# The form of a series

I have a problem with series. Consider a series $$\sum_{k=1}^\infty a_k.$$ Can I rewrite it as $$\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty a_{2k-1}+\sum_{k=1}^\infty a_{2k}\ ?$$ Or more generally, can I rewrite it as $$\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty a_{\sigma_k}+\sum_{k=1}^\infty a_{\delta_k},$$ where $\sigma,\delta$ is the selection of $k$? I'm considering the proof $$S_{2n}=\sum_{k=1}^n a_{2k-1}+\sum_{k=1}^n a_{2k},$$ then get the limit when $n\to \infty$, but I don't think it is right.

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Not always. Consider $1-1+1-1+1-\cdots$. – J. M. Apr 19 '12 at 7:06
If your series is conditionally convergent you cannot just rearrange terms like that. – user38268 Apr 19 '12 at 7:08
@J.M. Yeah, I know what you mean.Does that mean if the series is not absolutely convergent, you can't write it like that – 89085731 Apr 19 '12 at 7:21
@Take $a_k=\frac{(-1)^{k-1}}{k}$. Then $\sum a_k$ is convergent, but sum of evens, sum of odds are both divergent. – André Nicolas Apr 19 '12 at 13:53
\Sigma_{n=1}^\infty does not look the same as \sum_{n=1}^\infty. The latter is standard. In a "displayed" setting the latter makes the subscript and superscript appear below and above the \sum; the former fails to do that: $\displaystyle \Sigma_{n=1}^\infty$ versus $\displaystyle \sum_{n=1}^\infty$. Even in an "inline", as opposed to "displayed", setting, there's a difference: $\Sigma_{n=1}^\infty$ versus $\sum_{n=1}^\infty$. I edited the question accordingly. – Michael Hardy Apr 19 '12 at 15:07

In general: If you have two convergent series: $$\sum_{k=1}^{\infty} b_k \quad\text{and}\quad \sum_{k=1}^{\infty} c_k.$$ Then the series $$\sum_{k=1}^{\infty}(b_k + c_k)$$ is also convergent. So for example if $b_k$ and $c_k$ are given by $b_k = a_{2k-1}$ and $c_k = a{2k}$, then indeed you get a convergent series $$\sum_{k=1}^{\infty} a_{2k-1} + \sum_{k=1}^{\infty} a_{2k} = \sum_{k=1}^{\infty} a_{2k-1} + a_{2k} = \sum_{k=1}^{\infty} a_k.$$ Note that for this to be true, you do not need anything to be absolutely convergent.

Now if you rearrange the natural numbers $\mathbb{N}$ so that as $k$ varies in $\mathbb{N}$ then $\delta_k$ also varies through $\mathbb{N}$ (i.e. $\delta : k \mapsto \delta_k$ is a bijection, then in general it is not true that $$\sum_{k=1}^\infty a_k = \sum_{k=1}^{\infty} a_{\delta_k}$$ even if the series is convergent. But if the series is absolutely convergent, then this is true.

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As you've already noticed, the series must be absolutely convergent for this to be true. To see even better that all hell breaks loose when the series is only conditionally convergent, consider Riemann series theorem: if $\sum a_n$ is a conditionally convergent series, then for every real number $x$, (or even $x=\pm\infty$) there is a permutation $\sigma \in \mathfrak{S}(\mathbb{N})$ such that $\sum_{n=0}^\infty a_{\sigma(n)} = x$.

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All the answers say that I rearrange the terms. But i can't figure out how $\Sigma_{k=1}^{\infty}a_k=\Sigma_{k=1}^{\infty}a_{2k-1}+\Sigma_{k=1}^{\infty}a_{‌​2k}$ rearrange the original term. Can you help me – 89085731 Apr 19 '12 at 11:08
Because it's as if you summed first the even terms, then the odd terms (or the other way). You can't actually do that with a permutation of $\mathbb N$ (putting all the even terms then all the odd terms), but that's the gist of it. – Najib Idrissi Apr 19 '12 at 11:09

Yes ,it can write as that,because $\sum_{k=1}^{2n}a_k=\sum_{k=1}^n a_{2k-1}+\sum_{k=1}^n a_{2k}$, when $n \to \infty$,

then we have

$\sum_{k=1}^\infty a_k=\sum_{k=1}^\infty a_{2k-1}+\sum_{k=1}^\infty a_{2k}$, if when $n \to \infty$, $\sum_{k=1}^\infty a_{2k-1}$ and $\sum_{k=1}^\infty a_{2k}$ exist.

note: If $\lim_{n \to \infty}\sum_{k=1}^n a_k$ converges, then $\sum_{k=1}^\infty a_k=\lim_{n \to \infty}\sum_{k=1}^n a_k$,this is definition of series. But if when $n \to \infty$, $\sum_{k=1}^\infty a_{2k-1}$ and $\sum_{k=1}^\infty a_{2k}$ are not all exist. You can not write as that.

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I think your definition $\Sigma_{k=1}^{\infty}a_k=\lim_{n \to \infty}\sum_{k=1}^n a_k$ may not be right. Actually the series may not converge. – 89085731 Apr 19 '12 at 9:22
@KannappanSampath I think you should read through his answer. – 89085731 Apr 19 '12 at 9:42
@Gingerjin Right, I missed that. – user21436 Apr 19 '12 at 10:09
@Gingerjin thanks,correct it. – noname1014 Apr 19 '12 at 10:32