- #1
Walid-yahya
- 2
- 0
The sum of positive integers up to infinity: Was Sirinivasa right?
In summary, the article explores the concept of summing positive integers to infinity, a topic famously associated with mathematician Srinivasa Ramanujan. It examines Ramanujan's unconventional result that the sum of all positive integers (1 + 2 + 3 + ...) equals -1/12, a conclusion derived from analytic continuation and the Riemann zeta function. The discussion highlights the implications of this result in theoretical physics and mathematics, while clarifying that the sum does not converge in the traditional sense but can be interpreted within specific contexts. The article ultimately reflects on the blend of intuition and rigorous mathematics in Ramanujan's work.
- #2
- 19,753
- 25,757
What did you want to show? You could as well consider ##S_n=1+2+\ldots+n## and observe that ##S_{n+1}\geq S_{n}+1## for every ##n##, hence
$$
S:=\displaystyle{\lim_{n \to \infty}S_n}\geq \lim_{n \to \infty}(S_1 + n)=1+\lim_{n \to \infty}n = \infty .
$$
Things get interesting if you allow negative summands. In that case, re-orderings could result in different sums.
$$
S:=\displaystyle{\lim_{n \to \infty}S_n}\geq \lim_{n \to \infty}(S_1 + n)=1+\lim_{n \to \infty}n = \infty .
$$
Things get interesting if you allow negative summands. In that case, re-orderings could result in different sums.
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- #3
- 19,753
- 25,757
Here is an interesting note on series with alternating signs:
$$
\sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\pm\ldots=\log 2
$$
which can be rearranged such that
$$
\sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\pm\ldots=\log \sqrt{2}
$$
(My notation here is sloppy since it doesn't show the rearrangement. It is only to emphasize that rearrangements aren't automatically allowed. The reference is precise at this point.)
Reference: https://www.physicsforums.com/insig...rom-zeno-to-quantum-theory/#Domains-of-Series
$$
\sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\pm\ldots=\log 2
$$
which can be rearranged such that
$$
\sum_{n=1}^\infty \dfrac{(-1)^{n+1}}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\pm\ldots=\log \sqrt{2}
$$
(My notation here is sloppy since it doesn't show the rearrangement. It is only to emphasize that rearrangements aren't automatically allowed. The reference is precise at this point.)
Reference: https://www.physicsforums.com/insig...rom-zeno-to-quantum-theory/#Domains-of-Series
- #4
WWGD
Science Advisor
Gold Member
- 7,420
- 11,424
Well, you have a sequence that is clearly not Cauchy; that itself should do it as proof that it doesn't converge.
- #5
Walid-yahya
- 2
- 0
Great note dear It is clear that you have realized that the sum will reach infinity, and I place in your hands these papers for a new formula for this series.
- #6
- 19,753
- 25,757
You can play with rearrangements of the natural numbers as often as you like, but this is nothing we could discuss here. Furthermore, please use ##\LaTeX## (https://www.physicsforums.com/help/latexhelp/) instead of uploading pictures.
Your last sentence is nonsense and suggests an assessment as a personal speculation which we do not discuss here. It is the third shortest way to leave our community. The theory of cardinalities is not trivial and the term "wave" in your post is nonsense, particularly on a website dedicated to physics.
This thread is closed now.
Your last sentence is nonsense and suggests an assessment as a personal speculation which we do not discuss here. It is the third shortest way to leave our community. The theory of cardinalities is not trivial and the term "wave" in your post is nonsense, particularly on a website dedicated to physics.
This thread is closed now.
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