1
$\begingroup$

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?!

$\endgroup$
  • $\begingroup$ Need to show that each sum converges absolutely and your good. $\endgroup$ – Simply Beautiful Art Nov 10 '16 at 21:34
  • $\begingroup$ Do realize that this "proof" requires analytic continuation for $s<1$. For $s>1$, absolute convergence checks in every part. $\endgroup$ – Simply Beautiful Art Nov 10 '16 at 21:35
  • $\begingroup$ 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). $\endgroup$ – Nicolas FRANCOIS Nov 10 '16 at 21:36
  • $\begingroup$ @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 $s<1$? $\endgroup$ – Will Nov 10 '16 at 21:43
  • $\begingroup$ @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. $\endgroup$ – Will Nov 10 '16 at 21:50
1
$\begingroup$

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 :-)

$\endgroup$
  • $\begingroup$ 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?! $\endgroup$ – Will Nov 11 '16 at 11:57
  • $\begingroup$ Take integer parts of your boundaries. $\lfloor N/2 \rfloor$, the largest integer lower than $N/2$. $\endgroup$ – Nicolas FRANCOIS Nov 11 '16 at 20:01
  • $\begingroup$ 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. $\lfloor x\rfloor=\text{max}\lbrace m\in\mathbb{Z}\vert\;m\leq x\rbrace$?! $\endgroup$ – Will Nov 14 '16 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.