True or false: a sequence ${a_{n}}$ is convergent if and only if the subsequences $a_{2n}$ and $c_{2n + 1}$...
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This question already has an answer here:
Sufficiency to prove the convergence of a sequence using even and odd terms
4 answers
True or false: a sequence ${a_{n}}$ is convergent if and only if the
subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same
number
I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.
real-analysis sequences-and-series
asked Dec 14 '18 at 0:50
josephjoseph
515111
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marked as duplicate by Mike Earnest, RRL real-analysis
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Dec 14 '18 at 1:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Sufficiency to prove the convergence of a sequence using even and odd terms
4 answers
True or false: a sequence ${a_{n}}$ is convergent if and only if the
subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same
number
I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.
real-analysis sequences-and-series
asked Dec 14 '18 at 0:50
josephjoseph
515111
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marked as duplicate by Mike Earnest, RRL real-analysis
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Dec 14 '18 at 1:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$c_{2n+1}$? Did you mean $a_{2n+1}$?
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– Alex R.
Dec 14 '18 at 0:53
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The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
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– JMoravitz
Dec 14 '18 at 0:53
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yes, i mean $a_{2n + 1}$
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– joseph
Dec 14 '18 at 0:56
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Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
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– William Sun
Dec 14 '18 at 0:56
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You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
$endgroup$
– Will M.
Dec 14 '18 at 1:02
add a comment |
$begingroup$
This question already has an answer here:
Sufficiency to prove the convergence of a sequence using even and odd terms
4 answers
True or false: a sequence ${a_{n}}$ is convergent if and only if the
subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same
number
I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.
real-analysis sequences-and-series
asked Dec 14 '18 at 0:50
josephjoseph
515111
$endgroup$
This question already has an answer here:
Sufficiency to prove the convergence of a sequence using even and odd terms
4 answers
True or false: a sequence ${a_{n}}$ is convergent if and only if the
subsequences $a_{2n}$ and $c_{2n + 1}$ both converge to the same
number
I think that the answer is true. I know that the converse is true because the subsequences of a convergent sequence always converge to the same number. I'm not sure about the forward direction though. I couldn't come up with a counterexample.
This question already has an answer here:
Sufficiency to prove the convergence of a sequence using even and odd terms
4 answers
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Dec 14 '18 at 0:50
josephjoseph
515111
asked Dec 14 '18 at 0:50
josephjoseph
515111
asked Dec 14 '18 at 0:50
josephjoseph
515111
asked Dec 14 '18 at 0:50
josephjoseph
515111
asked Dec 14 '18 at 0:50
josephjoseph
515111
515111
marked as duplicate by Mike Earnest, RRL real-analysis
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Dec 14 '18 at 1:39
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Dec 14 '18 at 1:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
$c_{2n+1}$? Did you mean $a_{2n+1}$?
$endgroup$
– Alex R.
Dec 14 '18 at 0:53
$begingroup$
The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
$endgroup$
– JMoravitz
Dec 14 '18 at 0:53
$begingroup$
yes, i mean $a_{2n + 1}$
$endgroup$
– joseph
Dec 14 '18 at 0:56
$begingroup$
Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
$endgroup$
– William Sun
Dec 14 '18 at 0:56
$begingroup$
You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
$endgroup$
– Will M.
Dec 14 '18 at 1:02
add a comment |
$begingroup$
$c_{2n+1}$? Did you mean $a_{2n+1}$?
$endgroup$
– Alex R.
Dec 14 '18 at 0:53
$begingroup$
The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
$endgroup$
– JMoravitz
Dec 14 '18 at 0:53
$begingroup$
yes, i mean $a_{2n + 1}$
$endgroup$
– joseph
Dec 14 '18 at 0:56
$begingroup$
Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
$endgroup$
– William Sun
Dec 14 '18 at 0:56
$begingroup$
You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
$endgroup$
– Will M.
Dec 14 '18 at 1:02
$begingroup$
$c_{2n+1}$? Did you mean $a_{2n+1}$?
$endgroup$
– Alex R.
Dec 14 '18 at 0:53
$begingroup$
$c_{2n+1}$? Did you mean $a_{2n+1}$?
$endgroup$
– Alex R.
Dec 14 '18 at 0:53
$begingroup$
The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
$endgroup$
– JMoravitz
Dec 14 '18 at 0:53
$begingroup$
The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
$endgroup$
– JMoravitz
Dec 14 '18 at 0:53
$begingroup$
yes, i mean $a_{2n + 1}$
$endgroup$
– joseph
Dec 14 '18 at 0:56
$begingroup$
yes, i mean $a_{2n + 1}$
$endgroup$
– joseph
Dec 14 '18 at 0:56
$begingroup$
Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
$endgroup$
– William Sun
Dec 14 '18 at 0:56
$begingroup$
Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
$endgroup$
– William Sun
Dec 14 '18 at 0:56
$begingroup$
You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
$endgroup$
– Will M.
Dec 14 '18 at 1:02
$begingroup$
You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
$endgroup$
– Will M.
Dec 14 '18 at 1:02
add a comment |
1 Answer
1
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Suppose
$$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$
And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.
$$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$
Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.
answered Dec 14 '18 at 1:07
DonAntonioDonAntonio
177k1492225
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose
$$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$
And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.
$$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$
Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.
answered Dec 14 '18 at 1:07
DonAntonioDonAntonio
177k1492225
$endgroup$
add a comment |
$begingroup$
Suppose
$$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$
And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.
$$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$
Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.
answered Dec 14 '18 at 1:07
DonAntonioDonAntonio
177k1492225
$endgroup$
add a comment |
$begingroup$
Suppose
$$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$
And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.
$$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$
Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.
answered Dec 14 '18 at 1:07
DonAntonioDonAntonio
177k1492225
$endgroup$
Suppose
$$lim_{ntoinfty}a_{2n}=lim_{ntoinfty}a_{2n+1}=A$$
And let $;epsilon >0;$ be arbitrary, then there exist $;N_1,,N_2inBbb N;$ s.t.
$$begin{cases}|a_{2n}-A|<epsilon,,,,forall n>N_1\{}\|a_{2n+1}-A|<epsilon,,,,forall,n>N_2end{cases}$$
Let now $;M:=max{N_1,,N_2};$ , then for any $;n>M;$ , since $;n;$ is either even or odd....finish now with one line the proof.
answered Dec 14 '18 at 1:07
DonAntonioDonAntonio
177k1492225
answered Dec 14 '18 at 1:07
DonAntonioDonAntonio
177k1492225
answered Dec 14 '18 at 1:07
DonAntonioDonAntonio
177k1492225
answered Dec 14 '18 at 1:07
DonAntonioDonAntonio
177k1492225
177k1492225
add a comment |
add a comment |
$begingroup$
$c_{2n+1}$? Did you mean $a_{2n+1}$?
$endgroup$
– Alex R.
Dec 14 '18 at 0:53
$begingroup$
The forward direction is true as well. To see this, consider the contrapositive... Suppose that it is not true that $a_{2n}$ and $a_{2n+1}$ both converge to the same number. It could be that is a result of $a_{2n}$ not converging at all for example, in which case $a_n$ doesn't converge, or it could be that $a_{2n}$ converges to a number different than $a_{2n+1}$, in which case again $a_n$ wouldn't converge.
$endgroup$
– JMoravitz
Dec 14 '18 at 0:53
$begingroup$
yes, i mean $a_{2n + 1}$
$endgroup$
– joseph
Dec 14 '18 at 0:56
$begingroup$
Every neighborhood $U$ of the limit contains $a_{2n}$ for every $2n>N$($N$ is an integer). In the same way we find $M$ such that $U$ contains $a_{2n+1}$ for every $2n+1>M$. Taking the maximum of the two integers $M$, $N$, can you prove it now?
$endgroup$
– William Sun
Dec 14 '18 at 0:56
$begingroup$
You can't come up with a counterexample because is true. A proof can go in the lines of "It is immediate from definition." I am just kidding, it is really easy to prove, see @WilliamSun comment above mine.
$endgroup$
– Will M.
Dec 14 '18 at 1:02