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Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
Ah I see, thanks!
Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
By the way, why you have $\frac{1}{e^t-1}-\frac{1}{t}$ rather than simply $\frac{1}{e^t-1}$ in the first identity? Generating function is $\frac{1}{e^t-1}$.
Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
This well can be the case!
Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
@tired no, I used other considerations (non-rigorous).
Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
@tired I suspect the answer is $-2\gamma$. Need to be verified
Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
@tired I do not know how to use it to obtain closed form. I do not need a numerical result...
Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
@Clement C. how to sum it up using generalized summations?
Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
So what? The series has alternating sign. One can sum it up by gerneralization.
Sep
9
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
The series has alternating sign.
Sep
9
comment Is there any mathematical or physical situations that $1+2+3+\ldots\infty=-\frac{1}{12}$ shows itself?
You can see a geometrical interpretation for this here: mathoverflow.net/questions/215762/… In short, $(1+2+3+4+...)=-\frac{\omega_-^2}2$ where $\omega_-=(1+1+1+1+...)$ is the quantity of natural numbers.
Aug
29
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}$
Now I fixed the problem. Sorry. In this form it is exactly what I want.
Aug
29
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
@Daniel Fischer it came from there but modified. I just explained why there are no factorials.
Aug
29
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
@michaelrccurtis it came from Maclauren's expansion for $1/(x + 1/2)$
Aug
29
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
@Daniel Fischer well I would be OK with any of the summation methods, such as mean vanues etc.
Aug
29
comment What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
@michaelrccurtis Bernoulli numbers.
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
"would seem to contradict the definition" - it would not if $\varepsilon^2=0$ as in dual/parabolic numbers. The given definition as I figured out leads exactly to the dual numbers system.
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
Thank you for pointing out Euler's work. It seems this definition leads exactly to the system of dual numbers: en.wikipedia.org/wiki/Dual_number In dual numbers we have exactly $\exp(\varepsilon)=1+\varepsilon$ although the first form of the definition does not work because division by $\varepsilon$ is undefined.
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
@Hurkyl well I think this question can be closed because in dual numbers en.wikipedia.org/wiki/Dual_number we have exactly that: $\exp(\varepsilon)=1+\varepsilon$
Aug
23
comment What happens if to introduce infinite and infinitesimel quantities this way?
@Hurkyl what if to consider $\varepsilon$ not just an infinitesimal but as just some new algebraic element with this property?
Aug
23
comment Conventions adopted for extended reals
@mercio no, we get 6=5.