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Jul 27, 2020 at 19:01 comment added Ralph Furman The functional equation for $\zeta(s)$ gives a formula for $\zeta(n)$, $n=0,-1,-2,\cdots$ in terms of Bernoulli numbers. As you mentioned, this formula just gives 0 for negative even integers. The value you are interested in is equivalent to computing $\zeta(3), \zeta(5), \cdots$, but these are much more mysterious, with no closed formulae. In fact, despite there being a closed formula for even natural numbers, just proving that $\zeta(3)$ is irrational ended up a very subtle argument by Apery, and there are few irrationality results for the others.
Jul 26, 2020 at 6:26 comment added user155294 @user64494 Thank you.
Jul 26, 2020 at 5:51 comment added user64494 The command of Mathematica Limit[1/Zeta[s]/Gamma[s], s -> -2*n, Assumptions -> n \[Element] Integers && n > 0] produces $$ \frac{(2 n)!}{\zeta '(-2 n)}.$$
Jul 26, 2020 at 5:32 history asked user155294 CC BY-SA 4.0