Mixing
Proving convergence of a bizarre sequence
Clash Royale CLAN TAG#URR8PPP
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This was a question on an exam I recently took that I didn't do so great on, so I'm trying to understand this problem thoroughly. What flaws are there?
Suppose that $a_n$ and $b_n$ are two convergent sequences that both converge to the same $L in mathbbR$. Define a new sequence $c_n$ such that $c_n = a_n$ if $n$ is odd and $c_n = b_n$ if $n$ is even. Prove that $c_n$ converges to $L$.
Proof Since $a_n$ and $b_n$ both converge to $L$, we know that
$$lim sup a_n = lim inf a_n = lim sup b_n = lim inf b_n = L$$
Define $i_n = infc_k : k geq n$ and $s_n = supc_k : k geq n$. Then $lim i_n = lim inf c_n$ and $lim s_n = lim sup c_n$.
$L$ is a subsequential limit of $c_n$ and so $lim inf c_n leq L leq lim sup c_n$.
Suppose for contradiction that $lim inf c_n < L$. Then there exists an $n_0$ such that $L neq i_n_0$ and $i_n_0 < a_n_0$ or $i_n_0 < b_n_0$, but this contradicts the construction of $a_n$ and $b_n$ because one or both would not converge at all or would converge to $i_n_0$. Thus, $lim inf c_n = L$.
A similar argument can be used to show that $lim sup c_n = L$ by replacing $i_n_0$ with $s_n_0$, $lim inf c_n$ with $lim sup c_n$ and switching the inequality signs in the previous paragraph.
Since $lim inf c_n = L = lim sup c_n$, $c_n$ converges to $lim a_n = lim b_n = L$.
real-analysis sequences-and-series proof-verification limsup-and-liminf
asked 1 hour ago
Austin
517
add a comment |Â
up vote
1
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This was a question on an exam I recently took that I didn't do so great on, so I'm trying to understand this problem thoroughly. What flaws are there?
Suppose that $a_n$ and $b_n$ are two convergent sequences that both converge to the same $L in mathbbR$. Define a new sequence $c_n$ such that $c_n = a_n$ if $n$ is odd and $c_n = b_n$ if $n$ is even. Prove that $c_n$ converges to $L$.
Proof Since $a_n$ and $b_n$ both converge to $L$, we know that
$$lim sup a_n = lim inf a_n = lim sup b_n = lim inf b_n = L$$
Define $i_n = infc_k : k geq n$ and $s_n = supc_k : k geq n$. Then $lim i_n = lim inf c_n$ and $lim s_n = lim sup c_n$.
$L$ is a subsequential limit of $c_n$ and so $lim inf c_n leq L leq lim sup c_n$.
Suppose for contradiction that $lim inf c_n < L$. Then there exists an $n_0$ such that $L neq i_n_0$ and $i_n_0 < a_n_0$ or $i_n_0 < b_n_0$, but this contradicts the construction of $a_n$ and $b_n$ because one or both would not converge at all or would converge to $i_n_0$. Thus, $lim inf c_n = L$.
A similar argument can be used to show that $lim sup c_n = L$ by replacing $i_n_0$ with $s_n_0$, $lim inf c_n$ with $lim sup c_n$ and switching the inequality signs in the previous paragraph.
Since $lim inf c_n = L = lim sup c_n$, $c_n$ converges to $lim a_n = lim b_n = L$.
real-analysis sequences-and-series proof-verification limsup-and-liminf
asked 1 hour ago
Austin
517
1
While it might be correct, this approach seems way to convoluted. Isnt it actually you picj an epsilon so you know that tgere are n0 and m0 such that all element a_n and b_m with n>n0 and m>m0 are closer than epsilon to L. Then pick max n0,m0 and you are done.
– lalala
1 hour ago
@lalala. I noticed your comment after I was done typing up this sentiment. Full credit goes to you.
– Mason
1 hour ago
1
It would be nice if more than one of the answerers actually made an effort to address the asker's question of "what flaws are there" rather than just typing out the fastest proof of the claim that comes to mind.
– T. Bongers
1 hour ago
@T.Bongers I do appreciate all of the responses but I have to agree. ;)
– Austin
1 hour ago
@lalala That was also exacly my point, with the difference that I didn't give a full proof to let the asker prove by him/herself. Sometimes that is appreciated (do not give full answer!) sometime it is not, often depending upon who is answering!
– gimusi
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This was a question on an exam I recently took that I didn't do so great on, so I'm trying to understand this problem thoroughly. What flaws are there?
Suppose that $a_n$ and $b_n$ are two convergent sequences that both converge to the same $L in mathbbR$. Define a new sequence $c_n$ such that $c_n = a_n$ if $n$ is odd and $c_n = b_n$ if $n$ is even. Prove that $c_n$ converges to $L$.
Proof Since $a_n$ and $b_n$ both converge to $L$, we know that
$$lim sup a_n = lim inf a_n = lim sup b_n = lim inf b_n = L$$
Define $i_n = infc_k : k geq n$ and $s_n = supc_k : k geq n$. Then $lim i_n = lim inf c_n$ and $lim s_n = lim sup c_n$.
$L$ is a subsequential limit of $c_n$ and so $lim inf c_n leq L leq lim sup c_n$.
Suppose for contradiction that $lim inf c_n < L$. Then there exists an $n_0$ such that $L neq i_n_0$ and $i_n_0 < a_n_0$ or $i_n_0 < b_n_0$, but this contradicts the construction of $a_n$ and $b_n$ because one or both would not converge at all or would converge to $i_n_0$. Thus, $lim inf c_n = L$.
A similar argument can be used to show that $lim sup c_n = L$ by replacing $i_n_0$ with $s_n_0$, $lim inf c_n$ with $lim sup c_n$ and switching the inequality signs in the previous paragraph.
Since $lim inf c_n = L = lim sup c_n$, $c_n$ converges to $lim a_n = lim b_n = L$.
real-analysis sequences-and-series proof-verification limsup-and-liminf
asked 1 hour ago
Austin
517
This was a question on an exam I recently took that I didn't do so great on, so I'm trying to understand this problem thoroughly. What flaws are there?
Suppose that $a_n$ and $b_n$ are two convergent sequences that both converge to the same $L in mathbbR$. Define a new sequence $c_n$ such that $c_n = a_n$ if $n$ is odd and $c_n = b_n$ if $n$ is even. Prove that $c_n$ converges to $L$.
Proof Since $a_n$ and $b_n$ both converge to $L$, we know that
$$lim sup a_n = lim inf a_n = lim sup b_n = lim inf b_n = L$$
Define $i_n = infc_k : k geq n$ and $s_n = supc_k : k geq n$. Then $lim i_n = lim inf c_n$ and $lim s_n = lim sup c_n$.
$L$ is a subsequential limit of $c_n$ and so $lim inf c_n leq L leq lim sup c_n$.
Suppose for contradiction that $lim inf c_n < L$. Then there exists an $n_0$ such that $L neq i_n_0$ and $i_n_0 < a_n_0$ or $i_n_0 < b_n_0$, but this contradicts the construction of $a_n$ and $b_n$ because one or both would not converge at all or would converge to $i_n_0$. Thus, $lim inf c_n = L$.
A similar argument can be used to show that $lim sup c_n = L$ by replacing $i_n_0$ with $s_n_0$, $lim inf c_n$ with $lim sup c_n$ and switching the inequality signs in the previous paragraph.
Since $lim inf c_n = L = lim sup c_n$, $c_n$ converges to $lim a_n = lim b_n = L$.
real-analysis sequences-and-series proof-verification limsup-and-liminf
real-analysis sequences-and-series proof-verification limsup-and-liminf
asked 1 hour ago
Austin
517
asked 1 hour ago
Austin
517
asked 1 hour ago
Austin
517
asked 1 hour ago
Austin
517
asked 1 hour ago
Austin
517
517
1
While it might be correct, this approach seems way to convoluted. Isnt it actually you picj an epsilon so you know that tgere are n0 and m0 such that all element a_n and b_m with n>n0 and m>m0 are closer than epsilon to L. Then pick max n0,m0 and you are done.
– lalala
1 hour ago
@lalala. I noticed your comment after I was done typing up this sentiment. Full credit goes to you.
– Mason
1 hour ago
1
It would be nice if more than one of the answerers actually made an effort to address the asker's question of "what flaws are there" rather than just typing out the fastest proof of the claim that comes to mind.
– T. Bongers
1 hour ago
@T.Bongers I do appreciate all of the responses but I have to agree. ;)
– Austin
1 hour ago
@lalala That was also exacly my point, with the difference that I didn't give a full proof to let the asker prove by him/herself. Sometimes that is appreciated (do not give full answer!) sometime it is not, often depending upon who is answering!
– gimusi
1 hour ago
add a comment |Â
1
While it might be correct, this approach seems way to convoluted. Isnt it actually you picj an epsilon so you know that tgere are n0 and m0 such that all element a_n and b_m with n>n0 and m>m0 are closer than epsilon to L. Then pick max n0,m0 and you are done.
– lalala
1 hour ago
@lalala. I noticed your comment after I was done typing up this sentiment. Full credit goes to you.
– Mason
1 hour ago
1
It would be nice if more than one of the answerers actually made an effort to address the asker's question of "what flaws are there" rather than just typing out the fastest proof of the claim that comes to mind.
– T. Bongers
1 hour ago
@T.Bongers I do appreciate all of the responses but I have to agree. ;)
– Austin
1 hour ago
@lalala That was also exacly my point, with the difference that I didn't give a full proof to let the asker prove by him/herself. Sometimes that is appreciated (do not give full answer!) sometime it is not, often depending upon who is answering!
– gimusi
1 hour ago
1
1
While it might be correct, this approach seems way to convoluted. Isnt it actually you picj an epsilon so you know that tgere are n0 and m0 such that all element a_n and b_m with n>n0 and m>m0 are closer than epsilon to L. Then pick max n0,m0 and you are done.
– lalala
1 hour ago
While it might be correct, this approach seems way to convoluted. Isnt it actually you picj an epsilon so you know that tgere are n0 and m0 such that all element a_n and b_m with n>n0 and m>m0 are closer than epsilon to L. Then pick max n0,m0 and you are done.
– lalala
1 hour ago
@lalala. I noticed your comment after I was done typing up this sentiment. Full credit goes to you.
– Mason
1 hour ago
@lalala. I noticed your comment after I was done typing up this sentiment. Full credit goes to you.
– Mason
1 hour ago
1
1
It would be nice if more than one of the answerers actually made an effort to address the asker's question of "what flaws are there" rather than just typing out the fastest proof of the claim that comes to mind.
– T. Bongers
1 hour ago
It would be nice if more than one of the answerers actually made an effort to address the asker's question of "what flaws are there" rather than just typing out the fastest proof of the claim that comes to mind.
– T. Bongers
1 hour ago
@T.Bongers I do appreciate all of the responses but I have to agree. ;)
– Austin
1 hour ago
@T.Bongers I do appreciate all of the responses but I have to agree. ;)
– Austin
1 hour ago
@lalala That was also exacly my point, with the difference that I didn't give a full proof to let the asker prove by him/herself. Sometimes that is appreciated (do not give full answer!) sometime it is not, often depending upon who is answering!
– gimusi
1 hour ago
@lalala That was also exacly my point, with the difference that I didn't give a full proof to let the asker prove by him/herself. Sometimes that is appreciated (do not give full answer!) sometime it is not, often depending upon who is answering!
– gimusi
1 hour ago
add a comment |Â
5 Answers
5
active
oldest
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up vote
4
down vote
I am not sure whether your current proof is wrong but I am confident that it has some unneeded extra parts. This proof should be $3$ or $4$ sentences.
For a simpler approach try this without invoking $sup$ or $inf$. If you need help see below:
Let $epsilon>0 $ be given we need to show that $exists N_c$ such that $forall n>N_c$ we have $|c_n-L|<epsilon$ but because we know $a_n$ converges we can select $N_a$ such that $forall n>N_a$ we have $|a_n-L|<epsilon$ and $b_n$ converges we can select $N_b$ such that $forall n>N_b$ we have $|b_n-L|<epsilon$. So we can just take $N_c=maxN_a,N_b$ and then we are guaranteed to be within $epsilon$ of our target.
answered 1 hour ago
Mason
1,2721225
That was also my point! It can be proved directly from the definition without such kind of complicated approach!
– gimusi
1 hour ago
1
Maybe we can interpret T Bongers comment like this. The OP didn't really ask us to prove anything so the ideal thing to do is give the feedback that we had intended: The argument is unnecessarily convoluted and not 'spoil' the fun by showing too much.
– Mason
1 hour ago
@Mason Exactly: I read this as a feedback question, rather than a "let's show how quickly I can recall the proof" question. So (+1) for your answer.
– T. Bongers
1 hour ago
@Mason Indeed it was exactly my point! I admit that maybe my answer at first was not clear on that point but my goal was exactly your one. I'v eonly chosen to do not give a full solution to let the pleasure to the asker to find it by himself. You answer also is very good in my opinion. Bye
– gimusi
1 hour ago
add a comment |Â
up vote
2
down vote
The other answers give quick proofs, but if you want to use sups and infs, that's ok too. But the flaw in your proof, in my opinion, starts here: "Then there exists an $n_0$....". Let's pick it up at "Suppose for contradiction that $liminf c_n<L$", which is fine. Here is a sketch of what comes next. I will leave the details to you: if $liminf c_n$ is strictly less than $L$, then there is an infinite number of terms of $c_n$ less than $L$. And this means that-----
answered 1 hour ago
Matematleta
8,6092918
add a comment |Â
up vote
1
down vote
$$c_2n=b_2n$$
$$c_2n+1=a_2n+1$$
$$lim_nto+inftyb_2n=lim_nto+inftya_2n+1=L$$
thus
$$lim_nto+infty c_2n=lim_nto+infty c_2n+1=L$$
and
$$lim_nto+infty c_n=L$$
answered 1 hour ago
hamam_Abdallah
35.3k21532
3
This assumes a number of facts about subsequential limits that the asker may or may not know, and sort of trivializes the problem.
– T. Bongers
1 hour ago
add a comment |Â
up vote
-1
down vote
HINT
As noticed, I am also not sure whether your current proof is wrong but it seems too much complicated indeed we don't need to use $liminf$ and $limsup$ concept since it can be carried out directly by the definition of limit.
Indeed, since $c_n to L$ for $n$ even and $c_n to L$ for $n$ odd, since we are considering two subsequences wich cover all the possible values for $n$, we can easily shown that $c_n to L$.
answered 1 hour ago
gimusi
78.1k73889
This isn't a "hint," nor does it really address the question of "what flaws are there." Several of the other answers don't really address the question either, though.
– T. Bongers
1 hour ago
@T.Bongers Yes maybe I need to clarify that. My point is as that raised up by Mason. We don't need a so complicated proof but we can proceed in a simpler way directly by the definition of limit. I'm indeed answering to the main question the asker is trying to solve. Use a not effective method is the problem in my opinion. Note that the asker is not requiring to use a particular kind of proof and moreover he/she doesn't seem to be aware about that simple way.
– gimusi
1 hour ago
Then perhaps you could expand your answer to explain how you have addressed the only question contained in the body: "What flaws are there?"
– T. Bongers
1 hour ago
add a comment |Â
up vote
-1
down vote
Here's a simple proof
$textbfProof:$
Assume $epsilon>0$, and by convergence choose $N_1,N_2inmathbbN$ such that $a_n_1,b_n_2inxinmathbbC$ for all $n_1geq N_1$ and $n_2geq N_2$. Then choosing $N=max(N_1,N_2)$ we find $c_ninxinmathbbC$ for all $ngeq N$, therefore $c_nto L$, for $epsilon$ was arbitrary.$square$
answered 1 hour ago
Melody
595
1
The first word of your proof is "assume" but don't we know that there exist such values $N_1$ and $N_2$ from the premise? I think you may be rearranging the quantifiers... Usually we let $epsilon>0$ be given and then select $N$s that do the job we want.
– Mason
58 mins ago
Woops. We gotta choose $epsilon$ before we choose $N_1,N_2$. Corrected that. Thank you. After we choose $epsilon$, then we can choose $N_1,N_2$ from the $epsilon,N$ characterization of convergence.
– Melody
50 mins ago
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
I am not sure whether your current proof is wrong but I am confident that it has some unneeded extra parts. This proof should be $3$ or $4$ sentences.
For a simpler approach try this without invoking $sup$ or $inf$. If you need help see below:
Let $epsilon>0 $ be given we need to show that $exists N_c$ such that $forall n>N_c$ we have $|c_n-L|<epsilon$ but because we know $a_n$ converges we can select $N_a$ such that $forall n>N_a$ we have $|a_n-L|<epsilon$ and $b_n$ converges we can select $N_b$ such that $forall n>N_b$ we have $|b_n-L|<epsilon$. So we can just take $N_c=maxN_a,N_b$ and then we are guaranteed to be within $epsilon$ of our target.
answered 1 hour ago
Mason
1,2721225
That was also my point! It can be proved directly from the definition without such kind of complicated approach!
– gimusi
1 hour ago
1
Maybe we can interpret T Bongers comment like this. The OP didn't really ask us to prove anything so the ideal thing to do is give the feedback that we had intended: The argument is unnecessarily convoluted and not 'spoil' the fun by showing too much.
– Mason
1 hour ago
@Mason Exactly: I read this as a feedback question, rather than a "let's show how quickly I can recall the proof" question. So (+1) for your answer.
– T. Bongers
1 hour ago
@Mason Indeed it was exactly my point! I admit that maybe my answer at first was not clear on that point but my goal was exactly your one. I'v eonly chosen to do not give a full solution to let the pleasure to the asker to find it by himself. You answer also is very good in my opinion. Bye
– gimusi
1 hour ago
add a comment |Â
up vote
4
down vote
I am not sure whether your current proof is wrong but I am confident that it has some unneeded extra parts. This proof should be $3$ or $4$ sentences.
For a simpler approach try this without invoking $sup$ or $inf$. If you need help see below:
Let $epsilon>0 $ be given we need to show that $exists N_c$ such that $forall n>N_c$ we have $|c_n-L|<epsilon$ but because we know $a_n$ converges we can select $N_a$ such that $forall n>N_a$ we have $|a_n-L|<epsilon$ and $b_n$ converges we can select $N_b$ such that $forall n>N_b$ we have $|b_n-L|<epsilon$. So we can just take $N_c=maxN_a,N_b$ and then we are guaranteed to be within $epsilon$ of our target.
answered 1 hour ago
Mason
1,2721225
That was also my point! It can be proved directly from the definition without such kind of complicated approach!
– gimusi
1 hour ago
1
Maybe we can interpret T Bongers comment like this. The OP didn't really ask us to prove anything so the ideal thing to do is give the feedback that we had intended: The argument is unnecessarily convoluted and not 'spoil' the fun by showing too much.
– Mason
1 hour ago
@Mason Exactly: I read this as a feedback question, rather than a "let's show how quickly I can recall the proof" question. So (+1) for your answer.
– T. Bongers
1 hour ago
@Mason Indeed it was exactly my point! I admit that maybe my answer at first was not clear on that point but my goal was exactly your one. I'v eonly chosen to do not give a full solution to let the pleasure to the asker to find it by himself. You answer also is very good in my opinion. Bye
– gimusi
1 hour ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I am not sure whether your current proof is wrong but I am confident that it has some unneeded extra parts. This proof should be $3$ or $4$ sentences.
For a simpler approach try this without invoking $sup$ or $inf$. If you need help see below:
Let $epsilon>0 $ be given we need to show that $exists N_c$ such that $forall n>N_c$ we have $|c_n-L|<epsilon$ but because we know $a_n$ converges we can select $N_a$ such that $forall n>N_a$ we have $|a_n-L|<epsilon$ and $b_n$ converges we can select $N_b$ such that $forall n>N_b$ we have $|b_n-L|<epsilon$. So we can just take $N_c=maxN_a,N_b$ and then we are guaranteed to be within $epsilon$ of our target.
answered 1 hour ago
Mason
1,2721225
I am not sure whether your current proof is wrong but I am confident that it has some unneeded extra parts. This proof should be $3$ or $4$ sentences.
For a simpler approach try this without invoking $sup$ or $inf$. If you need help see below:
Let $epsilon>0 $ be given we need to show that $exists N_c$ such that $forall n>N_c$ we have $|c_n-L|<epsilon$ but because we know $a_n$ converges we can select $N_a$ such that $forall n>N_a$ we have $|a_n-L|<epsilon$ and $b_n$ converges we can select $N_b$ such that $forall n>N_b$ we have $|b_n-L|<epsilon$. So we can just take $N_c=maxN_a,N_b$ and then we are guaranteed to be within $epsilon$ of our target.
answered 1 hour ago
Mason
1,2721225
edited 1 hour ago
answered 1 hour ago
Mason
1,2721225
answered 1 hour ago
Mason
1,2721225
answered 1 hour ago
Mason
1,2721225
1,2721225
That was also my point! It can be proved directly from the definition without such kind of complicated approach!
– gimusi
1 hour ago
1
Maybe we can interpret T Bongers comment like this. The OP didn't really ask us to prove anything so the ideal thing to do is give the feedback that we had intended: The argument is unnecessarily convoluted and not 'spoil' the fun by showing too much.
– Mason
1 hour ago
@Mason Exactly: I read this as a feedback question, rather than a "let's show how quickly I can recall the proof" question. So (+1) for your answer.
– T. Bongers
1 hour ago
@Mason Indeed it was exactly my point! I admit that maybe my answer at first was not clear on that point but my goal was exactly your one. I'v eonly chosen to do not give a full solution to let the pleasure to the asker to find it by himself. You answer also is very good in my opinion. Bye
– gimusi
1 hour ago
add a comment |Â
That was also my point! It can be proved directly from the definition without such kind of complicated approach!
– gimusi
1 hour ago
1
Maybe we can interpret T Bongers comment like this. The OP didn't really ask us to prove anything so the ideal thing to do is give the feedback that we had intended: The argument is unnecessarily convoluted and not 'spoil' the fun by showing too much.
– Mason
1 hour ago
@Mason Exactly: I read this as a feedback question, rather than a "let's show how quickly I can recall the proof" question. So (+1) for your answer.
– T. Bongers
1 hour ago
@Mason Indeed it was exactly my point! I admit that maybe my answer at first was not clear on that point but my goal was exactly your one. I'v eonly chosen to do not give a full solution to let the pleasure to the asker to find it by himself. You answer also is very good in my opinion. Bye
– gimusi
1 hour ago
That was also my point! It can be proved directly from the definition without such kind of complicated approach!
– gimusi
1 hour ago
That was also my point! It can be proved directly from the definition without such kind of complicated approach!
– gimusi
1 hour ago
1
1
Maybe we can interpret T Bongers comment like this. The OP didn't really ask us to prove anything so the ideal thing to do is give the feedback that we had intended: The argument is unnecessarily convoluted and not 'spoil' the fun by showing too much.
– Mason
1 hour ago
Maybe we can interpret T Bongers comment like this. The OP didn't really ask us to prove anything so the ideal thing to do is give the feedback that we had intended: The argument is unnecessarily convoluted and not 'spoil' the fun by showing too much.
– Mason
1 hour ago
@Mason Exactly: I read this as a feedback question, rather than a "let's show how quickly I can recall the proof" question. So (+1) for your answer.
– T. Bongers
1 hour ago
@Mason Exactly: I read this as a feedback question, rather than a "let's show how quickly I can recall the proof" question. So (+1) for your answer.
– T. Bongers
1 hour ago
@Mason Indeed it was exactly my point! I admit that maybe my answer at first was not clear on that point but my goal was exactly your one. I'v eonly chosen to do not give a full solution to let the pleasure to the asker to find it by himself. You answer also is very good in my opinion. Bye
– gimusi
1 hour ago
@Mason Indeed it was exactly my point! I admit that maybe my answer at first was not clear on that point but my goal was exactly your one. I'v eonly chosen to do not give a full solution to let the pleasure to the asker to find it by himself. You answer also is very good in my opinion. Bye
– gimusi
1 hour ago
add a comment |Â
up vote
2
down vote
The other answers give quick proofs, but if you want to use sups and infs, that's ok too. But the flaw in your proof, in my opinion, starts here: "Then there exists an $n_0$....". Let's pick it up at "Suppose for contradiction that $liminf c_n<L$", which is fine. Here is a sketch of what comes next. I will leave the details to you: if $liminf c_n$ is strictly less than $L$, then there is an infinite number of terms of $c_n$ less than $L$. And this means that-----
answered 1 hour ago
Matematleta
8,6092918
add a comment |Â
up vote
2
down vote
The other answers give quick proofs, but if you want to use sups and infs, that's ok too. But the flaw in your proof, in my opinion, starts here: "Then there exists an $n_0$....". Let's pick it up at "Suppose for contradiction that $liminf c_n<L$", which is fine. Here is a sketch of what comes next. I will leave the details to you: if $liminf c_n$ is strictly less than $L$, then there is an infinite number of terms of $c_n$ less than $L$. And this means that-----
answered 1 hour ago
Matematleta
8,6092918
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The other answers give quick proofs, but if you want to use sups and infs, that's ok too. But the flaw in your proof, in my opinion, starts here: "Then there exists an $n_0$....". Let's pick it up at "Suppose for contradiction that $liminf c_n<L$", which is fine. Here is a sketch of what comes next. I will leave the details to you: if $liminf c_n$ is strictly less than $L$, then there is an infinite number of terms of $c_n$ less than $L$. And this means that-----
answered 1 hour ago
Matematleta
8,6092918
The other answers give quick proofs, but if you want to use sups and infs, that's ok too. But the flaw in your proof, in my opinion, starts here: "Then there exists an $n_0$....". Let's pick it up at "Suppose for contradiction that $liminf c_n<L$", which is fine. Here is a sketch of what comes next. I will leave the details to you: if $liminf c_n$ is strictly less than $L$, then there is an infinite number of terms of $c_n$ less than $L$. And this means that-----
answered 1 hour ago
Matematleta
8,6092918
answered 1 hour ago
Matematleta
8,6092918
answered 1 hour ago
Matematleta
8,6092918
answered 1 hour ago
Matematleta
8,6092918
8,6092918
add a comment |Â
add a comment |Â
up vote
1
down vote
$$c_2n=b_2n$$
$$c_2n+1=a_2n+1$$
$$lim_nto+inftyb_2n=lim_nto+inftya_2n+1=L$$
thus
$$lim_nto+infty c_2n=lim_nto+infty c_2n+1=L$$
and
$$lim_nto+infty c_n=L$$
answered 1 hour ago
hamam_Abdallah
35.3k21532
3
This assumes a number of facts about subsequential limits that the asker may or may not know, and sort of trivializes the problem.
– T. Bongers
1 hour ago
add a comment |Â
up vote
1
down vote
$$c_2n=b_2n$$
$$c_2n+1=a_2n+1$$
$$lim_nto+inftyb_2n=lim_nto+inftya_2n+1=L$$
thus
$$lim_nto+infty c_2n=lim_nto+infty c_2n+1=L$$
and
$$lim_nto+infty c_n=L$$
answered 1 hour ago
hamam_Abdallah
35.3k21532
3
This assumes a number of facts about subsequential limits that the asker may or may not know, and sort of trivializes the problem.
– T. Bongers
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$c_2n=b_2n$$
$$c_2n+1=a_2n+1$$
$$lim_nto+inftyb_2n=lim_nto+inftya_2n+1=L$$
thus
$$lim_nto+infty c_2n=lim_nto+infty c_2n+1=L$$
and
$$lim_nto+infty c_n=L$$
answered 1 hour ago
hamam_Abdallah
35.3k21532
$$c_2n=b_2n$$
$$c_2n+1=a_2n+1$$
$$lim_nto+inftyb_2n=lim_nto+inftya_2n+1=L$$
thus
$$lim_nto+infty c_2n=lim_nto+infty c_2n+1=L$$
and
$$lim_nto+infty c_n=L$$
answered 1 hour ago
hamam_Abdallah
35.3k21532
answered 1 hour ago
hamam_Abdallah
35.3k21532
answered 1 hour ago
hamam_Abdallah
35.3k21532
answered 1 hour ago
hamam_Abdallah
35.3k21532
35.3k21532
3
This assumes a number of facts about subsequential limits that the asker may or may not know, and sort of trivializes the problem.
– T. Bongers
1 hour ago
add a comment |Â
3
This assumes a number of facts about subsequential limits that the asker may or may not know, and sort of trivializes the problem.
– T. Bongers
1 hour ago
3
3
This assumes a number of facts about subsequential limits that the asker may or may not know, and sort of trivializes the problem.
– T. Bongers
1 hour ago
This assumes a number of facts about subsequential limits that the asker may or may not know, and sort of trivializes the problem.
– T. Bongers
1 hour ago
add a comment |Â
up vote
-1
down vote
HINT
As noticed, I am also not sure whether your current proof is wrong but it seems too much complicated indeed we don't need to use $liminf$ and $limsup$ concept since it can be carried out directly by the definition of limit.
Indeed, since $c_n to L$ for $n$ even and $c_n to L$ for $n$ odd, since we are considering two subsequences wich cover all the possible values for $n$, we can easily shown that $c_n to L$.
answered 1 hour ago
gimusi
78.1k73889
This isn't a "hint," nor does it really address the question of "what flaws are there." Several of the other answers don't really address the question either, though.
– T. Bongers
1 hour ago
@T.Bongers Yes maybe I need to clarify that. My point is as that raised up by Mason. We don't need a so complicated proof but we can proceed in a simpler way directly by the definition of limit. I'm indeed answering to the main question the asker is trying to solve. Use a not effective method is the problem in my opinion. Note that the asker is not requiring to use a particular kind of proof and moreover he/she doesn't seem to be aware about that simple way.
– gimusi
1 hour ago
Then perhaps you could expand your answer to explain how you have addressed the only question contained in the body: "What flaws are there?"
– T. Bongers
1 hour ago
add a comment |Â
up vote
-1
down vote
HINT
As noticed, I am also not sure whether your current proof is wrong but it seems too much complicated indeed we don't need to use $liminf$ and $limsup$ concept since it can be carried out directly by the definition of limit.
Indeed, since $c_n to L$ for $n$ even and $c_n to L$ for $n$ odd, since we are considering two subsequences wich cover all the possible values for $n$, we can easily shown that $c_n to L$.
answered 1 hour ago
gimusi
78.1k73889
This isn't a "hint," nor does it really address the question of "what flaws are there." Several of the other answers don't really address the question either, though.
– T. Bongers
1 hour ago
@T.Bongers Yes maybe I need to clarify that. My point is as that raised up by Mason. We don't need a so complicated proof but we can proceed in a simpler way directly by the definition of limit. I'm indeed answering to the main question the asker is trying to solve. Use a not effective method is the problem in my opinion. Note that the asker is not requiring to use a particular kind of proof and moreover he/she doesn't seem to be aware about that simple way.
– gimusi
1 hour ago
Then perhaps you could expand your answer to explain how you have addressed the only question contained in the body: "What flaws are there?"
– T. Bongers
1 hour ago
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
HINT
As noticed, I am also not sure whether your current proof is wrong but it seems too much complicated indeed we don't need to use $liminf$ and $limsup$ concept since it can be carried out directly by the definition of limit.
Indeed, since $c_n to L$ for $n$ even and $c_n to L$ for $n$ odd, since we are considering two subsequences wich cover all the possible values for $n$, we can easily shown that $c_n to L$.
answered 1 hour ago
gimusi
78.1k73889
HINT
As noticed, I am also not sure whether your current proof is wrong but it seems too much complicated indeed we don't need to use $liminf$ and $limsup$ concept since it can be carried out directly by the definition of limit.
Indeed, since $c_n to L$ for $n$ even and $c_n to L$ for $n$ odd, since we are considering two subsequences wich cover all the possible values for $n$, we can easily shown that $c_n to L$.
answered 1 hour ago
gimusi
78.1k73889
edited 1 hour ago
answered 1 hour ago
gimusi
78.1k73889
answered 1 hour ago
gimusi
78.1k73889
answered 1 hour ago
gimusi
78.1k73889
78.1k73889
This isn't a "hint," nor does it really address the question of "what flaws are there." Several of the other answers don't really address the question either, though.
– T. Bongers
1 hour ago
@T.Bongers Yes maybe I need to clarify that. My point is as that raised up by Mason. We don't need a so complicated proof but we can proceed in a simpler way directly by the definition of limit. I'm indeed answering to the main question the asker is trying to solve. Use a not effective method is the problem in my opinion. Note that the asker is not requiring to use a particular kind of proof and moreover he/she doesn't seem to be aware about that simple way.
– gimusi
1 hour ago
Then perhaps you could expand your answer to explain how you have addressed the only question contained in the body: "What flaws are there?"
– T. Bongers
1 hour ago
add a comment |Â
This isn't a "hint," nor does it really address the question of "what flaws are there." Several of the other answers don't really address the question either, though.
– T. Bongers
1 hour ago
@T.Bongers Yes maybe I need to clarify that. My point is as that raised up by Mason. We don't need a so complicated proof but we can proceed in a simpler way directly by the definition of limit. I'm indeed answering to the main question the asker is trying to solve. Use a not effective method is the problem in my opinion. Note that the asker is not requiring to use a particular kind of proof and moreover he/she doesn't seem to be aware about that simple way.
– gimusi
1 hour ago
Then perhaps you could expand your answer to explain how you have addressed the only question contained in the body: "What flaws are there?"
– T. Bongers
1 hour ago
This isn't a "hint," nor does it really address the question of "what flaws are there." Several of the other answers don't really address the question either, though.
– T. Bongers
1 hour ago
This isn't a "hint," nor does it really address the question of "what flaws are there." Several of the other answers don't really address the question either, though.
– T. Bongers
1 hour ago
@T.Bongers Yes maybe I need to clarify that. My point is as that raised up by Mason. We don't need a so complicated proof but we can proceed in a simpler way directly by the definition of limit. I'm indeed answering to the main question the asker is trying to solve. Use a not effective method is the problem in my opinion. Note that the asker is not requiring to use a particular kind of proof and moreover he/she doesn't seem to be aware about that simple way.
– gimusi
1 hour ago
@T.Bongers Yes maybe I need to clarify that. My point is as that raised up by Mason. We don't need a so complicated proof but we can proceed in a simpler way directly by the definition of limit. I'm indeed answering to the main question the asker is trying to solve. Use a not effective method is the problem in my opinion. Note that the asker is not requiring to use a particular kind of proof and moreover he/she doesn't seem to be aware about that simple way.
– gimusi
1 hour ago
Then perhaps you could expand your answer to explain how you have addressed the only question contained in the body: "What flaws are there?"
– T. Bongers
1 hour ago
Then perhaps you could expand your answer to explain how you have addressed the only question contained in the body: "What flaws are there?"
– T. Bongers
1 hour ago
add a comment |Â
up vote
-1
down vote
Here's a simple proof
$textbfProof:$
Assume $epsilon>0$, and by convergence choose $N_1,N_2inmathbbN$ such that $a_n_1,b_n_2inxinmathbbC$ for all $n_1geq N_1$ and $n_2geq N_2$. Then choosing $N=max(N_1,N_2)$ we find $c_ninxinmathbbC$ for all $ngeq N$, therefore $c_nto L$, for $epsilon$ was arbitrary.$square$
answered 1 hour ago
Melody
595
1
The first word of your proof is "assume" but don't we know that there exist such values $N_1$ and $N_2$ from the premise? I think you may be rearranging the quantifiers... Usually we let $epsilon>0$ be given and then select $N$s that do the job we want.
– Mason
58 mins ago
Woops. We gotta choose $epsilon$ before we choose $N_1,N_2$. Corrected that. Thank you. After we choose $epsilon$, then we can choose $N_1,N_2$ from the $epsilon,N$ characterization of convergence.
– Melody
50 mins ago
add a comment |Â
up vote
-1
down vote
Here's a simple proof
$textbfProof:$
Assume $epsilon>0$, and by convergence choose $N_1,N_2inmathbbN$ such that $a_n_1,b_n_2inxinmathbbC$ for all $n_1geq N_1$ and $n_2geq N_2$. Then choosing $N=max(N_1,N_2)$ we find $c_ninxinmathbbC$ for all $ngeq N$, therefore $c_nto L$, for $epsilon$ was arbitrary.$square$
answered 1 hour ago
Melody
595
1
The first word of your proof is "assume" but don't we know that there exist such values $N_1$ and $N_2$ from the premise? I think you may be rearranging the quantifiers... Usually we let $epsilon>0$ be given and then select $N$s that do the job we want.
– Mason
58 mins ago
Woops. We gotta choose $epsilon$ before we choose $N_1,N_2$. Corrected that. Thank you. After we choose $epsilon$, then we can choose $N_1,N_2$ from the $epsilon,N$ characterization of convergence.
– Melody
50 mins ago
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Here's a simple proof
$textbfProof:$
Assume $epsilon>0$, and by convergence choose $N_1,N_2inmathbbN$ such that $a_n_1,b_n_2inxinmathbbC$ for all $n_1geq N_1$ and $n_2geq N_2$. Then choosing $N=max(N_1,N_2)$ we find $c_ninxinmathbbC$ for all $ngeq N$, therefore $c_nto L$, for $epsilon$ was arbitrary.$square$
answered 1 hour ago
Melody
595
Here's a simple proof
$textbfProof:$
Assume $epsilon>0$, and by convergence choose $N_1,N_2inmathbbN$ such that $a_n_1,b_n_2inxinmathbbC$ for all $n_1geq N_1$ and $n_2geq N_2$. Then choosing $N=max(N_1,N_2)$ we find $c_ninxinmathbbC$ for all $ngeq N$, therefore $c_nto L$, for $epsilon$ was arbitrary.$square$
answered 1 hour ago
Melody
595
edited 51 mins ago
answered 1 hour ago
Melody
595
answered 1 hour ago
Melody
595
answered 1 hour ago
Melody
595
595
1
The first word of your proof is "assume" but don't we know that there exist such values $N_1$ and $N_2$ from the premise? I think you may be rearranging the quantifiers... Usually we let $epsilon>0$ be given and then select $N$s that do the job we want.
– Mason
58 mins ago
Woops. We gotta choose $epsilon$ before we choose $N_1,N_2$. Corrected that. Thank you. After we choose $epsilon$, then we can choose $N_1,N_2$ from the $epsilon,N$ characterization of convergence.
– Melody
50 mins ago
add a comment |Â
1
The first word of your proof is "assume" but don't we know that there exist such values $N_1$ and $N_2$ from the premise? I think you may be rearranging the quantifiers... Usually we let $epsilon>0$ be given and then select $N$s that do the job we want.
– Mason
58 mins ago
Woops. We gotta choose $epsilon$ before we choose $N_1,N_2$. Corrected that. Thank you. After we choose $epsilon$, then we can choose $N_1,N_2$ from the $epsilon,N$ characterization of convergence.
– Melody
50 mins ago
1
1
The first word of your proof is "assume" but don't we know that there exist such values $N_1$ and $N_2$ from the premise? I think you may be rearranging the quantifiers... Usually we let $epsilon>0$ be given and then select $N$s that do the job we want.
– Mason
58 mins ago
The first word of your proof is "assume" but don't we know that there exist such values $N_1$ and $N_2$ from the premise? I think you may be rearranging the quantifiers... Usually we let $epsilon>0$ be given and then select $N$s that do the job we want.
– Mason
58 mins ago
Woops. We gotta choose $epsilon$ before we choose $N_1,N_2$. Corrected that. Thank you. After we choose $epsilon$, then we can choose $N_1,N_2$ from the $epsilon,N$ characterization of convergence.
– Melody
50 mins ago
Woops. We gotta choose $epsilon$ before we choose $N_1,N_2$. Corrected that. Thank you. After we choose $epsilon$, then we can choose $N_1,N_2$ from the $epsilon,N$ characterization of convergence.
– Melody
50 mins ago
add a comment |Â
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1
While it might be correct, this approach seems way to convoluted. Isnt it actually you picj an epsilon so you know that tgere are n0 and m0 such that all element a_n and b_m with n>n0 and m>m0 are closer than epsilon to L. Then pick max n0,m0 and you are done.
– lalala
1 hour ago
@lalala. I noticed your comment after I was done typing up this sentiment. Full credit goes to you.
– Mason
1 hour ago
1
It would be nice if more than one of the answerers actually made an effort to address the asker's question of "what flaws are there" rather than just typing out the fastest proof of the claim that comes to mind.
– T. Bongers
1 hour ago
@T.Bongers I do appreciate all of the responses but I have to agree. ;)
– Austin
1 hour ago
@lalala That was also exacly my point, with the difference that I didn't give a full proof to let the asker prove by him/herself. Sometimes that is appreciated (do not give full answer!) sometime it is not, often depending upon who is answering!
– gimusi
1 hour ago