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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 $1/(x + 1/2)$