How can $\frac{(-1)^{n+1}}{n^s} = \frac{1}{(2n-1)^s}-\frac{1}{(2n)^s}$?

Question

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?

Answer 1

For every positive integer $n$ the $2n-1$-th and $2n$-th term of the series sum to $$\frac{(-1)^{2n}}{(2n-1)^s}+\frac{(-1)^{2n+1}}{(2n)^s}=\frac{1}{(2n-1)^s}-\frac{1}{(2n)^s}.$$ So it is a matter of rewriting the sum as $$\sum_{n\geq1}a_n=\sum_{n\geq1}(a_{2n-1}+a_{2n}).$$

Answer 2

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.