Why does this equality Dirichlet series hold? Following on from my question here, I have hit a second roadblock.
I am working (very slowly!) 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 top of page 6 in equation 14, the paper asserts that
$$\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} = \ldots = \frac{1}{s-1} \sum_{n=1}^\infty \biggl(\frac{n}{(n+1)^s}-\frac{n-s}{n^s}\biggr)$$
What are the logical steps that give this result? I assume that the expression $\frac{n}{(n+1)^s}-\frac{n-s}{n^s}$ is somehow arrived at by splitting odd and even $n$, but this gives me
$$\begin{aligned}
\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} &= \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \biggl( \frac{1}{2^sn^s}- \frac{1}{(2n-1)^s} \biggr)
\\ &= \sum_{n=1}^\infty \biggl( \frac{1}{n^s(2^s-2)}- \frac{1}{\bigl(n- \frac{1}{2}\bigr)^s(2^s-2)} \biggr)
\end{aligned}$$
But I can't see how to extract the factor $\frac{1}{s-1}$ to produce the desired result.
 A: For $\Re s >2$ so everything is absolutely convergent and then extending by analytic continuation wherever RHS converges:
$\frac{1}{s-1} \sum_{n=1}^\infty (\frac{n}{(n+1)^s}-\frac{n-s}{n^s})=\frac{1}{s-1} \sum_{n=1}^\infty (\frac{n+1}{(n+1)^s}-\frac{n}{n^s}-\frac{1}{(n+1)^s}+\frac{s}{n^s})=$
(by telescoping the first two terms so remaining with just a term of $-1$ and then inserting a $0=1-(1/1^s)$ to make the third term go from $1$)
(edited - more detail as per comment: note that for $n=1$ the first two terms inside the four term bracket are $2/2^s-1/1^s$ for $n=2$ the first two terms are $3/3^s-2/2^s$, for $n=3$ the first two terms are $4/4^s-3/3^s$; since we chose $\Re s >2$ and we have absolute convergence in the sense that the sum of absolute values of each individual term inside each summand is finite, so$\sum_{n=1}^\infty (|\frac{n+1}{(n+1)^s}|+|\frac{n}{n^s}|+|\frac{1}{(n+1)^s}|+|\frac{s}{n^s}|) < \infty$, we can move terms around at will so clearly all the terms noted above cancel out except the first which is the $-1$ we took out; similarly, the third term in the original 4 term bracket gives $-1/2^s-1/3^s-...$ so adding $1-1=0$ and taking the $1$ out gives $1-\zeta(s)$ there, while the last term is self evidently gives $s\zeta(s)$
$=\frac{1}{s-1}(-1+1+\sum_{n=1}^\infty (-\frac{1}{n^s}+\frac{s}{n^s}))= \sum_{n=1}^\infty \frac{1}{n^s}= \zeta(s)=\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$
which is the required equality
