Ramanujan summation Series [duplicate]
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This question already has an answer here:
Why does $1+2+3+cdots = -frac{1}{12}$?
16 answers
I am a class 12 student and I am Wondering about the Ramanujan summation Series that is 1+2+3+..... = -1/12. As per my knowledge Summation of positive number gives a positive number but in the Ramanujan summation Series value is negative. Can you please explain this in brief.
ramanujan-summation
asked Jan 8 at 6:40
saket kumarsaket kumar
416
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marked as duplicate by Lord Shark the Unknown, mrtaurho, Robert Z, Chris Culter, anomaly Jan 8 at 6:54
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:
Why does $1+2+3+cdots = -frac{1}{12}$?
16 answers
I am a class 12 student and I am Wondering about the Ramanujan summation Series that is 1+2+3+..... = -1/12. As per my knowledge Summation of positive number gives a positive number but in the Ramanujan summation Series value is negative. Can you please explain this in brief.
ramanujan-summation
asked Jan 8 at 6:40
saket kumarsaket kumar
416
$endgroup$
marked as duplicate by Lord Shark the Unknown, mrtaurho, Robert Z, Chris Culter, anomaly Jan 8 at 6:54
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.
$begingroup$
youtube.com/watch?v=jcKRGpMiVTw
$endgroup$
– Matti P.
Jan 8 at 6:42
$begingroup$
I watch this video but I can't understand so. Plz explain in brief
$endgroup$
– saket kumar
Jan 8 at 6:43
1
$begingroup$
The short of it is that Ramanujan's summation involves a certain manipulation that isn't quite immediate. The discussion might be a bit above your head but I had a similar question last month and looking at it might be fruitful to you, especially regarding the discussion of how he derives it: math.stackexchange.com/questions/3044810/…
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– Eevee Trainer
Jan 8 at 6:47
1
$begingroup$
Take a look at this here.
$endgroup$
– mrtaurho
Jan 8 at 6:47
add a comment |
$begingroup$
This question already has an answer here:
Why does $1+2+3+cdots = -frac{1}{12}$?
16 answers
I am a class 12 student and I am Wondering about the Ramanujan summation Series that is 1+2+3+..... = -1/12. As per my knowledge Summation of positive number gives a positive number but in the Ramanujan summation Series value is negative. Can you please explain this in brief.
ramanujan-summation
asked Jan 8 at 6:40
saket kumarsaket kumar
416
$endgroup$
This question already has an answer here:
Why does $1+2+3+cdots = -frac{1}{12}$?
16 answers
I am a class 12 student and I am Wondering about the Ramanujan summation Series that is 1+2+3+..... = -1/12. As per my knowledge Summation of positive number gives a positive number but in the Ramanujan summation Series value is negative. Can you please explain this in brief.
This question already has an answer here:
Why does $1+2+3+cdots = -frac{1}{12}$?
16 answers
ramanujan-summation
ramanujan-summation
asked Jan 8 at 6:40
saket kumarsaket kumar
416
asked Jan 8 at 6:40
saket kumarsaket kumar
416
asked Jan 8 at 6:40
saket kumarsaket kumar
416
asked Jan 8 at 6:40
saket kumarsaket kumar
416
asked Jan 8 at 6:40
saket kumarsaket kumar
416
416
marked as duplicate by Lord Shark the Unknown, mrtaurho, Robert Z, Chris Culter, anomaly Jan 8 at 6:54
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.
marked as duplicate by Lord Shark the Unknown, mrtaurho, Robert Z, Chris Culter, anomaly Jan 8 at 6:54
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.
$begingroup$
youtube.com/watch?v=jcKRGpMiVTw
$endgroup$
– Matti P.
Jan 8 at 6:42
$begingroup$
I watch this video but I can't understand so. Plz explain in brief
$endgroup$
– saket kumar
Jan 8 at 6:43
1
$begingroup$
The short of it is that Ramanujan's summation involves a certain manipulation that isn't quite immediate. The discussion might be a bit above your head but I had a similar question last month and looking at it might be fruitful to you, especially regarding the discussion of how he derives it: math.stackexchange.com/questions/3044810/…
$endgroup$
– Eevee Trainer
Jan 8 at 6:47
1
$begingroup$
Take a look at this here.
$endgroup$
– mrtaurho
Jan 8 at 6:47
add a comment |
$begingroup$
youtube.com/watch?v=jcKRGpMiVTw
$endgroup$
– Matti P.
Jan 8 at 6:42
$begingroup$
I watch this video but I can't understand so. Plz explain in brief
$endgroup$
– saket kumar
Jan 8 at 6:43
1
$begingroup$
The short of it is that Ramanujan's summation involves a certain manipulation that isn't quite immediate. The discussion might be a bit above your head but I had a similar question last month and looking at it might be fruitful to you, especially regarding the discussion of how he derives it: math.stackexchange.com/questions/3044810/…
$endgroup$
– Eevee Trainer
Jan 8 at 6:47
1
$begingroup$
Take a look at this here.
$endgroup$
– mrtaurho
Jan 8 at 6:47
$begingroup$
youtube.com/watch?v=jcKRGpMiVTw
$endgroup$
– Matti P.
Jan 8 at 6:42
$begingroup$
youtube.com/watch?v=jcKRGpMiVTw
$endgroup$
– Matti P.
Jan 8 at 6:42
$begingroup$
I watch this video but I can't understand so. Plz explain in brief
$endgroup$
– saket kumar
Jan 8 at 6:43
$begingroup$
I watch this video but I can't understand so. Plz explain in brief
$endgroup$
– saket kumar
Jan 8 at 6:43
1
1
$begingroup$
The short of it is that Ramanujan's summation involves a certain manipulation that isn't quite immediate. The discussion might be a bit above your head but I had a similar question last month and looking at it might be fruitful to you, especially regarding the discussion of how he derives it: math.stackexchange.com/questions/3044810/…
$endgroup$
– Eevee Trainer
Jan 8 at 6:47
$begingroup$
The short of it is that Ramanujan's summation involves a certain manipulation that isn't quite immediate. The discussion might be a bit above your head but I had a similar question last month and looking at it might be fruitful to you, especially regarding the discussion of how he derives it: math.stackexchange.com/questions/3044810/…
$endgroup$
– Eevee Trainer
Jan 8 at 6:47
1
1
$begingroup$
Take a look at this here.
$endgroup$
– mrtaurho
Jan 8 at 6:47
$begingroup$
Take a look at this here.
$endgroup$
– mrtaurho
Jan 8 at 6:47
add a comment |
1 Answer
1
active
oldest
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Some properties of addition can be proven for any finite number $n$ of terms by induction on $n$. If I mention that the sum of $n$ integers is an integer, or (to take an example mathematicians care more about) the sum of $n$ integers doesn't depend on how you order them, (a) you know I'm right and (b) if you tried writing down a proof, it'd use induction. But just because any finite size for the list of summed terms achieves such obvious outcomes, it doesn't mean infinite lists do, because induction can't be extended to those.
In fact, it turns out there isn't one unique way to define the sum of infinite series that reproduces the usual sum of finitely many terms in the case where only finitely many terms in the infinite series are non-zero. Ramanujan summation isn't the most widely used definition. The most widely used one is the $ntoinfty$ limit of the sum of the first $n$ terms, which in this case would give $infty$ instead of $-1/12$. (Note $infty$ isn't an integer either!)
But even the usual limit-of-partial-sums definition can have surprising outcomes, in the sense of induction-provable results failing. For example, the sum of finitely many rational numbers is rational; but every irrational number is expressible as a sum of infinitely many rational numbers (e.g. add multiples of powers of $10$ to gradually give the desired limit's decimal digits), so a sum of infinitely many rational numbers needn't be rational.
answered Jan 8 at 6:53
J.G.J.G.
23.9k22539
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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
$begingroup$
Some properties of addition can be proven for any finite number $n$ of terms by induction on $n$. If I mention that the sum of $n$ integers is an integer, or (to take an example mathematicians care more about) the sum of $n$ integers doesn't depend on how you order them, (a) you know I'm right and (b) if you tried writing down a proof, it'd use induction. But just because any finite size for the list of summed terms achieves such obvious outcomes, it doesn't mean infinite lists do, because induction can't be extended to those.
In fact, it turns out there isn't one unique way to define the sum of infinite series that reproduces the usual sum of finitely many terms in the case where only finitely many terms in the infinite series are non-zero. Ramanujan summation isn't the most widely used definition. The most widely used one is the $ntoinfty$ limit of the sum of the first $n$ terms, which in this case would give $infty$ instead of $-1/12$. (Note $infty$ isn't an integer either!)
But even the usual limit-of-partial-sums definition can have surprising outcomes, in the sense of induction-provable results failing. For example, the sum of finitely many rational numbers is rational; but every irrational number is expressible as a sum of infinitely many rational numbers (e.g. add multiples of powers of $10$ to gradually give the desired limit's decimal digits), so a sum of infinitely many rational numbers needn't be rational.
answered Jan 8 at 6:53
J.G.J.G.
23.9k22539
$endgroup$
add a comment |
$begingroup$
Some properties of addition can be proven for any finite number $n$ of terms by induction on $n$. If I mention that the sum of $n$ integers is an integer, or (to take an example mathematicians care more about) the sum of $n$ integers doesn't depend on how you order them, (a) you know I'm right and (b) if you tried writing down a proof, it'd use induction. But just because any finite size for the list of summed terms achieves such obvious outcomes, it doesn't mean infinite lists do, because induction can't be extended to those.
In fact, it turns out there isn't one unique way to define the sum of infinite series that reproduces the usual sum of finitely many terms in the case where only finitely many terms in the infinite series are non-zero. Ramanujan summation isn't the most widely used definition. The most widely used one is the $ntoinfty$ limit of the sum of the first $n$ terms, which in this case would give $infty$ instead of $-1/12$. (Note $infty$ isn't an integer either!)
But even the usual limit-of-partial-sums definition can have surprising outcomes, in the sense of induction-provable results failing. For example, the sum of finitely many rational numbers is rational; but every irrational number is expressible as a sum of infinitely many rational numbers (e.g. add multiples of powers of $10$ to gradually give the desired limit's decimal digits), so a sum of infinitely many rational numbers needn't be rational.
answered Jan 8 at 6:53
J.G.J.G.
23.9k22539
$endgroup$
add a comment |
$begingroup$
Some properties of addition can be proven for any finite number $n$ of terms by induction on $n$. If I mention that the sum of $n$ integers is an integer, or (to take an example mathematicians care more about) the sum of $n$ integers doesn't depend on how you order them, (a) you know I'm right and (b) if you tried writing down a proof, it'd use induction. But just because any finite size for the list of summed terms achieves such obvious outcomes, it doesn't mean infinite lists do, because induction can't be extended to those.
In fact, it turns out there isn't one unique way to define the sum of infinite series that reproduces the usual sum of finitely many terms in the case where only finitely many terms in the infinite series are non-zero. Ramanujan summation isn't the most widely used definition. The most widely used one is the $ntoinfty$ limit of the sum of the first $n$ terms, which in this case would give $infty$ instead of $-1/12$. (Note $infty$ isn't an integer either!)
But even the usual limit-of-partial-sums definition can have surprising outcomes, in the sense of induction-provable results failing. For example, the sum of finitely many rational numbers is rational; but every irrational number is expressible as a sum of infinitely many rational numbers (e.g. add multiples of powers of $10$ to gradually give the desired limit's decimal digits), so a sum of infinitely many rational numbers needn't be rational.
answered Jan 8 at 6:53
J.G.J.G.
23.9k22539
$endgroup$
Some properties of addition can be proven for any finite number $n$ of terms by induction on $n$. If I mention that the sum of $n$ integers is an integer, or (to take an example mathematicians care more about) the sum of $n$ integers doesn't depend on how you order them, (a) you know I'm right and (b) if you tried writing down a proof, it'd use induction. But just because any finite size for the list of summed terms achieves such obvious outcomes, it doesn't mean infinite lists do, because induction can't be extended to those.
In fact, it turns out there isn't one unique way to define the sum of infinite series that reproduces the usual sum of finitely many terms in the case where only finitely many terms in the infinite series are non-zero. Ramanujan summation isn't the most widely used definition. The most widely used one is the $ntoinfty$ limit of the sum of the first $n$ terms, which in this case would give $infty$ instead of $-1/12$. (Note $infty$ isn't an integer either!)
But even the usual limit-of-partial-sums definition can have surprising outcomes, in the sense of induction-provable results failing. For example, the sum of finitely many rational numbers is rational; but every irrational number is expressible as a sum of infinitely many rational numbers (e.g. add multiples of powers of $10$ to gradually give the desired limit's decimal digits), so a sum of infinitely many rational numbers needn't be rational.
answered Jan 8 at 6:53
J.G.J.G.
23.9k22539
answered Jan 8 at 6:53
J.G.J.G.
23.9k22539
answered Jan 8 at 6:53
J.G.J.G.
23.9k22539
answered Jan 8 at 6:53
J.G.J.G.
23.9k22539
23.9k22539
add a comment |
add a comment |
$begingroup$
youtube.com/watch?v=jcKRGpMiVTw
$endgroup$
– Matti P.
Jan 8 at 6:42
$begingroup$
I watch this video but I can't understand so. Plz explain in brief
$endgroup$
– saket kumar
Jan 8 at 6:43
1
$begingroup$
The short of it is that Ramanujan's summation involves a certain manipulation that isn't quite immediate. The discussion might be a bit above your head but I had a similar question last month and looking at it might be fruitful to you, especially regarding the discussion of how he derives it: math.stackexchange.com/questions/3044810/…
$endgroup$
– Eevee Trainer
Jan 8 at 6:47
1
$begingroup$
Take a look at this here.
$endgroup$
– mrtaurho
Jan 8 at 6:47