A question about properties of infinite sums

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?!

• Need to show that each sum converges absolutely and your good. – Simply Beautiful Art Nov 10 '16 at 21:34
• Do realize that this "proof" requires analytic continuation for $s<1$. For $s>1$, absolute convergence checks in every part. – Simply Beautiful Art Nov 10 '16 at 21:35
• There is nothing wrong with your calculations, (except maybe that you should prove that the alternate Riemann series is absolutely convergent ;-) If $\sum u_n$ is absolutely convergent, then every sub-series is absolutely convergent (easy to prove, for example by Cauchy criterion). – Nicolas FRANCOIS Nov 10 '16 at 21:36
• @SimpleArt So one should check that both the sum of even terms and the sum of odd terms individually are absolutely convergent?! How does one analytically continue to $s<1$? – Will Nov 10 '16 at 21:43
• @NicolasFRANCOIS I'm fairly new to this, how does one show that every sub series is absolutely convergent? I've read derivations of Euler's formula: $e^{ix}=\cos(x)+i\sin(x)$, where one splits up the Taylor expansion of $e^{ix}$ into its even and odd components, $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n)!}x^{2n}$ and $\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!}x^{2n+1}$ respectively. – Will Nov 10 '16 at 21:50

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.
• So if a given infinite series is absolutely convergent, is it always possible to rewrite it in terms of sums of its even and odd terms (since these sub-series will also be absolutely convergent)?! In the case of finite sums, how does one split them into sums of even and odd terms? Naively, I would have thought it would be something like $\sum_{k=0}^{N}a_{k}=\sum_{k=0}^{N/2}a_{2k}+\sum_{k=0}^{(N-1)/2}a_{2k+1}$, however, depending on which value of $N$ that I choose, the upper limit of one of the two sums will be non-integer, which doesn't make sense?! – Will Nov 11 '16 at 11:57
• Take integer parts of your boundaries. $\lfloor N/2 \rfloor$, the largest integer lower than $N/2$. – Nicolas FRANCOIS Nov 11 '16 at 20:01
• Ah I see, I didn't realise that there is notation for this. Is $\lfloor x\rfloor$ simply the so-called floor function, i.e. $\lfloor x\rfloor=\text{max}\lbrace m\in\mathbb{Z}\vert\;m\leq x\rbrace$?! – Will Nov 14 '16 at 12:51