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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? $\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}$
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
revised What is the sum of this series? $\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}$
added 3 characters in body; edited title
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
asked What is the sum of this series? $\sum_{n=1}^\infty\frac{\zeta(1-n)(-1)^{n+1}}{2^{n-1}}$
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
29
asked What is the sum of this series? $\sum_{n=1}^\infty\frac{2^n (-1)^{n+1} B_n}{n}$
Aug
26
revised definite integral of $x^2e^{-x^2}$
this is not related to Euler's constant
Aug
26
suggested approved edit on definite integral of $x^2e^{-x^2}$
Aug
23
revised Conventions adopted for extended reals
deleted 1 character in body