Let $k\geq 2$ be an integer, and $\zeta(k)=\sum_{n=1}^{\infty}n^{-k}$. Then we prove the following$$\zeta(k+1)=\frac{1}{k}\sum_{1\leq i_{1}\leq i_{2}\cdots \leq i_{k}}\frac{1}{i_{1}i_{2}\cdots i_{k}}\cdot\frac{1}{i_{k}},$$$$\zeta(k)=\frac{1}{k-1}\sum_{1\leq i_{1}\leq i_{2}\cdots \leq i_{k}}\frac{1}{i_{1}i_{2}\cdots i_{k}}\cdot\frac{1}{(i_{k}+1)},$$$$\zeta(k)=\frac{1}{2^{k}-2}\sum_{1\leq i_{1}\leq i_{2}\cdots \leq i_{k}}\frac{1}{i_{1}i_{2}\cdots i_{k}}\cdot \frac{\binom{2i_{k}}{i_{k}+1}}{i_{k}\cdot 2^{2i_{k}+1}}.,$$where the summation is over all $k$-tuples $(i_{1},i_{2},\cdots,i_{k})$ of natural numbers satisfying $1\leq i_{1}\leq i_{2}\leq \cdots\leq i_{k}$ for fixed $k$. Additionally, we also derive$$\eta(k)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{k}}=2^{k}\sum_{1\leq i_{1}\leq i_{2}\cdots \leq i_{k}}\frac{1}{i_{1}i_{2}\cdots i_{k}}\cdot \frac{1}{2^{(i_{k}+1)/{2}}}\cos\left(\frac{\pi(i_k+1)}{4}\right),$$$$\beta (k)=\sum _{{n=0}}^{\infty}{\frac {(-1)^{n}}{(2n+1)^{k}}}=\sum_{1\leq i_{1}\leq i_{2}\cdots \leq i_{k}}\frac{1}{i_{1}i_{2}\cdots i_{k}}\cdot \frac{1}{2^{(i_{k}+1)/{2}}}\sin\left(\frac{\pi(i_k+1)}{4}\right).$$