As I understand it, if a series is absolutely convergent, then one can legitimately rearrange terms in the series and the value of the sum will be unaffected.
This being the case, can one split a series into its even and odd components, i.e. $$\sum_{k=0}^{\infty}a_{k}=\sum_{k=0}^{\infty}a_{2k}+\sum_{k=0}^{\infty}a_{2k+1}\;\;\text{?}$$ For example, can one derive the analytic continuation of the Riemann zeta function, $\zeta(s)$ as follows: $$\zeta(s)=\sum_{k=0}^{\infty}\frac{1}{n^{s}}=\sum_{k=0}^{\infty}\frac{1}{(2n)^{s}}+\sum_{k=0}^{\infty}\frac{1}{(2n-1)^{s}}\qquad\qquad\qquad\\ =\sum_{k=0}^{\infty}\frac{2}{(2n)^{s}}+\sum_{k=0}^{\infty}\frac{(-1)^{2n-1}}{(2n)^{s}} +\sum_{k=0}^{\infty}\frac{(-1)^{(2n-1)-1}}{(2n-1)^{s}} \\ =\frac{1}{2^{s-1}}\sum_{k=0}^{\infty}\frac{1}{n^{s}} +\sum_{k=0}^{\infty}\frac{(-1)^{n-1}}{n^{s}}\qquad\qquad\qquad\qquad\\ \Rightarrow\qquad\left(1-\frac{1}{2^{s-1}}\right)\zeta(s)=\sum_{k=0}^{\infty}\frac{(-1)^{n-1}}{n^{s}}\qquad\qquad\qquad\qquad$$ I can't see anything wrong with what I've done, apart from the fact that I'm not quite sure whether one can legitimately split a series into its even and odd parts like this?!
If $\phi:\mathbb N\to\mathbb N$ is a strictly rising function, $$\sum_{k=0}^N \left|u_{\phi(k)} \right| \le \sum_{j=\phi(0)}^{\phi(N)} \left|u_{j}\right|\le \sum_{j=\phi(0)}^{\infty} \left|u_{j}\right|<+\infty$$ so every sub-series of an absolutely convergent series is absolutely convergent.
About the alternate Riemann series, it was a joke : if the "normal" series is absolutely convergent, the alternate one is also :-)