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 Sep 12 awarded Announcer Sep 10 revised Why does $e$ have multiple definitions? not related to Euler's constant 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 accepted How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$ 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 revised How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$ added 175 characters in body 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 asked How to sum up this series? $\sum_{n=1}^\infty\frac{(-1)^{n-1} B_n}{n}$ 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. Sep 9 answered Why does $1+2+3+\cdots = -\frac{1}{12}$? Sep 9 asked Series expansions of trigonometric functions using cosecant numbers Aug 29 asked What function is this? $\sum_{k=0}^\infty \frac{2^{2k}z^{2k-1}}{(2k)!}$ Aug 29 accepted What is the sum of this series? $\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}$