33560
$endgroup$
add a comment |
$begingroup$
Your proof is wrong because you only proved that, if $n$ is large enough, then $lvert x_n+1rvert<lvert x_nrvert$ and that is not enough to deduce that $lim_ntoinftyx_n=0$, even if you mention the monotone convergence theorem, which has nothing to do with this.
Since $limsup_nleftlvertfracx_n+1x_nrightrvert<1$, if $cinleft(limsup_nleftlvertfracx_n+1x_nrightrvert,1right)$, then $leftlvertfracx_n+1x_nrightrvert<c$, if $ngeqslant N$, for some $Ninmathbb N$. So, $lvert x_N+1rvertleqslant clvert x_nrvert$, $lvert x_N+1rvertleqslant clvert x_Nrvert$, $lvert x_N+2rvertleqslant c^2lvert x_Nrvert$, and so on. Therefore, by the squeeze theorem, $lim_nx_n=0$.
answered Mar 29 at 15:25
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
I do not see the assumption $limsup_nleftlvertfracx_n+1x_nrightrvert<1$ in the question.
$endgroup$
– Martin R
Mar 29 at 15:28
$begingroup$
I assumed that that's wht the OP meant by the expression $lim_nsup$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:29
$begingroup$
@MartinR I was actually thinking that $limsup_n left|fracx_n+1x_nright| < 1$ follows from $exists limsup_nfracx_n+1x_n < 1$. Do you mean that assumption is wrong?
$endgroup$
– roman
Mar 29 at 15:43
$begingroup$
@roman: Again, $x_n = (-2)^n$ is a counter-example.
$endgroup$
– Martin R
Mar 29 at 15:44
$begingroup$
@roman I suggest that you edit your question, replacing $limsup_nfracx_n+1x_n$ with $limsup_nleftlvertfracx_n+1x_nrightrvert$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:46
|
show 1 more comment
$begingroup$
The one weak step is when you go from the limit of the ratio $left|fracx_n+1x_nright| < 1$ to the statement that $|x_n|$ eventually becomes monotone decreasing.
That step is true, but at the level of a proof, I think you need to justify why it must hold.
answered Mar 29 at 15:25
Mark FischlerMark Fischler
34k12552
$endgroup$
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
The statement is wrong, a counter-example is $x_n = (-2)^n$ with
$limsup_ntoinftyx_n+1over x_n = -2 < 1$.
There are several flaws in the proof:
$y_n = fracx_n+1x_n$ is not necessarily convergent.
$limsup y_n = L$ does not imply $limsup |y_n| = |L|$, a counter-example is $y_n = -2 + (-1)^n$.
$L < 1$ does not imply $|L| in [0, 1)$, a counter-example is $L=2$.- If $|x_n|$ is monotone decreasing then it is convergent, but not necessarily against zero.
You'll have to require that $limsup_ntoinfty left|x_n+1over x_nright| < 1$ instead. Then (as José already pointed out),
$$
0 le |x_n| le c^n-n_0 |x_n_0|
$$
for some $c in [0, 1)$ and $n ge n_0$, and $x_n to 0 $ follows.
answered Mar 29 at 15:59
Martin RMartin R
30.7k33560
$endgroup$
add a comment |
$begingroup$
The statement is wrong, a counter-example is $x_n = (-2)^n$ with
$limsup_ntoinftyx_n+1over x_n = -2 < 1$.
There are several flaws in the proof:
$y_n = fracx_n+1x_n$ is not necessarily convergent.
$limsup y_n = L$ does not imply $limsup |y_n| = |L|$, a counter-example is $y_n = -2 + (-1)^n$.
$L < 1$ does not imply $|L| in [0, 1)$, a counter-example is $L=2$.- If $|x_n|$ is monotone decreasing then it is convergent, but not necessarily against zero.
You'll have to require that $limsup_ntoinfty left|x_n+1over x_nright| < 1$ instead. Then (as José already pointed out),
$$
0 le |x_n| le c^n-n_0 |x_n_0|
$$
for some $c in [0, 1)$ and $n ge n_0$, and $x_n to 0 $ follows.
answered Mar 29 at 15:59
Martin RMartin R
30.7k33560
$endgroup$
add a comment |
$begingroup$
The statement is wrong, a counter-example is $x_n = (-2)^n$ with
$limsup_ntoinftyx_n+1over x_n = -2 < 1$.
There are several flaws in the proof:
$y_n = fracx_n+1x_n$ is not necessarily convergent.
$limsup y_n = L$ does not imply $limsup |y_n| = |L|$, a counter-example is $y_n = -2 + (-1)^n$.
$L < 1$ does not imply $|L| in [0, 1)$, a counter-example is $L=2$.- If $|x_n|$ is monotone decreasing then it is convergent, but not necessarily against zero.
You'll have to require that $limsup_ntoinfty left|x_n+1over x_nright| < 1$ instead. Then (as José already pointed out),
$$
0 le |x_n| le c^n-n_0 |x_n_0|
$$
for some $c in [0, 1)$ and $n ge n_0$, and $x_n to 0 $ follows.
answered Mar 29 at 15:59
Martin RMartin R
30.7k33560
$endgroup$
The statement is wrong, a counter-example is $x_n = (-2)^n$ with
$limsup_ntoinftyx_n+1over x_n = -2 < 1$.
There are several flaws in the proof:
$y_n = fracx_n+1x_n$ is not necessarily convergent.
$limsup y_n = L$ does not imply $limsup |y_n| = |L|$, a counter-example is $y_n = -2 + (-1)^n$.
$L < 1$ does not imply $|L| in [0, 1)$, a counter-example is $L=2$.- If $|x_n|$ is monotone decreasing then it is convergent, but not necessarily against zero.
You'll have to require that $limsup_ntoinfty left|x_n+1over x_nright| < 1$ instead. Then (as José already pointed out),
$$
0 le |x_n| le c^n-n_0 |x_n_0|
$$
for some $c in [0, 1)$ and $n ge n_0$, and $x_n to 0 $ follows.
answered Mar 29 at 15:59
Martin RMartin R
30.7k33560
answered Mar 29 at 15:59
Martin RMartin R
30.7k33560
answered Mar 29 at 15:59
Martin RMartin R
30.7k33560
answered Mar 29 at 15:59
Martin RMartin R
30.7k33560
30.7k33560
add a comment |
add a comment |
$begingroup$
Your proof is wrong because you only proved that, if $n$ is large enough, then $lvert x_n+1rvert<lvert x_nrvert$ and that is not enough to deduce that $lim_ntoinftyx_n=0$, even if you mention the monotone convergence theorem, which has nothing to do with this.
Since $limsup_nleftlvertfracx_n+1x_nrightrvert<1$, if $cinleft(limsup_nleftlvertfracx_n+1x_nrightrvert,1right)$, then $leftlvertfracx_n+1x_nrightrvert<c$, if $ngeqslant N$, for some $Ninmathbb N$. So, $lvert x_N+1rvertleqslant clvert x_nrvert$, $lvert x_N+1rvertleqslant clvert x_Nrvert$, $lvert x_N+2rvertleqslant c^2lvert x_Nrvert$, and so on. Therefore, by the squeeze theorem, $lim_nx_n=0$.
answered Mar 29 at 15:25
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
I do not see the assumption $limsup_nleftlvertfracx_n+1x_nrightrvert<1$ in the question.
$endgroup$
– Martin R
Mar 29 at 15:28
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I assumed that that's wht the OP meant by the expression $lim_nsup$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:29
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@MartinR I was actually thinking that $limsup_n left|fracx_n+1x_nright| < 1$ follows from $exists limsup_nfracx_n+1x_n < 1$. Do you mean that assumption is wrong?
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– roman
Mar 29 at 15:43
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@roman: Again, $x_n = (-2)^n$ is a counter-example.
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– Martin R
Mar 29 at 15:44
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@roman I suggest that you edit your question, replacing $limsup_nfracx_n+1x_n$ with $limsup_nleftlvertfracx_n+1x_nrightrvert$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:46
|
show 1 more comment
$begingroup$
Your proof is wrong because you only proved that, if $n$ is large enough, then $lvert x_n+1rvert<lvert x_nrvert$ and that is not enough to deduce that $lim_ntoinftyx_n=0$, even if you mention the monotone convergence theorem, which has nothing to do with this.
Since $limsup_nleftlvertfracx_n+1x_nrightrvert<1$, if $cinleft(limsup_nleftlvertfracx_n+1x_nrightrvert,1right)$, then $leftlvertfracx_n+1x_nrightrvert<c$, if $ngeqslant N$, for some $Ninmathbb N$. So, $lvert x_N+1rvertleqslant clvert x_nrvert$, $lvert x_N+1rvertleqslant clvert x_Nrvert$, $lvert x_N+2rvertleqslant c^2lvert x_Nrvert$, and so on. Therefore, by the squeeze theorem, $lim_nx_n=0$.
answered Mar 29 at 15:25
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
I do not see the assumption $limsup_nleftlvertfracx_n+1x_nrightrvert<1$ in the question.
$endgroup$
– Martin R
Mar 29 at 15:28
$begingroup$
I assumed that that's wht the OP meant by the expression $lim_nsup$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:29
$begingroup$
@MartinR I was actually thinking that $limsup_n left|fracx_n+1x_nright| < 1$ follows from $exists limsup_nfracx_n+1x_n < 1$. Do you mean that assumption is wrong?
$endgroup$
– roman
Mar 29 at 15:43
$begingroup$
@roman: Again, $x_n = (-2)^n$ is a counter-example.
$endgroup$
– Martin R
Mar 29 at 15:44
$begingroup$
@roman I suggest that you edit your question, replacing $limsup_nfracx_n+1x_n$ with $limsup_nleftlvertfracx_n+1x_nrightrvert$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:46
|
show 1 more comment
$begingroup$
Your proof is wrong because you only proved that, if $n$ is large enough, then $lvert x_n+1rvert<lvert x_nrvert$ and that is not enough to deduce that $lim_ntoinftyx_n=0$, even if you mention the monotone convergence theorem, which has nothing to do with this.
Since $limsup_nleftlvertfracx_n+1x_nrightrvert<1$, if $cinleft(limsup_nleftlvertfracx_n+1x_nrightrvert,1right)$, then $leftlvertfracx_n+1x_nrightrvert<c$, if $ngeqslant N$, for some $Ninmathbb N$. So, $lvert x_N+1rvertleqslant clvert x_nrvert$, $lvert x_N+1rvertleqslant clvert x_Nrvert$, $lvert x_N+2rvertleqslant c^2lvert x_Nrvert$, and so on. Therefore, by the squeeze theorem, $lim_nx_n=0$.
answered Mar 29 at 15:25
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
Your proof is wrong because you only proved that, if $n$ is large enough, then $lvert x_n+1rvert<lvert x_nrvert$ and that is not enough to deduce that $lim_ntoinftyx_n=0$, even if you mention the monotone convergence theorem, which has nothing to do with this.
Since $limsup_nleftlvertfracx_n+1x_nrightrvert<1$, if $cinleft(limsup_nleftlvertfracx_n+1x_nrightrvert,1right)$, then $leftlvertfracx_n+1x_nrightrvert<c$, if $ngeqslant N$, for some $Ninmathbb N$. So, $lvert x_N+1rvertleqslant clvert x_nrvert$, $lvert x_N+1rvertleqslant clvert x_Nrvert$, $lvert x_N+2rvertleqslant c^2lvert x_Nrvert$, and so on. Therefore, by the squeeze theorem, $lim_nx_n=0$.
answered Mar 29 at 15:25
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 29 at 15:25
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 29 at 15:25
José Carlos SantosJosé Carlos Santos
173k23133241
answered Mar 29 at 15:25
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
$begingroup$
I do not see the assumption $limsup_nleftlvertfracx_n+1x_nrightrvert<1$ in the question.
$endgroup$
– Martin R
Mar 29 at 15:28
$begingroup$
I assumed that that's wht the OP meant by the expression $lim_nsup$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:29
$begingroup$
@MartinR I was actually thinking that $limsup_n left|fracx_n+1x_nright| < 1$ follows from $exists limsup_nfracx_n+1x_n < 1$. Do you mean that assumption is wrong?
$endgroup$
– roman
Mar 29 at 15:43
$begingroup$
@roman: Again, $x_n = (-2)^n$ is a counter-example.
$endgroup$
– Martin R
Mar 29 at 15:44
$begingroup$
@roman I suggest that you edit your question, replacing $limsup_nfracx_n+1x_n$ with $limsup_nleftlvertfracx_n+1x_nrightrvert$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:46
|
show 1 more comment
$begingroup$
I do not see the assumption $limsup_nleftlvertfracx_n+1x_nrightrvert<1$ in the question.
$endgroup$
– Martin R
Mar 29 at 15:28
$begingroup$
I assumed that that's wht the OP meant by the expression $lim_nsup$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:29
$begingroup$
@MartinR I was actually thinking that $limsup_n left|fracx_n+1x_nright| < 1$ follows from $exists limsup_nfracx_n+1x_n < 1$. Do you mean that assumption is wrong?
$endgroup$
– roman
Mar 29 at 15:43
$begingroup$
@roman: Again, $x_n = (-2)^n$ is a counter-example.
$endgroup$
– Martin R
Mar 29 at 15:44
$begingroup$
@roman I suggest that you edit your question, replacing $limsup_nfracx_n+1x_n$ with $limsup_nleftlvertfracx_n+1x_nrightrvert$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:46
$begingroup$
I do not see the assumption $limsup_nleftlvertfracx_n+1x_nrightrvert<1$ in the question.
$endgroup$
– Martin R
Mar 29 at 15:28
$begingroup$
I do not see the assumption $limsup_nleftlvertfracx_n+1x_nrightrvert<1$ in the question.
$endgroup$
– Martin R
Mar 29 at 15:28
$begingroup$
I assumed that that's wht the OP meant by the expression $lim_nsup$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:29
$begingroup$
I assumed that that's wht the OP meant by the expression $lim_nsup$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:29
$begingroup$
@MartinR I was actually thinking that $limsup_n left|fracx_n+1x_nright| < 1$ follows from $exists limsup_nfracx_n+1x_n < 1$. Do you mean that assumption is wrong?
$endgroup$
– roman
Mar 29 at 15:43
$begingroup$
@MartinR I was actually thinking that $limsup_n left|fracx_n+1x_nright| < 1$ follows from $exists limsup_nfracx_n+1x_n < 1$. Do you mean that assumption is wrong?
$endgroup$
– roman
Mar 29 at 15:43
$begingroup$
@roman: Again, $x_n = (-2)^n$ is a counter-example.
$endgroup$
– Martin R
Mar 29 at 15:44
$begingroup$
@roman: Again, $x_n = (-2)^n$ is a counter-example.
$endgroup$
– Martin R
Mar 29 at 15:44
$begingroup$
@roman I suggest that you edit your question, replacing $limsup_nfracx_n+1x_n$ with $limsup_nleftlvertfracx_n+1x_nrightrvert$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:46
$begingroup$
@roman I suggest that you edit your question, replacing $limsup_nfracx_n+1x_n$ with $limsup_nleftlvertfracx_n+1x_nrightrvert$.
$endgroup$
– José Carlos Santos
Mar 29 at 15:46
|
show 1 more comment
$begingroup$
The one weak step is when you go from the limit of the ratio $left|fracx_n+1x_nright| < 1$ to the statement that $|x_n|$ eventually becomes monotone decreasing.
That step is true, but at the level of a proof, I think you need to justify why it must hold.
answered Mar 29 at 15:25
Mark FischlerMark Fischler
34k12552
$endgroup$
add a comment |
$begingroup$
The one weak step is when you go from the limit of the ratio $left|fracx_n+1x_nright| < 1$ to the statement that $|x_n|$ eventually becomes monotone decreasing.
That step is true, but at the level of a proof, I think you need to justify why it must hold.
answered Mar 29 at 15:25
Mark FischlerMark Fischler
34k12552
$endgroup$
add a comment |
$begingroup$
The one weak step is when you go from the limit of the ratio $left|fracx_n+1x_nright| < 1$ to the statement that $|x_n|$ eventually becomes monotone decreasing.
That step is true, but at the level of a proof, I think you need to justify why it must hold.
answered Mar 29 at 15:25
Mark FischlerMark Fischler
34k12552
$endgroup$
The one weak step is when you go from the limit of the ratio $left|fracx_n+1x_nright| < 1$ to the statement that $|x_n|$ eventually becomes monotone decreasing.
That step is true, but at the level of a proof, I think you need to justify why it must hold.
answered Mar 29 at 15:25
Mark FischlerMark Fischler
34k12552
answered Mar 29 at 15:25
Mark FischlerMark Fischler
34k12552
answered Mar 29 at 15:25
Mark FischlerMark Fischler
34k12552
answered Mar 29 at 15:25
Mark FischlerMark Fischler
34k12552
34k12552
add a comment |
add a comment |
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$begingroup$
That is wrong, a counter-example is $x_n = (-2)^n$ with $fracx_n+1x_n = -2 < 1$. Or do you assume $x_n > 0$?
$endgroup$
– Martin R
Mar 29 at 15:24
$begingroup$
@MartinR Nope, I did not assume that. By your counterexample, the above is indeed wrong.
$endgroup$
– roman
Mar 29 at 15:30
$begingroup$
Note that $L < 1$ does not imply $|L| in [0, 1)$.
$endgroup$
– Martin R
Mar 29 at 15:32
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Your last edit changes the problem completely so that my answer does not apply anymore. I suggest that you revert to the original version and ask a new question for the updated problem. Alternatively please unaccept my answer so that I can delete it.
$endgroup$
– Martin R
Mar 29 at 17:21
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@MartinR But your answer is the one I was looking for, why do you want to delete it? As it does apply to the problem. Thanks to you and José Carlos Santos I've figured out the problem was incorrectly stated in the book. And that is the only thing reflected in the update to the question along with some thought from your answer. Of course, I can unaccept your answer if you still insist.
$endgroup$
– roman
Mar 29 at 17:27