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Nov
7
comment Why can $2^3$ be defined but $0^0$ cannot
$0^0$ can be defined and has been by many people.
Nov
7
comment Is $\frac00=\infty$? And what is $\frac10$? Are they same? Does it hold true for any constant $a$ in $\frac{a}0$
Obviously a wrong question linked as duplicate.
Nov
7
comment Is term “real number” equivalent to “group of algorithms generating stream of digits”?
@Asaf Karagila maybe somebody would like to improve it to make it better.
Nov
7
comment a Function with several periods
@Aniket it has no smallest period but has other periods.
Nov
7
comment a Function with several periods
OK, discrete, and what?
Oct
16
comment Is Aleph 0 a natural number?
@Will R I would say number is a common property of sets of various objects whose elements can be put in bijectional relations. Of course this would not work for non-integer numbers though
Sep
10
comment How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$
This is the reason why I needed this series, by the way: mathoverflow.net/questions/216252/…
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 $1/(x + 1/2)$