これ を示す。
$$\sum_{n=0}^\infty\frac{(-1)^{n+1}}{n^k}=\sum_{1\leq a_1\leq\cdots\leq a_k}\frac{1}{a_1\cdots a_k2^{a_k}}$$
\begin{align} \sum_{1\leq a_1\leq\cdots\leq a_k}\frac{1}{a_1\cdots a_k2^{a_k}}&= \sum_{1\leq a_1\leq\cdots\leq a_k}\frac{1}{a_2\cdots a_k2^{a_k}}\int_0^1 t^{a_1-1}dt \\ &=\int_0^1 \sum_{1\leq a_2\leq\cdots\leq a_k} \frac{1}{a_2\cdots a_k2^{a_k}} \sum_{a_1=1}^{a_2} t^{a_1-1}dt \\ &=\int_0^1 \sum_{1\leq a_2\leq\cdots\leq a_k} \frac{1}{a_2\cdots a_k2^{a_k}} \frac{1-t^{a_2}}{1-t} dt \\ &=\int_0^1\frac{dt}{1-t}\sum_{1\leq a_2\leq\cdots\leq a_k} \frac{1-t^{a_2}}{a_2\cdots a_k2^{a_k}} \\ &=\int_0^1\frac{dt}{1-t}\sum_{1\leq a_2\leq\cdots\leq a_k} \frac{1}{a_3\cdots a_k2^{a_k}}\int_t^1 u^{a_2-1}du \\ &=\int_0^1\frac{dt}{1-t_1}\int_{t_1}^1\frac{dt_2}{1-t_2}\cdots\int_{t_{k-2}}^1\frac{dt_{k-1}}{1-t_{k-1}}\sum_{1\leq a_k}\frac{1-t_{k-1}^{a_k}}{a_k2^{a_k}} \\ &=\int_0^1\frac{dt}{1-t_1}\cdots\int_{t_{k-2}}^1\frac{dt_{k-1}}{1-t_{k-1}}\int_{t_{k-1}}^1\sum_{1\leq a_k}\frac{t_{k}^{a_k-1}}{2^{a_k}}dt_k \\ &=\int_0^1\frac{dt}{1-t_1}\cdots\int_{t_{k-2}}^1\frac{dt_{k-1}}{1-t_{k-1}}\int_{t_{k-1}}^1\frac{dt_k}{2-t_k} \\ &=\int_{0< t_1<\cdots< t_k<1}\frac{dt_1}{1-t_1}\cdots\frac{dt_{k-1}}{1-t_{k-1}}\frac{dt_k}{2-t_k} \\ &=\int_{0< u_k<\cdots< u_1<1}\frac{du_1}{u_1}\cdots\frac{du_{k-1}}{u_{k-1}}\frac{du_k}{1+u_k} \qquad(t_i\mapsto1-u_i) \\ &=-\int_{0< u_k<\cdots< u_1<1}\frac{du_1}{u_1}\cdots\frac{du_{k-1}}{u_{k-1}}\frac{du_k}{-1-u_k} \\ &=-\zeta(\ol{k}) \\ &=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^k} \end{align}