A question on sub-sequential limits of a partial sum of a convergent sequence.Let $a = liminf x_n$, $b =...
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A question on sub-sequential limits of a partial sum of a convergent sequence.
Let $a = liminf x_n$, $b = limsup x_n$. If $lim (x_{n+1} - x_n) = 0$ then every element of $(a,b)$ is a subsequential limit of $(x_n)$A series converges absolutely if and only if every subseries convergesQuestion about bounded sequence with two sub-sequential limits.Convergence of sequence of real numbers and (sub)subsequenceProve that a bounded sequence has two convergent subsequences.Two subsequences with different limits $implies$ not convergentShow that if all convergent sub-sequences of a sequence ${s_n}$ converge to 0 and ${s_n}$ is bounded, then ${s_n}$ converges to $0$Using sequential criterion for limits find the limitsFind the subsequential limits for ${x_n}=left{1,{1over 10},{2over 10},cdots,{9over 10},{1over 10^2},cdots{10^n-1over 10^n},cdotsright}$Proof verification. If $x_n$ is a monotone sequence and it has a convergent subsequence $x_{n_k}$, then $x_n$ is convergent to the same limit.Proof verification. Show $x_{n+1} = 1 + {bover x_n}$ diverges if $b < -{1over 4}$
$begingroup$
Let:
$$
lim_{ntoinfty} x_n = 0
$$
Define a sequence ${S_n}$:
$$
S_n = x_1 + x_2 + cdots +x_n\
ninBbb N
$$
Is it possible for $S_n$ to have only two sub-sequential limits if:
$x_n$ is in reals, $x_n in Bbb R$
$x_n$ is in complex numbers, $x_n inBbb C$
I've started with supposing that $S_n$ has two distinct finite sub-sequential limits, namely:
$$
begin{cases}
lim_{n_k to infty} S_{n_k} = a\
lim_{n_p to infty} S_{n_p} = b\
a, b in Bbb R\
ane b
end{cases}
$$
I've shown earlier that:
$$
lim_{ntoinfty} S_n = L in Bbb R implies lim_{ntoinfty}x_n = 0
$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$
lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0 \
lim_{n_p to infty} S_{n_p} = b implies lim_{n_p toinfty} x_{n_p} = 0
$$
By initial conditions we are given that:
$$
lim_{ntoinfty}x_n = 0
$$
Thus any subsequence of $x_n$ converges to the same limit, namely:
$$
lim_{n_k toinfty}x_{n_k} = lim_{n_p toinfty}x_{n_p} = lim_{n toinfty}x_{n} = 0
$$
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$. In general it seems like $S_n$ may have infinitely many sub-sequential limits and even in this case $x_n$ will still converge to $0$.
My first question is about the correctness of the argument above. Is the above enough to consider the question answered? If not what would be the right way?
The second question is how do I handle the complex numbers case? I've never taken any course on complex analysis and only have some basic understanding of the complex plane.
real-analysis calculus sequences-and-series limits proof-verification
$endgroup$
add a comment |
$begingroup$
Let:
$$
lim_{ntoinfty} x_n = 0
$$
Define a sequence ${S_n}$:
$$
S_n = x_1 + x_2 + cdots +x_n\
ninBbb N
$$
Is it possible for $S_n$ to have only two sub-sequential limits if:
$x_n$ is in reals, $x_n in Bbb R$
$x_n$ is in complex numbers, $x_n inBbb C$
I've started with supposing that $S_n$ has two distinct finite sub-sequential limits, namely:
$$
begin{cases}
lim_{n_k to infty} S_{n_k} = a\
lim_{n_p to infty} S_{n_p} = b\
a, b in Bbb R\
ane b
end{cases}
$$
I've shown earlier that:
$$
lim_{ntoinfty} S_n = L in Bbb R implies lim_{ntoinfty}x_n = 0
$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$
lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0 \
lim_{n_p to infty} S_{n_p} = b implies lim_{n_p toinfty} x_{n_p} = 0
$$
By initial conditions we are given that:
$$
lim_{ntoinfty}x_n = 0
$$
Thus any subsequence of $x_n$ converges to the same limit, namely:
$$
lim_{n_k toinfty}x_{n_k} = lim_{n_p toinfty}x_{n_p} = lim_{n toinfty}x_{n} = 0
$$
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$. In general it seems like $S_n$ may have infinitely many sub-sequential limits and even in this case $x_n$ will still converge to $0$.
My first question is about the correctness of the argument above. Is the above enough to consider the question answered? If not what would be the right way?
The second question is how do I handle the complex numbers case? I've never taken any course on complex analysis and only have some basic understanding of the complex plane.
real-analysis calculus sequences-and-series limits proof-verification
$endgroup$
$begingroup$
This question might be useful: Let $a = liminf x_n$, $b = limsup x_n$. If $lim (x_{n+1} - x_n) = 0$ then every element of $(a,b)$ is a subsequential limit of $(x_n)$.
$endgroup$
– Martin Sleziak
Mar 13 at 17:05
$begingroup$
FYI - wherever you have $lim_{n_k to infty}$, what you actually mean is $lim_{kto infty}$. It is $k$ that is varying. $n_k$ only changes as function of $k$.
$endgroup$
– Paul Sinclair
Mar 14 at 1:43
add a comment |
$begingroup$
Let:
$$
lim_{ntoinfty} x_n = 0
$$
Define a sequence ${S_n}$:
$$
S_n = x_1 + x_2 + cdots +x_n\
ninBbb N
$$
Is it possible for $S_n$ to have only two sub-sequential limits if:
$x_n$ is in reals, $x_n in Bbb R$
$x_n$ is in complex numbers, $x_n inBbb C$
I've started with supposing that $S_n$ has two distinct finite sub-sequential limits, namely:
$$
begin{cases}
lim_{n_k to infty} S_{n_k} = a\
lim_{n_p to infty} S_{n_p} = b\
a, b in Bbb R\
ane b
end{cases}
$$
I've shown earlier that:
$$
lim_{ntoinfty} S_n = L in Bbb R implies lim_{ntoinfty}x_n = 0
$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$
lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0 \
lim_{n_p to infty} S_{n_p} = b implies lim_{n_p toinfty} x_{n_p} = 0
$$
By initial conditions we are given that:
$$
lim_{ntoinfty}x_n = 0
$$
Thus any subsequence of $x_n$ converges to the same limit, namely:
$$
lim_{n_k toinfty}x_{n_k} = lim_{n_p toinfty}x_{n_p} = lim_{n toinfty}x_{n} = 0
$$
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$. In general it seems like $S_n$ may have infinitely many sub-sequential limits and even in this case $x_n$ will still converge to $0$.
My first question is about the correctness of the argument above. Is the above enough to consider the question answered? If not what would be the right way?
The second question is how do I handle the complex numbers case? I've never taken any course on complex analysis and only have some basic understanding of the complex plane.
real-analysis calculus sequences-and-series limits proof-verification
$endgroup$
Let:
$$
lim_{ntoinfty} x_n = 0
$$
Define a sequence ${S_n}$:
$$
S_n = x_1 + x_2 + cdots +x_n\
ninBbb N
$$
Is it possible for $S_n$ to have only two sub-sequential limits if:
$x_n$ is in reals, $x_n in Bbb R$
$x_n$ is in complex numbers, $x_n inBbb C$
I've started with supposing that $S_n$ has two distinct finite sub-sequential limits, namely:
$$
begin{cases}
lim_{n_k to infty} S_{n_k} = a\
lim_{n_p to infty} S_{n_p} = b\
a, b in Bbb R\
ane b
end{cases}
$$
I've shown earlier that:
$$
lim_{ntoinfty} S_n = L in Bbb R implies lim_{ntoinfty}x_n = 0
$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$
lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0 \
lim_{n_p to infty} S_{n_p} = b implies lim_{n_p toinfty} x_{n_p} = 0
$$
By initial conditions we are given that:
$$
lim_{ntoinfty}x_n = 0
$$
Thus any subsequence of $x_n$ converges to the same limit, namely:
$$
lim_{n_k toinfty}x_{n_k} = lim_{n_p toinfty}x_{n_p} = lim_{n toinfty}x_{n} = 0
$$
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$. In general it seems like $S_n$ may have infinitely many sub-sequential limits and even in this case $x_n$ will still converge to $0$.
My first question is about the correctness of the argument above. Is the above enough to consider the question answered? If not what would be the right way?
The second question is how do I handle the complex numbers case? I've never taken any course on complex analysis and only have some basic understanding of the complex plane.
real-analysis calculus sequences-and-series limits proof-verification
real-analysis calculus sequences-and-series limits proof-verification
asked Mar 13 at 16:28
romanroman
2,39321225
2,39321225
$begingroup$
This question might be useful: Let $a = liminf x_n$, $b = limsup x_n$. If $lim (x_{n+1} - x_n) = 0$ then every element of $(a,b)$ is a subsequential limit of $(x_n)$.
$endgroup$
– Martin Sleziak
Mar 13 at 17:05
$begingroup$
FYI - wherever you have $lim_{n_k to infty}$, what you actually mean is $lim_{kto infty}$. It is $k$ that is varying. $n_k$ only changes as function of $k$.
$endgroup$
– Paul Sinclair
Mar 14 at 1:43
add a comment |
$begingroup$
This question might be useful: Let $a = liminf x_n$, $b = limsup x_n$. If $lim (x_{n+1} - x_n) = 0$ then every element of $(a,b)$ is a subsequential limit of $(x_n)$.
$endgroup$
– Martin Sleziak
Mar 13 at 17:05
$begingroup$
FYI - wherever you have $lim_{n_k to infty}$, what you actually mean is $lim_{kto infty}$. It is $k$ that is varying. $n_k$ only changes as function of $k$.
$endgroup$
– Paul Sinclair
Mar 14 at 1:43
$begingroup$
This question might be useful: Let $a = liminf x_n$, $b = limsup x_n$. If $lim (x_{n+1} - x_n) = 0$ then every element of $(a,b)$ is a subsequential limit of $(x_n)$.
$endgroup$
– Martin Sleziak
Mar 13 at 17:05
$begingroup$
This question might be useful: Let $a = liminf x_n$, $b = limsup x_n$. If $lim (x_{n+1} - x_n) = 0$ then every element of $(a,b)$ is a subsequential limit of $(x_n)$.
$endgroup$
– Martin Sleziak
Mar 13 at 17:05
$begingroup$
FYI - wherever you have $lim_{n_k to infty}$, what you actually mean is $lim_{kto infty}$. It is $k$ that is varying. $n_k$ only changes as function of $k$.
$endgroup$
– Paul Sinclair
Mar 14 at 1:43
$begingroup$
FYI - wherever you have $lim_{n_k to infty}$, what you actually mean is $lim_{kto infty}$. It is $k$ that is varying. $n_k$ only changes as function of $k$.
$endgroup$
– Paul Sinclair
Mar 14 at 1:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I've shown earlier that:
$$lim_{n to infty} S_n = L in Bbb R implies lim_{n toinfty} x_n = 0$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0$$
What you had shown earlier can be more generally stated as: For a sequence ${y_k}$, $$lim_{Ktoinfty} sum_{k=1}^K y_k = L in Bbb R implies lim_{k to infty} y_k = 0$$
where I've changed all the names to protect the innocent. I.e., to remind you that the sequences are not necessarily the same as the ones you are talking about in the rest of your post.
Now, if you want to prove $lim_{n_k toinfty} x_{n_k} = 0$ with this, then $y_k = x_{n_k}$, so what you need to know is that $lim_{Ktoinfty}sum_{k=1}^K x_{n_k}$ converges. But you have $lim_{Ktoinfty} S_{n_K} = 0$. Note that
$$sum_{k=1}^K x_{n_k} = x_{n_1} + x_{n_2} + x_{n_3} + ... + x_{n_K}$$
while
$$S_{n_k} = x_1 + x_2 + x_3 + ... + x_{n_k}$$
These are not the same. So you need to do some more work here.
Thus any subsequence of $x_n$ converges to the same limit
Not "any". You didn't start out by assuming that $x_{n_k}$ or $x_{n_p}$ were arbitrary subsequences. Even if you correct the earlier argument (and it is a relatively easy fix), all you've shown is that $lim_k x_{n_k} = 0$ for convergent subsequences of $S_n$.
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$.
Where did that come from? It doesn't follow logically from what you've said before at all.
A more general problem is that this "proof" is not even headed in a direction you need to go. You went through this and then declared in the end that $lim_n x_n = 0$. but that is a statement you already knew. Arriving at it means you've gone nowhere.
What you need to consider is this: If subsequences of $S_n$ converge to both $a$ and $b$ with $b > a$, this means that when you start summing $x_n$, eventually the sum will get close to $a$. As you keep summing, it may drift away from $a$, but eventually, it'll end up coming back to $a$ over and over again, all the way to infinity. But the same can be said about $b$. The sum must keep coming near to it as well. So it has to go near $a$, then go near $b$, then later it must go near $a$ again, and then back to $b$, and so on. It must oscilate back and forth between $a$ and $b$ forever. but since $lim_n x_n = 0$, it has to keep doing this oscilation in smaller and smaller steps. So what happens if $c$ is between $a$ and $b$?
As for $Bbb C$, there is nothing in this problem that makes any use of the algebraic properties of $Bbb C$. As far as you need to be concerned here, $Bbb C$ is just another name for the plane $Bbb R^2$. Can you see how a second dimension might offer a way for $S_n$ to get from $a$ to $b$ and back without covering the same country in the middle every time?
$endgroup$
$begingroup$
Thank you for such a detailed answer. There is still a thing I don't get. If we assume there are two subsequences of $S_n$, namely $S_{n_k}$ and $S_{n_p}$ which both converge to different limits, why would the sum of $x_n$ come close to $a$ and then to $b$ back and forth? Could you please elaborate on that? Based on the rest of your answer, it looks like $c$ is going to be between $a$ and $b$ with the distance between them converging to $0$. Which would eventually mean $a = c = b$, meaning that it's not possible for $S_n$ to have two separate sub-sequential limits. Right?
$endgroup$
– roman
Mar 14 at 8:42
$begingroup$
As for the case of complex plane there must be a circle with some radius $b-a$ for $b>a$ which keeps shrinking into a single point $c$ as $ntoinfty$, which it terms also mean all subsequences of $S_n$ must converge to the same number if I understood everything correctly
$endgroup$
– roman
Mar 14 at 8:56
1
$begingroup$
That $S_{n_k} to a$ means that there are infinite number of values of $n$ (the $n_k$ for high enough $k$) with $S_n$ close to $a$. In particular, for any $N$, there has to be $n > N$ with $S_n$ close to $a$. The same is true for $b$. So find an $n$ with $S_n$ close to $a$. There is therefore some higher $n$ with $S_n$ close to $b$. And some higher $n$ yet with $S_n$ close to $a$, and so on.
$endgroup$
– Paul Sinclair
Mar 14 at 16:01
$begingroup$
Second, $a$ and $b$ are fixed points. They do not depend on $n$. They do not get closer together, they do not shrink down to some point. They are simply two goal posts that $S_n$ moves back and forth between in steps as $n$ increases. The size of those steps is $S_n - S_{n-1} = x_n$, so the steps eventually become smaller and smaller. So the distance $S_n$ can maintain from any point $c$ between $a$ and $b$ also gets smaller and smaller. However, in 2 dimensions, $S_n$ is not constrained to cover the same territory on every trip.
$endgroup$
– Paul Sinclair
Mar 14 at 16:08
add a comment |
Your Answer
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1 Answer
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$begingroup$
I've shown earlier that:
$$lim_{n to infty} S_n = L in Bbb R implies lim_{n toinfty} x_n = 0$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0$$
What you had shown earlier can be more generally stated as: For a sequence ${y_k}$, $$lim_{Ktoinfty} sum_{k=1}^K y_k = L in Bbb R implies lim_{k to infty} y_k = 0$$
where I've changed all the names to protect the innocent. I.e., to remind you that the sequences are not necessarily the same as the ones you are talking about in the rest of your post.
Now, if you want to prove $lim_{n_k toinfty} x_{n_k} = 0$ with this, then $y_k = x_{n_k}$, so what you need to know is that $lim_{Ktoinfty}sum_{k=1}^K x_{n_k}$ converges. But you have $lim_{Ktoinfty} S_{n_K} = 0$. Note that
$$sum_{k=1}^K x_{n_k} = x_{n_1} + x_{n_2} + x_{n_3} + ... + x_{n_K}$$
while
$$S_{n_k} = x_1 + x_2 + x_3 + ... + x_{n_k}$$
These are not the same. So you need to do some more work here.
Thus any subsequence of $x_n$ converges to the same limit
Not "any". You didn't start out by assuming that $x_{n_k}$ or $x_{n_p}$ were arbitrary subsequences. Even if you correct the earlier argument (and it is a relatively easy fix), all you've shown is that $lim_k x_{n_k} = 0$ for convergent subsequences of $S_n$.
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$.
Where did that come from? It doesn't follow logically from what you've said before at all.
A more general problem is that this "proof" is not even headed in a direction you need to go. You went through this and then declared in the end that $lim_n x_n = 0$. but that is a statement you already knew. Arriving at it means you've gone nowhere.
What you need to consider is this: If subsequences of $S_n$ converge to both $a$ and $b$ with $b > a$, this means that when you start summing $x_n$, eventually the sum will get close to $a$. As you keep summing, it may drift away from $a$, but eventually, it'll end up coming back to $a$ over and over again, all the way to infinity. But the same can be said about $b$. The sum must keep coming near to it as well. So it has to go near $a$, then go near $b$, then later it must go near $a$ again, and then back to $b$, and so on. It must oscilate back and forth between $a$ and $b$ forever. but since $lim_n x_n = 0$, it has to keep doing this oscilation in smaller and smaller steps. So what happens if $c$ is between $a$ and $b$?
As for $Bbb C$, there is nothing in this problem that makes any use of the algebraic properties of $Bbb C$. As far as you need to be concerned here, $Bbb C$ is just another name for the plane $Bbb R^2$. Can you see how a second dimension might offer a way for $S_n$ to get from $a$ to $b$ and back without covering the same country in the middle every time?
$endgroup$
$begingroup$
Thank you for such a detailed answer. There is still a thing I don't get. If we assume there are two subsequences of $S_n$, namely $S_{n_k}$ and $S_{n_p}$ which both converge to different limits, why would the sum of $x_n$ come close to $a$ and then to $b$ back and forth? Could you please elaborate on that? Based on the rest of your answer, it looks like $c$ is going to be between $a$ and $b$ with the distance between them converging to $0$. Which would eventually mean $a = c = b$, meaning that it's not possible for $S_n$ to have two separate sub-sequential limits. Right?
$endgroup$
– roman
Mar 14 at 8:42
$begingroup$
As for the case of complex plane there must be a circle with some radius $b-a$ for $b>a$ which keeps shrinking into a single point $c$ as $ntoinfty$, which it terms also mean all subsequences of $S_n$ must converge to the same number if I understood everything correctly
$endgroup$
– roman
Mar 14 at 8:56
1
$begingroup$
That $S_{n_k} to a$ means that there are infinite number of values of $n$ (the $n_k$ for high enough $k$) with $S_n$ close to $a$. In particular, for any $N$, there has to be $n > N$ with $S_n$ close to $a$. The same is true for $b$. So find an $n$ with $S_n$ close to $a$. There is therefore some higher $n$ with $S_n$ close to $b$. And some higher $n$ yet with $S_n$ close to $a$, and so on.
$endgroup$
– Paul Sinclair
Mar 14 at 16:01
$begingroup$
Second, $a$ and $b$ are fixed points. They do not depend on $n$. They do not get closer together, they do not shrink down to some point. They are simply two goal posts that $S_n$ moves back and forth between in steps as $n$ increases. The size of those steps is $S_n - S_{n-1} = x_n$, so the steps eventually become smaller and smaller. So the distance $S_n$ can maintain from any point $c$ between $a$ and $b$ also gets smaller and smaller. However, in 2 dimensions, $S_n$ is not constrained to cover the same territory on every trip.
$endgroup$
– Paul Sinclair
Mar 14 at 16:08
add a comment |
$begingroup$
I've shown earlier that:
$$lim_{n to infty} S_n = L in Bbb R implies lim_{n toinfty} x_n = 0$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0$$
What you had shown earlier can be more generally stated as: For a sequence ${y_k}$, $$lim_{Ktoinfty} sum_{k=1}^K y_k = L in Bbb R implies lim_{k to infty} y_k = 0$$
where I've changed all the names to protect the innocent. I.e., to remind you that the sequences are not necessarily the same as the ones you are talking about in the rest of your post.
Now, if you want to prove $lim_{n_k toinfty} x_{n_k} = 0$ with this, then $y_k = x_{n_k}$, so what you need to know is that $lim_{Ktoinfty}sum_{k=1}^K x_{n_k}$ converges. But you have $lim_{Ktoinfty} S_{n_K} = 0$. Note that
$$sum_{k=1}^K x_{n_k} = x_{n_1} + x_{n_2} + x_{n_3} + ... + x_{n_K}$$
while
$$S_{n_k} = x_1 + x_2 + x_3 + ... + x_{n_k}$$
These are not the same. So you need to do some more work here.
Thus any subsequence of $x_n$ converges to the same limit
Not "any". You didn't start out by assuming that $x_{n_k}$ or $x_{n_p}$ were arbitrary subsequences. Even if you correct the earlier argument (and it is a relatively easy fix), all you've shown is that $lim_k x_{n_k} = 0$ for convergent subsequences of $S_n$.
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$.
Where did that come from? It doesn't follow logically from what you've said before at all.
A more general problem is that this "proof" is not even headed in a direction you need to go. You went through this and then declared in the end that $lim_n x_n = 0$. but that is a statement you already knew. Arriving at it means you've gone nowhere.
What you need to consider is this: If subsequences of $S_n$ converge to both $a$ and $b$ with $b > a$, this means that when you start summing $x_n$, eventually the sum will get close to $a$. As you keep summing, it may drift away from $a$, but eventually, it'll end up coming back to $a$ over and over again, all the way to infinity. But the same can be said about $b$. The sum must keep coming near to it as well. So it has to go near $a$, then go near $b$, then later it must go near $a$ again, and then back to $b$, and so on. It must oscilate back and forth between $a$ and $b$ forever. but since $lim_n x_n = 0$, it has to keep doing this oscilation in smaller and smaller steps. So what happens if $c$ is between $a$ and $b$?
As for $Bbb C$, there is nothing in this problem that makes any use of the algebraic properties of $Bbb C$. As far as you need to be concerned here, $Bbb C$ is just another name for the plane $Bbb R^2$. Can you see how a second dimension might offer a way for $S_n$ to get from $a$ to $b$ and back without covering the same country in the middle every time?
$endgroup$
$begingroup$
Thank you for such a detailed answer. There is still a thing I don't get. If we assume there are two subsequences of $S_n$, namely $S_{n_k}$ and $S_{n_p}$ which both converge to different limits, why would the sum of $x_n$ come close to $a$ and then to $b$ back and forth? Could you please elaborate on that? Based on the rest of your answer, it looks like $c$ is going to be between $a$ and $b$ with the distance between them converging to $0$. Which would eventually mean $a = c = b$, meaning that it's not possible for $S_n$ to have two separate sub-sequential limits. Right?
$endgroup$
– roman
Mar 14 at 8:42
$begingroup$
As for the case of complex plane there must be a circle with some radius $b-a$ for $b>a$ which keeps shrinking into a single point $c$ as $ntoinfty$, which it terms also mean all subsequences of $S_n$ must converge to the same number if I understood everything correctly
$endgroup$
– roman
Mar 14 at 8:56
1
$begingroup$
That $S_{n_k} to a$ means that there are infinite number of values of $n$ (the $n_k$ for high enough $k$) with $S_n$ close to $a$. In particular, for any $N$, there has to be $n > N$ with $S_n$ close to $a$. The same is true for $b$. So find an $n$ with $S_n$ close to $a$. There is therefore some higher $n$ with $S_n$ close to $b$. And some higher $n$ yet with $S_n$ close to $a$, and so on.
$endgroup$
– Paul Sinclair
Mar 14 at 16:01
$begingroup$
Second, $a$ and $b$ are fixed points. They do not depend on $n$. They do not get closer together, they do not shrink down to some point. They are simply two goal posts that $S_n$ moves back and forth between in steps as $n$ increases. The size of those steps is $S_n - S_{n-1} = x_n$, so the steps eventually become smaller and smaller. So the distance $S_n$ can maintain from any point $c$ between $a$ and $b$ also gets smaller and smaller. However, in 2 dimensions, $S_n$ is not constrained to cover the same territory on every trip.
$endgroup$
– Paul Sinclair
Mar 14 at 16:08
add a comment |
$begingroup$
I've shown earlier that:
$$lim_{n to infty} S_n = L in Bbb R implies lim_{n toinfty} x_n = 0$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0$$
What you had shown earlier can be more generally stated as: For a sequence ${y_k}$, $$lim_{Ktoinfty} sum_{k=1}^K y_k = L in Bbb R implies lim_{k to infty} y_k = 0$$
where I've changed all the names to protect the innocent. I.e., to remind you that the sequences are not necessarily the same as the ones you are talking about in the rest of your post.
Now, if you want to prove $lim_{n_k toinfty} x_{n_k} = 0$ with this, then $y_k = x_{n_k}$, so what you need to know is that $lim_{Ktoinfty}sum_{k=1}^K x_{n_k}$ converges. But you have $lim_{Ktoinfty} S_{n_K} = 0$. Note that
$$sum_{k=1}^K x_{n_k} = x_{n_1} + x_{n_2} + x_{n_3} + ... + x_{n_K}$$
while
$$S_{n_k} = x_1 + x_2 + x_3 + ... + x_{n_k}$$
These are not the same. So you need to do some more work here.
Thus any subsequence of $x_n$ converges to the same limit
Not "any". You didn't start out by assuming that $x_{n_k}$ or $x_{n_p}$ were arbitrary subsequences. Even if you correct the earlier argument (and it is a relatively easy fix), all you've shown is that $lim_k x_{n_k} = 0$ for convergent subsequences of $S_n$.
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$.
Where did that come from? It doesn't follow logically from what you've said before at all.
A more general problem is that this "proof" is not even headed in a direction you need to go. You went through this and then declared in the end that $lim_n x_n = 0$. but that is a statement you already knew. Arriving at it means you've gone nowhere.
What you need to consider is this: If subsequences of $S_n$ converge to both $a$ and $b$ with $b > a$, this means that when you start summing $x_n$, eventually the sum will get close to $a$. As you keep summing, it may drift away from $a$, but eventually, it'll end up coming back to $a$ over and over again, all the way to infinity. But the same can be said about $b$. The sum must keep coming near to it as well. So it has to go near $a$, then go near $b$, then later it must go near $a$ again, and then back to $b$, and so on. It must oscilate back and forth between $a$ and $b$ forever. but since $lim_n x_n = 0$, it has to keep doing this oscilation in smaller and smaller steps. So what happens if $c$ is between $a$ and $b$?
As for $Bbb C$, there is nothing in this problem that makes any use of the algebraic properties of $Bbb C$. As far as you need to be concerned here, $Bbb C$ is just another name for the plane $Bbb R^2$. Can you see how a second dimension might offer a way for $S_n$ to get from $a$ to $b$ and back without covering the same country in the middle every time?
$endgroup$
I've shown earlier that:
$$lim_{n to infty} S_n = L in Bbb R implies lim_{n toinfty} x_n = 0$$
So if we use this fact for both subsequences of $S_n$ we may obtain:
$$lim_{n_k to infty} S_{n_k} = a implies lim_{n_k toinfty} x_{n_k} = 0$$
What you had shown earlier can be more generally stated as: For a sequence ${y_k}$, $$lim_{Ktoinfty} sum_{k=1}^K y_k = L in Bbb R implies lim_{k to infty} y_k = 0$$
where I've changed all the names to protect the innocent. I.e., to remind you that the sequences are not necessarily the same as the ones you are talking about in the rest of your post.
Now, if you want to prove $lim_{n_k toinfty} x_{n_k} = 0$ with this, then $y_k = x_{n_k}$, so what you need to know is that $lim_{Ktoinfty}sum_{k=1}^K x_{n_k}$ converges. But you have $lim_{Ktoinfty} S_{n_K} = 0$. Note that
$$sum_{k=1}^K x_{n_k} = x_{n_1} + x_{n_2} + x_{n_3} + ... + x_{n_K}$$
while
$$S_{n_k} = x_1 + x_2 + x_3 + ... + x_{n_k}$$
These are not the same. So you need to do some more work here.
Thus any subsequence of $x_n$ converges to the same limit
Not "any". You didn't start out by assuming that $x_{n_k}$ or $x_{n_p}$ were arbitrary subsequences. Even if you correct the earlier argument (and it is a relatively easy fix), all you've shown is that $lim_k x_{n_k} = 0$ for convergent subsequences of $S_n$.
So in case two subsequences of $S_n$ converge to some number then it implies $x_n$ converges to $0$.
Where did that come from? It doesn't follow logically from what you've said before at all.
A more general problem is that this "proof" is not even headed in a direction you need to go. You went through this and then declared in the end that $lim_n x_n = 0$. but that is a statement you already knew. Arriving at it means you've gone nowhere.
What you need to consider is this: If subsequences of $S_n$ converge to both $a$ and $b$ with $b > a$, this means that when you start summing $x_n$, eventually the sum will get close to $a$. As you keep summing, it may drift away from $a$, but eventually, it'll end up coming back to $a$ over and over again, all the way to infinity. But the same can be said about $b$. The sum must keep coming near to it as well. So it has to go near $a$, then go near $b$, then later it must go near $a$ again, and then back to $b$, and so on. It must oscilate back and forth between $a$ and $b$ forever. but since $lim_n x_n = 0$, it has to keep doing this oscilation in smaller and smaller steps. So what happens if $c$ is between $a$ and $b$?
As for $Bbb C$, there is nothing in this problem that makes any use of the algebraic properties of $Bbb C$. As far as you need to be concerned here, $Bbb C$ is just another name for the plane $Bbb R^2$. Can you see how a second dimension might offer a way for $S_n$ to get from $a$ to $b$ and back without covering the same country in the middle every time?
answered Mar 14 at 2:48
Paul SinclairPaul Sinclair
20.5k21543
20.5k21543
$begingroup$
Thank you for such a detailed answer. There is still a thing I don't get. If we assume there are two subsequences of $S_n$, namely $S_{n_k}$ and $S_{n_p}$ which both converge to different limits, why would the sum of $x_n$ come close to $a$ and then to $b$ back and forth? Could you please elaborate on that? Based on the rest of your answer, it looks like $c$ is going to be between $a$ and $b$ with the distance between them converging to $0$. Which would eventually mean $a = c = b$, meaning that it's not possible for $S_n$ to have two separate sub-sequential limits. Right?
$endgroup$
– roman
Mar 14 at 8:42
$begingroup$
As for the case of complex plane there must be a circle with some radius $b-a$ for $b>a$ which keeps shrinking into a single point $c$ as $ntoinfty$, which it terms also mean all subsequences of $S_n$ must converge to the same number if I understood everything correctly
$endgroup$
– roman
Mar 14 at 8:56
1
$begingroup$
That $S_{n_k} to a$ means that there are infinite number of values of $n$ (the $n_k$ for high enough $k$) with $S_n$ close to $a$. In particular, for any $N$, there has to be $n > N$ with $S_n$ close to $a$. The same is true for $b$. So find an $n$ with $S_n$ close to $a$. There is therefore some higher $n$ with $S_n$ close to $b$. And some higher $n$ yet with $S_n$ close to $a$, and so on.
$endgroup$
– Paul Sinclair
Mar 14 at 16:01
$begingroup$
Second, $a$ and $b$ are fixed points. They do not depend on $n$. They do not get closer together, they do not shrink down to some point. They are simply two goal posts that $S_n$ moves back and forth between in steps as $n$ increases. The size of those steps is $S_n - S_{n-1} = x_n$, so the steps eventually become smaller and smaller. So the distance $S_n$ can maintain from any point $c$ between $a$ and $b$ also gets smaller and smaller. However, in 2 dimensions, $S_n$ is not constrained to cover the same territory on every trip.
$endgroup$
– Paul Sinclair
Mar 14 at 16:08
add a comment |
$begingroup$
Thank you for such a detailed answer. There is still a thing I don't get. If we assume there are two subsequences of $S_n$, namely $S_{n_k}$ and $S_{n_p}$ which both converge to different limits, why would the sum of $x_n$ come close to $a$ and then to $b$ back and forth? Could you please elaborate on that? Based on the rest of your answer, it looks like $c$ is going to be between $a$ and $b$ with the distance between them converging to $0$. Which would eventually mean $a = c = b$, meaning that it's not possible for $S_n$ to have two separate sub-sequential limits. Right?
$endgroup$
– roman
Mar 14 at 8:42
$begingroup$
As for the case of complex plane there must be a circle with some radius $b-a$ for $b>a$ which keeps shrinking into a single point $c$ as $ntoinfty$, which it terms also mean all subsequences of $S_n$ must converge to the same number if I understood everything correctly
$endgroup$
– roman
Mar 14 at 8:56
1
$begingroup$
That $S_{n_k} to a$ means that there are infinite number of values of $n$ (the $n_k$ for high enough $k$) with $S_n$ close to $a$. In particular, for any $N$, there has to be $n > N$ with $S_n$ close to $a$. The same is true for $b$. So find an $n$ with $S_n$ close to $a$. There is therefore some higher $n$ with $S_n$ close to $b$. And some higher $n$ yet with $S_n$ close to $a$, and so on.
$endgroup$
– Paul Sinclair
Mar 14 at 16:01
$begingroup$
Second, $a$ and $b$ are fixed points. They do not depend on $n$. They do not get closer together, they do not shrink down to some point. They are simply two goal posts that $S_n$ moves back and forth between in steps as $n$ increases. The size of those steps is $S_n - S_{n-1} = x_n$, so the steps eventually become smaller and smaller. So the distance $S_n$ can maintain from any point $c$ between $a$ and $b$ also gets smaller and smaller. However, in 2 dimensions, $S_n$ is not constrained to cover the same territory on every trip.
$endgroup$
– Paul Sinclair
Mar 14 at 16:08
$begingroup$
Thank you for such a detailed answer. There is still a thing I don't get. If we assume there are two subsequences of $S_n$, namely $S_{n_k}$ and $S_{n_p}$ which both converge to different limits, why would the sum of $x_n$ come close to $a$ and then to $b$ back and forth? Could you please elaborate on that? Based on the rest of your answer, it looks like $c$ is going to be between $a$ and $b$ with the distance between them converging to $0$. Which would eventually mean $a = c = b$, meaning that it's not possible for $S_n$ to have two separate sub-sequential limits. Right?
$endgroup$
– roman
Mar 14 at 8:42
$begingroup$
Thank you for such a detailed answer. There is still a thing I don't get. If we assume there are two subsequences of $S_n$, namely $S_{n_k}$ and $S_{n_p}$ which both converge to different limits, why would the sum of $x_n$ come close to $a$ and then to $b$ back and forth? Could you please elaborate on that? Based on the rest of your answer, it looks like $c$ is going to be between $a$ and $b$ with the distance between them converging to $0$. Which would eventually mean $a = c = b$, meaning that it's not possible for $S_n$ to have two separate sub-sequential limits. Right?
$endgroup$
– roman
Mar 14 at 8:42
$begingroup$
As for the case of complex plane there must be a circle with some radius $b-a$ for $b>a$ which keeps shrinking into a single point $c$ as $ntoinfty$, which it terms also mean all subsequences of $S_n$ must converge to the same number if I understood everything correctly
$endgroup$
– roman
Mar 14 at 8:56
$begingroup$
As for the case of complex plane there must be a circle with some radius $b-a$ for $b>a$ which keeps shrinking into a single point $c$ as $ntoinfty$, which it terms also mean all subsequences of $S_n$ must converge to the same number if I understood everything correctly
$endgroup$
– roman
Mar 14 at 8:56
1
1
$begingroup$
That $S_{n_k} to a$ means that there are infinite number of values of $n$ (the $n_k$ for high enough $k$) with $S_n$ close to $a$. In particular, for any $N$, there has to be $n > N$ with $S_n$ close to $a$. The same is true for $b$. So find an $n$ with $S_n$ close to $a$. There is therefore some higher $n$ with $S_n$ close to $b$. And some higher $n$ yet with $S_n$ close to $a$, and so on.
$endgroup$
– Paul Sinclair
Mar 14 at 16:01
$begingroup$
That $S_{n_k} to a$ means that there are infinite number of values of $n$ (the $n_k$ for high enough $k$) with $S_n$ close to $a$. In particular, for any $N$, there has to be $n > N$ with $S_n$ close to $a$. The same is true for $b$. So find an $n$ with $S_n$ close to $a$. There is therefore some higher $n$ with $S_n$ close to $b$. And some higher $n$ yet with $S_n$ close to $a$, and so on.
$endgroup$
– Paul Sinclair
Mar 14 at 16:01
$begingroup$
Second, $a$ and $b$ are fixed points. They do not depend on $n$. They do not get closer together, they do not shrink down to some point. They are simply two goal posts that $S_n$ moves back and forth between in steps as $n$ increases. The size of those steps is $S_n - S_{n-1} = x_n$, so the steps eventually become smaller and smaller. So the distance $S_n$ can maintain from any point $c$ between $a$ and $b$ also gets smaller and smaller. However, in 2 dimensions, $S_n$ is not constrained to cover the same territory on every trip.
$endgroup$
– Paul Sinclair
Mar 14 at 16:08
$begingroup$
Second, $a$ and $b$ are fixed points. They do not depend on $n$. They do not get closer together, they do not shrink down to some point. They are simply two goal posts that $S_n$ moves back and forth between in steps as $n$ increases. The size of those steps is $S_n - S_{n-1} = x_n$, so the steps eventually become smaller and smaller. So the distance $S_n$ can maintain from any point $c$ between $a$ and $b$ also gets smaller and smaller. However, in 2 dimensions, $S_n$ is not constrained to cover the same territory on every trip.
$endgroup$
– Paul Sinclair
Mar 14 at 16:08
add a comment |
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This question might be useful: Let $a = liminf x_n$, $b = limsup x_n$. If $lim (x_{n+1} - x_n) = 0$ then every element of $(a,b)$ is a subsequential limit of $(x_n)$.
$endgroup$
– Martin Sleziak
Mar 13 at 17:05
$begingroup$
FYI - wherever you have $lim_{n_k to infty}$, what you actually mean is $lim_{kto infty}$. It is $k$ that is varying. $n_k$ only changes as function of $k$.
$endgroup$
– Paul Sinclair
Mar 14 at 1:43