values of the Riemann zeta function in terms of Bernoulli numbers

Theorem.

Let $k$ be an even integer and let $B_k$ be the $k$th Bernoulli number. Let $\zeta \mathit{}\mathrm{(}s\mathrm{)}$ be the Riemann zeta function. Then:

$$\zeta (k)=\frac{2^{k-1}|B_k|{\pi }^k}{k!}$$

Moreover, by using the functional equation (http://planetmath.org/RiemannZetaFunction), one calculates for all $n\mathrm{\ge }\mathrm{1}$:

$$\zeta (1-n)=\frac{{(-1)}^{n+1}{B}_n}{n}$$

which shows that $\zeta \mathit{}\mathrm{(}\mathrm{1}\mathrm{-}n\mathrm{)}\mathrm{=}\mathrm{0}$ for $n\mathrm{\ge }\mathrm{3}$ odd. For $k\mathrm{\ge }\mathrm{2}$ even, one has:

$$\zeta (1-k)=-\frac{B_k}{k}.$$

Remark.

The zeroes of the zeta function shown above, $\zeta (1-n)=0$ for $n\ge 3$ odd, are usually called the trivial zeroes of the Riemann zeta function, while the non-trivial zeroes are those in the critical strip.