Proof verification. Show that if ${x_n}$ diverges then there must be a sequence ${p_n}subsetBbb N$ such that...
Let ${x_n}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers ${p_n} subset Bbb N$ such that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Suppose that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) = 0
$$
Clearly $n+p_n > n$. Denote $n+p_n = m_n$:
$$
lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon
$$
which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.
Therefore for $x_n$ to be divergent there must exist some sequence ${p_n}$ for which:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!
calculus limits proof-verification convergence
asked Dec 26 at 15:53
roman
1,93421221
add a comment |
Let ${x_n}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers ${p_n} subset Bbb N$ such that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Suppose that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) = 0
$$
Clearly $n+p_n > n$. Denote $n+p_n = m_n$:
$$
lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon
$$
which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.
Therefore for $x_n$ to be divergent there must exist some sequence ${p_n}$ for which:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!
calculus limits proof-verification convergence
asked Dec 26 at 15:53
roman
1,93421221
1
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
– Michał Miśkiewicz
Dec 26 at 18:03
A visualization of the idea from @RafayAshary answer. Just for the case
– roman
Dec 26 at 18:36
add a comment |
Let ${x_n}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers ${p_n} subset Bbb N$ such that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Suppose that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) = 0
$$
Clearly $n+p_n > n$. Denote $n+p_n = m_n$:
$$
lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon
$$
which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.
Therefore for $x_n$ to be divergent there must exist some sequence ${p_n}$ for which:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!
calculus limits proof-verification convergence
asked Dec 26 at 15:53
roman
1,93421221
Let ${x_n}$ denote a non-convergent sequence. Show that there exists a sequence of natural numbers ${p_n} subset Bbb N$ such that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Suppose that:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) = 0
$$
Clearly $n+p_n > n$. Denote $n+p_n = m_n$:
$$
lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon
$$
which denotes a Cauchy Criterion for the sequence $x_n$. If $x_n$ is fundamental then it must converge to some limit, but from the problem statement $x_n$ is divergent and hence we've arrived at a contradiction.
Therefore for $x_n$ to be divergent there must exist some sequence ${p_n}$ for which:
$$
lim_{ntoinfty}(x_{n+p_n} - x_n) ne 0
$$
Please let me know whether there is anything wrong with the proof or whether it's fine. Thank you!
calculus limits proof-verification convergence
calculus limits proof-verification convergence
asked Dec 26 at 15:53
roman
1,93421221
asked Dec 26 at 15:53
roman
1,93421221
edited Dec 26 at 16:11
asked Dec 26 at 15:53
roman
1,93421221
asked Dec 26 at 15:53
roman
1,93421221
asked Dec 26 at 15:53
roman
1,93421221
1,93421221
1
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
– Michał Miśkiewicz
Dec 26 at 18:03
A visualization of the idea from @RafayAshary answer. Just for the case
– roman
Dec 26 at 18:36
add a comment |
1
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
– Michał Miśkiewicz
Dec 26 at 18:03
A visualization of the idea from @RafayAshary answer. Just for the case
– roman
Dec 26 at 18:36
1
1
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
– Michał Miśkiewicz
Dec 26 at 18:03
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
– Michał Miśkiewicz
Dec 26 at 18:03
A visualization of the idea from @RafayAshary answer. Just for the case
– roman
Dec 26 at 18:36
A visualization of the idea from @RafayAshary answer. Just for the case
– roman
Dec 26 at 18:36
add a comment |
1 Answer
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I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 at 17:56
Rafay Ashary
83618
2
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
– John Doe
Dec 26 at 18:01
Yes, my apologies :)
– Rafay Ashary
Dec 26 at 18:19
That's a nice way to approach the proof, thank you!
– roman
Dec 26 at 18:37
add a comment |
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I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 at 17:56
Rafay Ashary
83618
2
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
– John Doe
Dec 26 at 18:01
Yes, my apologies :)
– Rafay Ashary
Dec 26 at 18:19
That's a nice way to approach the proof, thank you!
– roman
Dec 26 at 18:37
add a comment |
I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 at 17:56
Rafay Ashary
83618
2
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
– John Doe
Dec 26 at 18:01
Yes, my apologies :)
– Rafay Ashary
Dec 26 at 18:19
That's a nice way to approach the proof, thank you!
– roman
Dec 26 at 18:37
add a comment |
I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 at 17:56
Rafay Ashary
83618
I don't see why $$lim_{ntoinfty}(x_{m_n} - x_n) = 0 iff forall epsilon > 0 exists N inBbb N: forall m_n > n > N implies |x_{m_n} - x_n| < epsilon$$
Implies that ${x_n}_{ninmathbb N}$ converges. (What if $m_n=n+1$ and $x_n=sum_{i=1}^n tfrac{1}{i}$?)
An alternative approach is to take $x^+=limsup_{ntoinfty}(x_n)$ and $x^-=liminf_{ntoinfty}(x_n)$. Then by assumption (specifically the non-convergence of $x_n$) we have that $x^+neq x^-$. Let $x^+-x^-=delta>0$. By definition the two sets $I^+={ninmathbb N:x_n>x^+-tfrac{1}{3}delta}$ and $I^-={ninmathbb N:x_n<x^-+tfrac{1}{3}delta}$ are disjoint and infinite. So for each $nin I^-$ we may choose $n+p_nin I^+$, in which case $$x_{n+p_n}-x_n>(x^+-tfrac{1}{3}delta)-(x^-+tfrac{1}{3}delta)=tfrac{1}{3}delta$$ And this situation occurs infinitely often, so $lim_{ntoinfty}(x_{n+p_n}-x)$ cannot possibly exist $Box$
answered Dec 26 at 17:56
Rafay Ashary
83618
edited Dec 26 at 18:19
answered Dec 26 at 17:56
Rafay Ashary
83618
answered Dec 26 at 17:56
Rafay Ashary
83618
answered Dec 26 at 17:56
Rafay Ashary
83618
83618
2
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
– John Doe
Dec 26 at 18:01
Yes, my apologies :)
– Rafay Ashary
Dec 26 at 18:19
That's a nice way to approach the proof, thank you!
– roman
Dec 26 at 18:37
add a comment |
2
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
– John Doe
Dec 26 at 18:01
Yes, my apologies :)
– Rafay Ashary
Dec 26 at 18:19
That's a nice way to approach the proof, thank you!
– roman
Dec 26 at 18:37
2
2
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
– John Doe
Dec 26 at 18:01
in your counter example, you mean $$x_n=sum_{i=1}^n frac{1}{i}$$
– John Doe
Dec 26 at 18:01
Yes, my apologies :)
– Rafay Ashary
Dec 26 at 18:19
Yes, my apologies :)
– Rafay Ashary
Dec 26 at 18:19
That's a nice way to approach the proof, thank you!
– roman
Dec 26 at 18:37
That's a nice way to approach the proof, thank you!
– roman
Dec 26 at 18:37
add a comment |
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1
As Rafay Ashary noted, there's a problem with the reformulation. Something like "for every sequence $p_n$" should appear on both sides of the equivalence. Here, it seems to appear in the wrong place in the RHS, and it doesn't appear in the LHS at all.
– Michał Miśkiewicz
Dec 26 at 18:03
A visualization of the idea from @RafayAshary answer. Just for the case
– roman
Dec 26 at 18:36