Mixing
$sum_{n=1}^{infty}frac{(-1)^n}{n}$ converges to any arbitrary real number? [closed]
$begingroup$
I've been reading about Riemann series theorem and I think I understand the proof and the idea. But still, I have the feeling that something is wrong with it. For instance, if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?
sequences-and-series convergence
asked Jan 27 at 19:07
ShayShay
193
$endgroup$
closed as off-topic by Did, max_zorn, Ali Caglayan, clathratus, RRL Jan 28 at 18:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, max_zorn, Ali Caglayan, clathratus, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I've been reading about Riemann series theorem and I think I understand the proof and the idea. But still, I have the feeling that something is wrong with it. For instance, if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?
sequences-and-series convergence
asked Jan 27 at 19:07
ShayShay
193
$endgroup$
closed as off-topic by Did, max_zorn, Ali Caglayan, clathratus, RRL Jan 28 at 18:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, max_zorn, Ali Caglayan, clathratus, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
When changing $sigma$ a permutation of the integers then $sum_{n=1}^infty frac{(-1)^{sigma(n)}}{sigma(n)}$ can diverge or converge to any real number. $sum_{n=1}^{infty}frac{(-1)^n}{n}$ is with the trivial permutation $sigma(n) = n$, it converges and its value is well-defined, unique
$endgroup$
– reuns
Jan 27 at 19:10
$begingroup$
The series $sum_{n=1}^infty (-1)^n/n$ has exactly one value (i.e. the limit of its partial sums). The sums of its permutations (i.e. the rearranged sums discussed in the theorem) are (counterintuitively) not necessarily equal to $sum_{n=1}^infty (-1)^n/n$.
$endgroup$
– Omnomnomnom
Jan 27 at 19:10
$begingroup$
Can you explain further how a permutation in the terms of the series makes a difference in its sum?
$endgroup$
– Shay
Jan 27 at 19:15
3
$begingroup$
"if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?" Yes it implies $0=1.$ Fortunately neither of the "ifs" is correct.
$endgroup$
– zhw.
Jan 27 at 19:29
1
$begingroup$
Downvoters: this is a genuine confusion many people have, and it's clearly expressed. That makes it a useful question.
$endgroup$
– timtfj
Jan 27 at 23:32
add a comment |
$begingroup$
I've been reading about Riemann series theorem and I think I understand the proof and the idea. But still, I have the feeling that something is wrong with it. For instance, if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?
sequences-and-series convergence
asked Jan 27 at 19:07
ShayShay
193
$endgroup$
I've been reading about Riemann series theorem and I think I understand the proof and the idea. But still, I have the feeling that something is wrong with it. For instance, if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?
sequences-and-series convergence
sequences-and-series convergence
asked Jan 27 at 19:07
ShayShay
193
asked Jan 27 at 19:07
ShayShay
193
asked Jan 27 at 19:07
ShayShay
193
asked Jan 27 at 19:07
ShayShay
193
asked Jan 27 at 19:07
ShayShay
193
193
closed as off-topic by Did, max_zorn, Ali Caglayan, clathratus, RRL Jan 28 at 18:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, max_zorn, Ali Caglayan, clathratus, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, max_zorn, Ali Caglayan, clathratus, RRL Jan 28 at 18:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, max_zorn, Ali Caglayan, clathratus, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
When changing $sigma$ a permutation of the integers then $sum_{n=1}^infty frac{(-1)^{sigma(n)}}{sigma(n)}$ can diverge or converge to any real number. $sum_{n=1}^{infty}frac{(-1)^n}{n}$ is with the trivial permutation $sigma(n) = n$, it converges and its value is well-defined, unique
$endgroup$
– reuns
Jan 27 at 19:10
$begingroup$
The series $sum_{n=1}^infty (-1)^n/n$ has exactly one value (i.e. the limit of its partial sums). The sums of its permutations (i.e. the rearranged sums discussed in the theorem) are (counterintuitively) not necessarily equal to $sum_{n=1}^infty (-1)^n/n$.
$endgroup$
– Omnomnomnom
Jan 27 at 19:10
$begingroup$
Can you explain further how a permutation in the terms of the series makes a difference in its sum?
$endgroup$
– Shay
Jan 27 at 19:15
3
$begingroup$
"if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?" Yes it implies $0=1.$ Fortunately neither of the "ifs" is correct.
$endgroup$
– zhw.
Jan 27 at 19:29
1
$begingroup$
Downvoters: this is a genuine confusion many people have, and it's clearly expressed. That makes it a useful question.
$endgroup$
– timtfj
Jan 27 at 23:32
add a comment |
1
$begingroup$
When changing $sigma$ a permutation of the integers then $sum_{n=1}^infty frac{(-1)^{sigma(n)}}{sigma(n)}$ can diverge or converge to any real number. $sum_{n=1}^{infty}frac{(-1)^n}{n}$ is with the trivial permutation $sigma(n) = n$, it converges and its value is well-defined, unique
$endgroup$
– reuns
Jan 27 at 19:10
$begingroup$
The series $sum_{n=1}^infty (-1)^n/n$ has exactly one value (i.e. the limit of its partial sums). The sums of its permutations (i.e. the rearranged sums discussed in the theorem) are (counterintuitively) not necessarily equal to $sum_{n=1}^infty (-1)^n/n$.
$endgroup$
– Omnomnomnom
Jan 27 at 19:10
$begingroup$
Can you explain further how a permutation in the terms of the series makes a difference in its sum?
$endgroup$
– Shay
Jan 27 at 19:15
3
$begingroup$
"if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?" Yes it implies $0=1.$ Fortunately neither of the "ifs" is correct.
$endgroup$
– zhw.
Jan 27 at 19:29
1
$begingroup$
Downvoters: this is a genuine confusion many people have, and it's clearly expressed. That makes it a useful question.
$endgroup$
– timtfj
Jan 27 at 23:32
1
1
$begingroup$
When changing $sigma$ a permutation of the integers then $sum_{n=1}^infty frac{(-1)^{sigma(n)}}{sigma(n)}$ can diverge or converge to any real number. $sum_{n=1}^{infty}frac{(-1)^n}{n}$ is with the trivial permutation $sigma(n) = n$, it converges and its value is well-defined, unique
$endgroup$
– reuns
Jan 27 at 19:10
$begingroup$
When changing $sigma$ a permutation of the integers then $sum_{n=1}^infty frac{(-1)^{sigma(n)}}{sigma(n)}$ can diverge or converge to any real number. $sum_{n=1}^{infty}frac{(-1)^n}{n}$ is with the trivial permutation $sigma(n) = n$, it converges and its value is well-defined, unique
$endgroup$
– reuns
Jan 27 at 19:10
$begingroup$
The series $sum_{n=1}^infty (-1)^n/n$ has exactly one value (i.e. the limit of its partial sums). The sums of its permutations (i.e. the rearranged sums discussed in the theorem) are (counterintuitively) not necessarily equal to $sum_{n=1}^infty (-1)^n/n$.
$endgroup$
– Omnomnomnom
Jan 27 at 19:10
$begingroup$
The series $sum_{n=1}^infty (-1)^n/n$ has exactly one value (i.e. the limit of its partial sums). The sums of its permutations (i.e. the rearranged sums discussed in the theorem) are (counterintuitively) not necessarily equal to $sum_{n=1}^infty (-1)^n/n$.
$endgroup$
– Omnomnomnom
Jan 27 at 19:10
$begingroup$
Can you explain further how a permutation in the terms of the series makes a difference in its sum?
$endgroup$
– Shay
Jan 27 at 19:15
$begingroup$
Can you explain further how a permutation in the terms of the series makes a difference in its sum?
$endgroup$
– Shay
Jan 27 at 19:15
3
3
$begingroup$
"if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?" Yes it implies $0=1.$ Fortunately neither of the "ifs" is correct.
$endgroup$
– zhw.
Jan 27 at 19:29
$begingroup$
"if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?" Yes it implies $0=1.$ Fortunately neither of the "ifs" is correct.
$endgroup$
– zhw.
Jan 27 at 19:29
1
1
$begingroup$
Downvoters: this is a genuine confusion many people have, and it's clearly expressed. That makes it a useful question.
$endgroup$
– timtfj
Jan 27 at 23:32
$begingroup$
Downvoters: this is a genuine confusion many people have, and it's clearly expressed. That makes it a useful question.
$endgroup$
– timtfj
Jan 27 at 23:32
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An explanation, rather than a complete proof, of how the reordered series can converge to any predefined limit.
Consider a conditionally convergent series such as the one in your question. The series of positive terms $sum p_n$ diverges, as does the series of negative terms $sum q_m$.
Choose $a$ to be whatever limit you want the reordered series—not the original series—to converge to. For the sake of argument let's use $a>0$.
Since $sum p_n$ diverges, you can take just enough terms from the start of the sequence $(p_n)$ to add up to $>a$. Next, since $sum q_m$ also diverges, you can follow these with just enough terms from $(q_n)$ to get the sum back below $a$. Switch back to taking positive terms to get above $a$ again.
Continue like this, alternating between groups of positive and negative terms. Each time, the amount by which you overshoot $a$ is no more than the size of the last term used.
Now, since the original series is convergent, the sequences $(p_n)$ and $(q_m)$ both converge to $0$. So the size of the "overshoot" also converges to $0$. In other words, the reordered series we're constructing converges to $a$.
The divergerce of $sum p_n$ and $sum q_n$ guarantees you never get stuck one side of $a$, and convergence of $(p_n)$ and $(q_m)$ to $0$ guarantees that the reordered series converges.
The crucial point is that the reordered series isn't the same series, and its limit is the limit of the process of adding more terms—not the sum of "all the positive terms and all the negative terms", which is $infty-infty$ and hence undefined.
Edit: I think a big source of confusion here is the notation $sum_{n=0}^{infty}{a_n}$ which looks as though it should mean "Collect all the infinitely many terms $a_n$ together and find their sum". But it's defined not to mean that because the sum doesn't always exist. Instead it's defined as the limit of a sequence of finite sums containing more and more terms. Changing the order of the terms changes the sequence of sums, which can sometimes change the limit.
answered Jan 27 at 21:45
timtfjtimtfj
2,478420
$endgroup$
1
$begingroup$
"it's defined as the limit of a sequence of finite sums containing more and more terms." Aaaaaaah. Something just clicked! I understand now. Thank you so much!
$endgroup$
– Shay
Jan 28 at 9:10
$begingroup$
@Shay Excellent!
$endgroup$
– timtfj
Jan 28 at 11:27
add a comment |
$begingroup$
The Riemann series theorem has nothing to do with the sum of the series $displaystylesum_{n=1}^inftyfrac{(-1)^n}n$ which is equal to $-log2$ (but it's not equal to $0$ or to $1$).
If you apply the theorem to this series, what it tells you is that, for any $alphainmathbb R$, there is some bijection $bcolonmathbb{N}longrightarrowmathbb N$ such that $displaystylesum_{n=1}^inftyfrac{(-1)^{b(n)}}{b(n)}=alpha$.
answered Jan 27 at 19:16
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
1
$begingroup$
Can you explain how a permutation of the series' terms changes its sum? It is really counterintuitive for me.
$endgroup$
– Shay
Jan 27 at 19:19
$begingroup$
Take a look at the proof.
$endgroup$
– José Carlos Santos
Jan 27 at 19:23
1
$begingroup$
Yes, I understand the proof, and I accept it's true. Even then, I still don't understand why $sum a_{sigma(n)}$ would be different from $sum a_n$. They all have the same terms, it's just a permutation. Infinity is weird.
$endgroup$
– Shay
Jan 27 at 19:44
2
$begingroup$
@Shay Yes, the problem is the infinity, and, probably, the definition of the series. To avoid the problem it would be sufficient to require a canonical order of the terms (e.g. $forall m>n: |a_m|le|a_n|$). Actually most series are in this form and I know no case where some other order has a practical importance.
$endgroup$
– user
Jan 27 at 20:21
1
$begingroup$
@shay It's because the literal "sum" would be $infty-infty$ so is undefined. $sum_{n=0}^{infty}$ is shorthand for "take the limit of the sequence of finite sums", not "add up all the terms". Changing the order of the terms changes the sequence of partial sums.
$endgroup$
– timtfj
Jan 27 at 22:44
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An explanation, rather than a complete proof, of how the reordered series can converge to any predefined limit.
Consider a conditionally convergent series such as the one in your question. The series of positive terms $sum p_n$ diverges, as does the series of negative terms $sum q_m$.
Choose $a$ to be whatever limit you want the reordered series—not the original series—to converge to. For the sake of argument let's use $a>0$.
Since $sum p_n$ diverges, you can take just enough terms from the start of the sequence $(p_n)$ to add up to $>a$. Next, since $sum q_m$ also diverges, you can follow these with just enough terms from $(q_n)$ to get the sum back below $a$. Switch back to taking positive terms to get above $a$ again.
Continue like this, alternating between groups of positive and negative terms. Each time, the amount by which you overshoot $a$ is no more than the size of the last term used.
Now, since the original series is convergent, the sequences $(p_n)$ and $(q_m)$ both converge to $0$. So the size of the "overshoot" also converges to $0$. In other words, the reordered series we're constructing converges to $a$.
The divergerce of $sum p_n$ and $sum q_n$ guarantees you never get stuck one side of $a$, and convergence of $(p_n)$ and $(q_m)$ to $0$ guarantees that the reordered series converges.
The crucial point is that the reordered series isn't the same series, and its limit is the limit of the process of adding more terms—not the sum of "all the positive terms and all the negative terms", which is $infty-infty$ and hence undefined.
Edit: I think a big source of confusion here is the notation $sum_{n=0}^{infty}{a_n}$ which looks as though it should mean "Collect all the infinitely many terms $a_n$ together and find their sum". But it's defined not to mean that because the sum doesn't always exist. Instead it's defined as the limit of a sequence of finite sums containing more and more terms. Changing the order of the terms changes the sequence of sums, which can sometimes change the limit.
answered Jan 27 at 21:45
timtfjtimtfj
2,478420
$endgroup$
1
$begingroup$
"it's defined as the limit of a sequence of finite sums containing more and more terms." Aaaaaaah. Something just clicked! I understand now. Thank you so much!
$endgroup$
– Shay
Jan 28 at 9:10
$begingroup$
@Shay Excellent!
$endgroup$
– timtfj
Jan 28 at 11:27
add a comment |
$begingroup$
An explanation, rather than a complete proof, of how the reordered series can converge to any predefined limit.
Consider a conditionally convergent series such as the one in your question. The series of positive terms $sum p_n$ diverges, as does the series of negative terms $sum q_m$.
Choose $a$ to be whatever limit you want the reordered series—not the original series—to converge to. For the sake of argument let's use $a>0$.
Since $sum p_n$ diverges, you can take just enough terms from the start of the sequence $(p_n)$ to add up to $>a$. Next, since $sum q_m$ also diverges, you can follow these with just enough terms from $(q_n)$ to get the sum back below $a$. Switch back to taking positive terms to get above $a$ again.
Continue like this, alternating between groups of positive and negative terms. Each time, the amount by which you overshoot $a$ is no more than the size of the last term used.
Now, since the original series is convergent, the sequences $(p_n)$ and $(q_m)$ both converge to $0$. So the size of the "overshoot" also converges to $0$. In other words, the reordered series we're constructing converges to $a$.
The divergerce of $sum p_n$ and $sum q_n$ guarantees you never get stuck one side of $a$, and convergence of $(p_n)$ and $(q_m)$ to $0$ guarantees that the reordered series converges.
The crucial point is that the reordered series isn't the same series, and its limit is the limit of the process of adding more terms—not the sum of "all the positive terms and all the negative terms", which is $infty-infty$ and hence undefined.
Edit: I think a big source of confusion here is the notation $sum_{n=0}^{infty}{a_n}$ which looks as though it should mean "Collect all the infinitely many terms $a_n$ together and find their sum". But it's defined not to mean that because the sum doesn't always exist. Instead it's defined as the limit of a sequence of finite sums containing more and more terms. Changing the order of the terms changes the sequence of sums, which can sometimes change the limit.
answered Jan 27 at 21:45
timtfjtimtfj
2,478420
$endgroup$
1
$begingroup$
"it's defined as the limit of a sequence of finite sums containing more and more terms." Aaaaaaah. Something just clicked! I understand now. Thank you so much!
$endgroup$
– Shay
Jan 28 at 9:10
$begingroup$
@Shay Excellent!
$endgroup$
– timtfj
Jan 28 at 11:27
add a comment |
$begingroup$
An explanation, rather than a complete proof, of how the reordered series can converge to any predefined limit.
Consider a conditionally convergent series such as the one in your question. The series of positive terms $sum p_n$ diverges, as does the series of negative terms $sum q_m$.
Choose $a$ to be whatever limit you want the reordered series—not the original series—to converge to. For the sake of argument let's use $a>0$.
Since $sum p_n$ diverges, you can take just enough terms from the start of the sequence $(p_n)$ to add up to $>a$. Next, since $sum q_m$ also diverges, you can follow these with just enough terms from $(q_n)$ to get the sum back below $a$. Switch back to taking positive terms to get above $a$ again.
Continue like this, alternating between groups of positive and negative terms. Each time, the amount by which you overshoot $a$ is no more than the size of the last term used.
Now, since the original series is convergent, the sequences $(p_n)$ and $(q_m)$ both converge to $0$. So the size of the "overshoot" also converges to $0$. In other words, the reordered series we're constructing converges to $a$.
The divergerce of $sum p_n$ and $sum q_n$ guarantees you never get stuck one side of $a$, and convergence of $(p_n)$ and $(q_m)$ to $0$ guarantees that the reordered series converges.
The crucial point is that the reordered series isn't the same series, and its limit is the limit of the process of adding more terms—not the sum of "all the positive terms and all the negative terms", which is $infty-infty$ and hence undefined.
Edit: I think a big source of confusion here is the notation $sum_{n=0}^{infty}{a_n}$ which looks as though it should mean "Collect all the infinitely many terms $a_n$ together and find their sum". But it's defined not to mean that because the sum doesn't always exist. Instead it's defined as the limit of a sequence of finite sums containing more and more terms. Changing the order of the terms changes the sequence of sums, which can sometimes change the limit.
answered Jan 27 at 21:45
timtfjtimtfj
2,478420
$endgroup$
An explanation, rather than a complete proof, of how the reordered series can converge to any predefined limit.
Consider a conditionally convergent series such as the one in your question. The series of positive terms $sum p_n$ diverges, as does the series of negative terms $sum q_m$.
Choose $a$ to be whatever limit you want the reordered series—not the original series—to converge to. For the sake of argument let's use $a>0$.
Since $sum p_n$ diverges, you can take just enough terms from the start of the sequence $(p_n)$ to add up to $>a$. Next, since $sum q_m$ also diverges, you can follow these with just enough terms from $(q_n)$ to get the sum back below $a$. Switch back to taking positive terms to get above $a$ again.
Continue like this, alternating between groups of positive and negative terms. Each time, the amount by which you overshoot $a$ is no more than the size of the last term used.
Now, since the original series is convergent, the sequences $(p_n)$ and $(q_m)$ both converge to $0$. So the size of the "overshoot" also converges to $0$. In other words, the reordered series we're constructing converges to $a$.
The divergerce of $sum p_n$ and $sum q_n$ guarantees you never get stuck one side of $a$, and convergence of $(p_n)$ and $(q_m)$ to $0$ guarantees that the reordered series converges.
The crucial point is that the reordered series isn't the same series, and its limit is the limit of the process of adding more terms—not the sum of "all the positive terms and all the negative terms", which is $infty-infty$ and hence undefined.
Edit: I think a big source of confusion here is the notation $sum_{n=0}^{infty}{a_n}$ which looks as though it should mean "Collect all the infinitely many terms $a_n$ together and find their sum". But it's defined not to mean that because the sum doesn't always exist. Instead it's defined as the limit of a sequence of finite sums containing more and more terms. Changing the order of the terms changes the sequence of sums, which can sometimes change the limit.
answered Jan 27 at 21:45
timtfjtimtfj
2,478420
edited Jan 27 at 23:15
answered Jan 27 at 21:45
timtfjtimtfj
2,478420
answered Jan 27 at 21:45
timtfjtimtfj
2,478420
answered Jan 27 at 21:45
timtfjtimtfj
2,478420
2,478420
1
$begingroup$
"it's defined as the limit of a sequence of finite sums containing more and more terms." Aaaaaaah. Something just clicked! I understand now. Thank you so much!
$endgroup$
– Shay
Jan 28 at 9:10
$begingroup$
@Shay Excellent!
$endgroup$
– timtfj
Jan 28 at 11:27
add a comment |
1
$begingroup$
"it's defined as the limit of a sequence of finite sums containing more and more terms." Aaaaaaah. Something just clicked! I understand now. Thank you so much!
$endgroup$
– Shay
Jan 28 at 9:10
$begingroup$
@Shay Excellent!
$endgroup$
– timtfj
Jan 28 at 11:27
1
1
$begingroup$
"it's defined as the limit of a sequence of finite sums containing more and more terms." Aaaaaaah. Something just clicked! I understand now. Thank you so much!
$endgroup$
– Shay
Jan 28 at 9:10
$begingroup$
"it's defined as the limit of a sequence of finite sums containing more and more terms." Aaaaaaah. Something just clicked! I understand now. Thank you so much!
$endgroup$
– Shay
Jan 28 at 9:10
$begingroup$
@Shay Excellent!
$endgroup$
– timtfj
Jan 28 at 11:27
$begingroup$
@Shay Excellent!
$endgroup$
– timtfj
Jan 28 at 11:27
add a comment |
$begingroup$
The Riemann series theorem has nothing to do with the sum of the series $displaystylesum_{n=1}^inftyfrac{(-1)^n}n$ which is equal to $-log2$ (but it's not equal to $0$ or to $1$).
If you apply the theorem to this series, what it tells you is that, for any $alphainmathbb R$, there is some bijection $bcolonmathbb{N}longrightarrowmathbb N$ such that $displaystylesum_{n=1}^inftyfrac{(-1)^{b(n)}}{b(n)}=alpha$.
answered Jan 27 at 19:16
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
1
$begingroup$
Can you explain how a permutation of the series' terms changes its sum? It is really counterintuitive for me.
$endgroup$
– Shay
Jan 27 at 19:19
$begingroup$
Take a look at the proof.
$endgroup$
– José Carlos Santos
Jan 27 at 19:23
1
$begingroup$
Yes, I understand the proof, and I accept it's true. Even then, I still don't understand why $sum a_{sigma(n)}$ would be different from $sum a_n$. They all have the same terms, it's just a permutation. Infinity is weird.
$endgroup$
– Shay
Jan 27 at 19:44
2
$begingroup$
@Shay Yes, the problem is the infinity, and, probably, the definition of the series. To avoid the problem it would be sufficient to require a canonical order of the terms (e.g. $forall m>n: |a_m|le|a_n|$). Actually most series are in this form and I know no case where some other order has a practical importance.
$endgroup$
– user
Jan 27 at 20:21
1
$begingroup$
@shay It's because the literal "sum" would be $infty-infty$ so is undefined. $sum_{n=0}^{infty}$ is shorthand for "take the limit of the sequence of finite sums", not "add up all the terms". Changing the order of the terms changes the sequence of partial sums.
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– timtfj
Jan 27 at 22:44
add a comment |
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The Riemann series theorem has nothing to do with the sum of the series $displaystylesum_{n=1}^inftyfrac{(-1)^n}n$ which is equal to $-log2$ (but it's not equal to $0$ or to $1$).
If you apply the theorem to this series, what it tells you is that, for any $alphainmathbb R$, there is some bijection $bcolonmathbb{N}longrightarrowmathbb N$ such that $displaystylesum_{n=1}^inftyfrac{(-1)^{b(n)}}{b(n)}=alpha$.
answered Jan 27 at 19:16
José Carlos SantosJosé Carlos Santos
170k23132238
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1
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Can you explain how a permutation of the series' terms changes its sum? It is really counterintuitive for me.
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– Shay
Jan 27 at 19:19
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Take a look at the proof.
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– José Carlos Santos
Jan 27 at 19:23
1
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Yes, I understand the proof, and I accept it's true. Even then, I still don't understand why $sum a_{sigma(n)}$ would be different from $sum a_n$. They all have the same terms, it's just a permutation. Infinity is weird.
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– Shay
Jan 27 at 19:44
2
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@Shay Yes, the problem is the infinity, and, probably, the definition of the series. To avoid the problem it would be sufficient to require a canonical order of the terms (e.g. $forall m>n: |a_m|le|a_n|$). Actually most series are in this form and I know no case where some other order has a practical importance.
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– user
Jan 27 at 20:21
1
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@shay It's because the literal "sum" would be $infty-infty$ so is undefined. $sum_{n=0}^{infty}$ is shorthand for "take the limit of the sequence of finite sums", not "add up all the terms". Changing the order of the terms changes the sequence of partial sums.
$endgroup$
– timtfj
Jan 27 at 22:44
add a comment |
$begingroup$
The Riemann series theorem has nothing to do with the sum of the series $displaystylesum_{n=1}^inftyfrac{(-1)^n}n$ which is equal to $-log2$ (but it's not equal to $0$ or to $1$).
If you apply the theorem to this series, what it tells you is that, for any $alphainmathbb R$, there is some bijection $bcolonmathbb{N}longrightarrowmathbb N$ such that $displaystylesum_{n=1}^inftyfrac{(-1)^{b(n)}}{b(n)}=alpha$.
answered Jan 27 at 19:16
José Carlos SantosJosé Carlos Santos
170k23132238
$endgroup$
The Riemann series theorem has nothing to do with the sum of the series $displaystylesum_{n=1}^inftyfrac{(-1)^n}n$ which is equal to $-log2$ (but it's not equal to $0$ or to $1$).
If you apply the theorem to this series, what it tells you is that, for any $alphainmathbb R$, there is some bijection $bcolonmathbb{N}longrightarrowmathbb N$ such that $displaystylesum_{n=1}^inftyfrac{(-1)^{b(n)}}{b(n)}=alpha$.
answered Jan 27 at 19:16
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 19:16
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 19:16
José Carlos SantosJosé Carlos Santos
170k23132238
answered Jan 27 at 19:16
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
1
$begingroup$
Can you explain how a permutation of the series' terms changes its sum? It is really counterintuitive for me.
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– Shay
Jan 27 at 19:19
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Take a look at the proof.
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– José Carlos Santos
Jan 27 at 19:23
1
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Yes, I understand the proof, and I accept it's true. Even then, I still don't understand why $sum a_{sigma(n)}$ would be different from $sum a_n$. They all have the same terms, it's just a permutation. Infinity is weird.
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– Shay
Jan 27 at 19:44
2
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@Shay Yes, the problem is the infinity, and, probably, the definition of the series. To avoid the problem it would be sufficient to require a canonical order of the terms (e.g. $forall m>n: |a_m|le|a_n|$). Actually most series are in this form and I know no case where some other order has a practical importance.
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– user
Jan 27 at 20:21
1
$begingroup$
@shay It's because the literal "sum" would be $infty-infty$ so is undefined. $sum_{n=0}^{infty}$ is shorthand for "take the limit of the sequence of finite sums", not "add up all the terms". Changing the order of the terms changes the sequence of partial sums.
$endgroup$
– timtfj
Jan 27 at 22:44
add a comment |
1
$begingroup$
Can you explain how a permutation of the series' terms changes its sum? It is really counterintuitive for me.
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– Shay
Jan 27 at 19:19
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Take a look at the proof.
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– José Carlos Santos
Jan 27 at 19:23
1
$begingroup$
Yes, I understand the proof, and I accept it's true. Even then, I still don't understand why $sum a_{sigma(n)}$ would be different from $sum a_n$. They all have the same terms, it's just a permutation. Infinity is weird.
$endgroup$
– Shay
Jan 27 at 19:44
2
$begingroup$
@Shay Yes, the problem is the infinity, and, probably, the definition of the series. To avoid the problem it would be sufficient to require a canonical order of the terms (e.g. $forall m>n: |a_m|le|a_n|$). Actually most series are in this form and I know no case where some other order has a practical importance.
$endgroup$
– user
Jan 27 at 20:21
1
$begingroup$
@shay It's because the literal "sum" would be $infty-infty$ so is undefined. $sum_{n=0}^{infty}$ is shorthand for "take the limit of the sequence of finite sums", not "add up all the terms". Changing the order of the terms changes the sequence of partial sums.
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– timtfj
Jan 27 at 22:44
1
1
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Can you explain how a permutation of the series' terms changes its sum? It is really counterintuitive for me.
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– Shay
Jan 27 at 19:19
$begingroup$
Can you explain how a permutation of the series' terms changes its sum? It is really counterintuitive for me.
$endgroup$
– Shay
Jan 27 at 19:19
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Take a look at the proof.
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– José Carlos Santos
Jan 27 at 19:23
$begingroup$
Take a look at the proof.
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– José Carlos Santos
Jan 27 at 19:23
1
1
$begingroup$
Yes, I understand the proof, and I accept it's true. Even then, I still don't understand why $sum a_{sigma(n)}$ would be different from $sum a_n$. They all have the same terms, it's just a permutation. Infinity is weird.
$endgroup$
– Shay
Jan 27 at 19:44
$begingroup$
Yes, I understand the proof, and I accept it's true. Even then, I still don't understand why $sum a_{sigma(n)}$ would be different from $sum a_n$. They all have the same terms, it's just a permutation. Infinity is weird.
$endgroup$
– Shay
Jan 27 at 19:44
2
2
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@Shay Yes, the problem is the infinity, and, probably, the definition of the series. To avoid the problem it would be sufficient to require a canonical order of the terms (e.g. $forall m>n: |a_m|le|a_n|$). Actually most series are in this form and I know no case where some other order has a practical importance.
$endgroup$
– user
Jan 27 at 20:21
$begingroup$
@Shay Yes, the problem is the infinity, and, probably, the definition of the series. To avoid the problem it would be sufficient to require a canonical order of the terms (e.g. $forall m>n: |a_m|le|a_n|$). Actually most series are in this form and I know no case where some other order has a practical importance.
$endgroup$
– user
Jan 27 at 20:21
1
1
$begingroup$
@shay It's because the literal "sum" would be $infty-infty$ so is undefined. $sum_{n=0}^{infty}$ is shorthand for "take the limit of the sequence of finite sums", not "add up all the terms". Changing the order of the terms changes the sequence of partial sums.
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– timtfj
Jan 27 at 22:44
$begingroup$
@shay It's because the literal "sum" would be $infty-infty$ so is undefined. $sum_{n=0}^{infty}$ is shorthand for "take the limit of the sequence of finite sums", not "add up all the terms". Changing the order of the terms changes the sequence of partial sums.
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– timtfj
Jan 27 at 22:44
add a comment |
1
$begingroup$
When changing $sigma$ a permutation of the integers then $sum_{n=1}^infty frac{(-1)^{sigma(n)}}{sigma(n)}$ can diverge or converge to any real number. $sum_{n=1}^{infty}frac{(-1)^n}{n}$ is with the trivial permutation $sigma(n) = n$, it converges and its value is well-defined, unique
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– reuns
Jan 27 at 19:10
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The series $sum_{n=1}^infty (-1)^n/n$ has exactly one value (i.e. the limit of its partial sums). The sums of its permutations (i.e. the rearranged sums discussed in the theorem) are (counterintuitively) not necessarily equal to $sum_{n=1}^infty (-1)^n/n$.
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– Omnomnomnom
Jan 27 at 19:10
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Can you explain further how a permutation in the terms of the series makes a difference in its sum?
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– Shay
Jan 27 at 19:15
3
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"if $sum_{n=1}^{infty}frac{(-1)^n}{n}=0$ and $sum_{n=1}^{infty}frac{(-1)^n}{n}=1$ doesn't this mean $0=1$?" Yes it implies $0=1.$ Fortunately neither of the "ifs" is correct.
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– zhw.
Jan 27 at 19:29
1
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Downvoters: this is a genuine confusion many people have, and it's clearly expressed. That makes it a useful question.
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– timtfj
Jan 27 at 23:32