Arrangement of the following term

Question

How

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{2m}}$$

can be rewritten as

$$\sum_{k=1}^{\infty}\frac{(-1)^{2k-2}}{(2k-1)^{2m}}+\sum_{k=1}^{\infty}\frac{(-1)^{2k-1}}{(2k)^{2m}}$$

???

Answer 1

Every positive integer $n$ is either even, i.e., of the form $n=2k$ for $k \geq 1$, or odd, i.e., of the form $n=2k-1$ for $k \geq 1$.

More formally, if $f:\mathbb{N}\rightarrow\mathbb{R}$, then $$\sum_{n=1}^\infty f(n)=\sum_{\text{odd } n=1}^\infty f(n)+\sum_{\text{even } n=1}^\infty f(n)$$ since addition of real numbers is commutative and hence $$\sum_{n=1}^\infty f(n)=\sum_{k = 1}^\infty f(2k-1)+\sum_{k=1}^\infty f(2k).$$