/sci/, let's calculate this : 1^x + 2^x + 3^x + 4^x + ... + n^x
all we need to do is express it as a simple formula.
PS : making it as a sequel of exponential numbers doesn't help, but an interesting approach would be to integer it among Relatives number
>>8589921
[math]\zeta(x)[/math]
DUH
2n^x.
JUST HOW MUCH THREADS ABOUT ZETA CAN /SCI PRODUCE PER SECOND HOLY FUCK
>>8589921
Fuck off, Tao, we're not here to do your homework
(-1/12)^x
This is the trivial Barnett function.