Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$. Does $x_n$ converge to $1$?(edited) Let $(x_n_n)$ be a sequence of positive real numbers which has no convergent subsequence. Prove lim $x_n=+infty$Prove $x_n$ converges to $a$ iff every subsequence of $x_n$ also converges to $a$.A sequence converges if and only if every subsequence converges?If $lim_x to infty x_n = liminf_n to infty x_n = limsup_n to infty x_n= -infty$ does it exist a convergent subsequence of $x_n$?Given subsequences converge, prove that the sequence converges.If all proper subsequences converge to same limit then the sequence converges.Show that if all convergent subsequences of a bounded sequence converge to $l$, the sequence itself must also converge to $l$.Let $(a_n)_n in mathbb N$ be a sequence such that all non-trivial (*) subsequence converges, does $(a_n)$ converge?Convergence of sequence where every subsequence of specific type convergesProof 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.
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Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$. Does $x_n$ converge to $1$?
(edited) Let $(x_n_n)$ be a sequence of positive real numbers which has no convergent subsequence. Prove lim $x_n=+infty$Prove $x_n$ converges to $a$ iff every subsequence of $x_n$ also converges to $a$.A sequence converges if and only if every subsequence converges?If $lim_x to infty x_n = liminf_n to infty x_n = limsup_n to infty x_n= -infty$ does it exist a convergent subsequence of $x_n$?Given subsequences converge, prove that the sequence converges.If all proper subsequences converge to same limit then the sequence converges.Show that if all convergent subsequences of a bounded sequence converge to $l$, the sequence itself must also converge to $l$.Let $(a_n)_n in mathbb N$ be a sequence such that all non-trivial (*) subsequence converges, does $(a_n)$ converge?Convergence of sequence where every subsequence of specific type convergesProof 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.
$begingroup$
Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$.
Does $x_n$ converge to $1$?
I'm a bit confused by this problem since no constraints on $p$ are given. The only reasonable way seems to consider $p$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $k$:
$$
k = 2:x_pk = x_2, x_4, dots, x_2p \
k = 3:x_pk = x_3, x_4, dots, x_3p \
cdots\
k = n:x_pk = x_n, x_2n, dots, x_np \
$$
All of the sequences above converge to $1$, namely:
$$
lim_ptoinftyx_2p = 1\
lim_ptoinftyx_3p = 1\
cdots\
lim_ptoinftyx_np = 1\
$$
At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $x_n$, and since $k$ is an arbitrary natural number we may consider the following set:
$$
X = bigcuplimits_k=1^infty x_pk
$$
Moreover, since every sequence in the union converges to $1$ then it must follow that $X$ also converges to one. But $X$ is essentially equivalent to $x_n$ which would imply that:
$$
lim_ntoinftyx_n = 1
$$
I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete?
Thank you!
real-analysis sequences-and-series limits proof-verification
asked Mar 15 at 17:09
romanroman
2,39321225
$endgroup$
add a comment |
$begingroup$
Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$.
Does $x_n$ converge to $1$?
I'm a bit confused by this problem since no constraints on $p$ are given. The only reasonable way seems to consider $p$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $k$:
$$
k = 2:x_pk = x_2, x_4, dots, x_2p \
k = 3:x_pk = x_3, x_4, dots, x_3p \
cdots\
k = n:x_pk = x_n, x_2n, dots, x_np \
$$
All of the sequences above converge to $1$, namely:
$$
lim_ptoinftyx_2p = 1\
lim_ptoinftyx_3p = 1\
cdots\
lim_ptoinftyx_np = 1\
$$
At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $x_n$, and since $k$ is an arbitrary natural number we may consider the following set:
$$
X = bigcuplimits_k=1^infty x_pk
$$
Moreover, since every sequence in the union converges to $1$ then it must follow that $X$ also converges to one. But $X$ is essentially equivalent to $x_n$ which would imply that:
$$
lim_ntoinftyx_n = 1
$$
I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete?
Thank you!
real-analysis sequences-and-series limits proof-verification
asked Mar 15 at 17:09
romanroman
2,39321225
$endgroup$
add a comment |
$begingroup$
Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$.
Does $x_n$ converge to $1$?
I'm a bit confused by this problem since no constraints on $p$ are given. The only reasonable way seems to consider $p$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $k$:
$$
k = 2:x_pk = x_2, x_4, dots, x_2p \
k = 3:x_pk = x_3, x_4, dots, x_3p \
cdots\
k = n:x_pk = x_n, x_2n, dots, x_np \
$$
All of the sequences above converge to $1$, namely:
$$
lim_ptoinftyx_2p = 1\
lim_ptoinftyx_3p = 1\
cdots\
lim_ptoinftyx_np = 1\
$$
At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $x_n$, and since $k$ is an arbitrary natural number we may consider the following set:
$$
X = bigcuplimits_k=1^infty x_pk
$$
Moreover, since every sequence in the union converges to $1$ then it must follow that $X$ also converges to one. But $X$ is essentially equivalent to $x_n$ which would imply that:
$$
lim_ntoinftyx_n = 1
$$
I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete?
Thank you!
real-analysis sequences-and-series limits proof-verification
asked Mar 15 at 17:09
romanroman
2,39321225
$endgroup$
Let $x_n$ be a sequence such that $forall k > 1 in Bbb N$ its subsequence $x_pk$ converges to $1$.
Does $x_n$ converge to $1$?
I'm a bit confused by this problem since no constraints on $p$ are given. The only reasonable way seems to consider $p$ a natural number. Intuitively the statement seems to be true. Let's try to consider different values for $k$:
$$
k = 2:x_pk = x_2, x_4, dots, x_2p \
k = 3:x_pk = x_3, x_4, dots, x_3p \
cdots\
k = n:x_pk = x_n, x_2n, dots, x_np \
$$
All of the sequences above converge to $1$, namely:
$$
lim_ptoinftyx_2p = 1\
lim_ptoinftyx_3p = 1\
cdots\
lim_ptoinftyx_np = 1\
$$
At this point, I was thinking about the union of all the sequences above. All of them are subsequences of $x_n$, and since $k$ is an arbitrary natural number we may consider the following set:
$$
X = bigcuplimits_k=1^infty x_pk
$$
Moreover, since every sequence in the union converges to $1$ then it must follow that $X$ also converges to one. But $X$ is essentially equivalent to $x_n$ which would imply that:
$$
lim_ntoinftyx_n = 1
$$
I'm not sure the reasoning above might be applied to the problem. So I would like to ask for verification or a valid solution in case mine is wrong. Also, in case the idea in mine is fine then is it rigorous enough to consider the proof complete?
Thank you!
real-analysis sequences-and-series limits proof-verification
real-analysis sequences-and-series limits proof-verification
asked Mar 15 at 17:09
romanroman
2,39321225
asked Mar 15 at 17:09
romanroman
2,39321225
edited Mar 15 at 17:28
roman
asked Mar 15 at 17:09
romanroman
2,39321225
asked Mar 15 at 17:09
romanroman
2,39321225
asked Mar 15 at 17:09
romanroman
2,39321225
2,39321225
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.
answered Mar 15 at 17:11
mathworker21mathworker21
9,4561928
$endgroup$
$begingroup$
Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
$endgroup$
– roman
Mar 15 at 17:30
1
$begingroup$
I don't get the relevance
$endgroup$
– mathworker21
Mar 15 at 17:47
$begingroup$
You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
$endgroup$
– roman
Mar 15 at 17:51
$begingroup$
$x_n$ is not convergent
$endgroup$
– mathworker21
Mar 15 at 18:23
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.
answered Mar 15 at 17:11
mathworker21mathworker21
9,4561928
$endgroup$
$begingroup$
Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
$endgroup$
– roman
Mar 15 at 17:30
1
$begingroup$
I don't get the relevance
$endgroup$
– mathworker21
Mar 15 at 17:47
$begingroup$
You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
$endgroup$
– roman
Mar 15 at 17:51
$begingroup$
$x_n$ is not convergent
$endgroup$
– mathworker21
Mar 15 at 18:23
add a comment |
$begingroup$
Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.
answered Mar 15 at 17:11
mathworker21mathworker21
9,4561928
$endgroup$
$begingroup$
Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
$endgroup$
– roman
Mar 15 at 17:30
1
$begingroup$
I don't get the relevance
$endgroup$
– mathworker21
Mar 15 at 17:47
$begingroup$
You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
$endgroup$
– roman
Mar 15 at 17:51
$begingroup$
$x_n$ is not convergent
$endgroup$
– mathworker21
Mar 15 at 18:23
add a comment |
$begingroup$
Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.
answered Mar 15 at 17:11
mathworker21mathworker21
9,4561928
$endgroup$
Of course not (in general). Take $x_n$ to be $0$ if $n$ is prime and $1$ otherwise.
answered Mar 15 at 17:11
mathworker21mathworker21
9,4561928
answered Mar 15 at 17:11
mathworker21mathworker21
9,4561928
answered Mar 15 at 17:11
mathworker21mathworker21
9,4561928
answered Mar 15 at 17:11
mathworker21mathworker21
9,4561928
9,4561928
$begingroup$
Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
$endgroup$
– roman
Mar 15 at 17:30
1
$begingroup$
I don't get the relevance
$endgroup$
– mathworker21
Mar 15 at 17:47
$begingroup$
You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
$endgroup$
– roman
Mar 15 at 17:51
$begingroup$
$x_n$ is not convergent
$endgroup$
– mathworker21
Mar 15 at 18:23
add a comment |
$begingroup$
Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
$endgroup$
– roman
Mar 15 at 17:30
1
$begingroup$
I don't get the relevance
$endgroup$
– mathworker21
Mar 15 at 17:47
$begingroup$
You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
$endgroup$
– roman
Mar 15 at 17:51
$begingroup$
$x_n$ is not convergent
$endgroup$
– mathworker21
Mar 15 at 18:23
$begingroup$
Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
$endgroup$
– roman
Mar 15 at 17:30
$begingroup$
Thank you for your answer. Sorry for the confusion but I've missed adding an important notice about $x_pk$. Which is $x_pk$ is a subsequence of $x_n$.
$endgroup$
– roman
Mar 15 at 17:30
1
1
$begingroup$
I don't get the relevance
$endgroup$
– mathworker21
Mar 15 at 17:47
$begingroup$
I don't get the relevance
$endgroup$
– mathworker21
Mar 15 at 17:47
$begingroup$
You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
$endgroup$
– roman
Mar 15 at 17:51
$begingroup$
You are right, after thinking more carefully on this $x_n$ is indeed not necessarily convergent, even given the fact it has convergent subsequences.
$endgroup$
– roman
Mar 15 at 17:51
$begingroup$
$x_n$ is not convergent
$endgroup$
– mathworker21
Mar 15 at 18:23
$begingroup$
$x_n$ is not convergent
$endgroup$
– mathworker21
Mar 15 at 18:23
add a comment |
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