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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? $\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}$