How can $\frac{(-1)^{n+1}}{n^s} = \frac{1}{(2n-1)^s}-\frac{1}{(2n)^s}$? I am working through a paper here that demonstrates Riemann's analytic continuation of the zeta function $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$ to the complex plane (except for the pole at $s=1$). At the bottom of page 5 in equation 13, the paper asserts (in the middle of a chain of equations) that
$$\begin{aligned} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}+\frac{2}{2^s}\sum_{n=1}^\infty \frac{1}{n^s} &= \sum_{n=1}^\infty \bigl(\frac{1}{(2n-1)^s}-\frac{1}{(2n)^s}+\frac{2}{(2n)^s}\bigr) \end{aligned}$$
Could someone please explain this step? This much is immediately obvious:
$$\begin{aligned} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}+\frac{2}{2^s}\sum_{n=1}^\infty \frac{1}{n^s} &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}+\sum_{n=1}^\infty \frac{2}{(2n)^s} \\ &= \sum_{n=1}^\infty \biggl(\frac{(-1)^{n+1}}{n^s}+\frac{2}{(2n)^s}\biggr) \end{aligned}$$
But I am not at all clear why it should be the case that
$$\begin{aligned} \frac{(-1)^{n+1}}{n^s} &= \frac{1}{(2n-1)^s}-\frac{1}{(2n)^s} \end{aligned}$$
as the equation seems to imply. Clearly, I am missing something fairly fundamental, or have made some embarrassingly stupid error. Can anyone explain?

When $n=2k$ you have
$$
\frac{(-1)^{n+1}}{n^s}=\frac{(-1)^{2k+1}}{(2k)^s}=-\frac{1}{(2k)^s}
$$
otherwise $n=2k-1$ and you have
$$
\frac{(-1)^{n+1}}{n^s}=\frac{(-1)^{2k}}{(2k-1)^s}=\frac{1}{(2k-1)^s}
$$
so you can rewrite your series as follows
$$
\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}{n^s}
=\sum_{k=1}^{+\infty}\frac{1}{(2k-1)^s}
-\frac{1}{(2k)^s}
$$
and relabeling $k$ as $n$ your final computation reads as
\begin{align}
\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}+\frac{2}{2^s}\sum_{n=1}^\infty \frac{1}{n^s}
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}+\sum_{n=1}^\infty \frac{2}{(2n)^s} \\ &= \sum_{n=1}^\infty \frac{1}{(2n-1)^s}-\frac{1}{(2n)^s}+\frac{2}{(2n)^s}\\
&= \sum_{n=1}^\infty \frac{1}{(2n-1)^s}+\frac{1}{(2n)^s}\\
&= \sum_{n=1}^\infty \frac{1}{n^s}=\zeta(s)
\end{align}
The step you miss is my second inequality; you basically separate the odd and even values of $n$, in order to "master" the sign.
