Solve this math problem for the glory of Satan.

Question

compassionate · Answer

$$\zeta(s)=\sum_{n=1}^{\infty}\frac{ 1 }{ n^5 }$$

hero · Answer

anonymous · Answer

what the hell?

experimentx · Answer

anonymous · Answer

compassionate · Answer

>Everyone cheated.

experimentx · Answer

this isn't easy ... 
probably relies on Riemann hypothesis that hasn't been proven yet.

compassionate · Answer

No, try solving for all positive integers.

experimentx · Answer

I don't know ... and i would also like to know.

experimentx · Answer

according to the page i posted above ... it says it's unproved theorem.

compassionate · Answer

By analytic continuation, and thus, by analytic continuation,
$$\zeta(-1)=\sum_{n=1}^{\infty}n = -1/2$$

unklerhaukus · Answer

oh ,

compassionate · Answer

Hail Satan for this math problem.

experimentx · Answer

try posting it here http://math.stackexchange.com/ I'm just starting complex analysis

lgbasallote · Answer

interesting....so many people willing to solve for the glory of satan...interesting

anonymous · Answer

lets not

experimentx · Answer

lemme ask people on chat.

anonymous · Answer

i'll ask @rezas  for this problem but not for the glory of satan !!!

experimentx · Answer

those guys on MSE ... they say there isn't elementary expression for odd values of Riemann zeta function

compassionate · Answer

Everyone who attempts to solve this is doing it for the glory of Satan.

anonymous · Answer

i've evaluated $$\sum \frac{1}{n^3}$$with fourier series be4 
i'll go for$$\sum \frac{1}{n^5}$$later..

experimentx · Answer

experimentx · Answer

is that 2 or 3 ?

experimentx · Answer

i guess the problems is same for all 
$$1 \over n^{2k+1}$$

experimentx · Answer

I thought this kinda form would appear with Fourier series http://www.wolframalpha.com/input/?i=sum [1%2F%282n%2B1%29^3%2C+{n%2C+0%2C+Infinity}]

experimentx · Answer

$$ {1 \over (2n+1)^3}$$
relies on zeta after all

experimentx · Answer

for now ... i give up :(((((