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If $a_k\in\mathbb{C}$ ($k\in\mathbb{Z}$), here are two equivalent definitions for $$ \sum_{k=-\infty}^{\infty}a_k $$ For reference, the two definitions are:

  • $$\sum_{k=-\infty}^{\infty}a_k=L\\ \Updownarrow\\\forall\epsilon>0,\exists N,m,n> N\implies\left|\sum_{k=-m}^na_k-L\right|<\epsilon$$
  • $$ \sum_{k=-\infty}^{\infty}a_k=L\\ \Updownarrow\\ \sum_{k=0}^{\infty}a_k\text{ and }\sum_{k=1}^{\infty}a_{-k}\text{ both exist and }\sum_{k=0}^{\infty}a_k+\sum_{k=1}^{\infty}a_{-k}=L $$

Questions:

  1. Is the notation $\displaystyle\sum_{k\in\mathbb{Z}}a_k$ usually defined to mean $\displaystyle\sum_{k=-\infty}^{\infty}a_k$ (according to one of the two equivalent definitions given)?

  2. Is the notation $\displaystyle\sum_{k\in\mathbb{Z}}a_k$ a particular case of a definition of a summation of complex numbers over arbitrary index sets (see here)?

  3. If the answer to question 2. is yes, then are the definitions I just gave consistent with the general definition?

Community
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NeedForHelp
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  • So, to put it in an example, your concern would be if a series like $\sum_{n\in\mathbb{Z},n\neq0}\text{sign}(n)\frac{1}{n}$ is convergent right? I'm interested... –  Feb 02 '17 at 00:18

2 Answers2

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I am not sure if there is a unanimous consensus on the meaning of the notation $\sum_{k\in\Bbb{Z}} a_k$, but it is often defines as one of the following equivalent notion:

  1. $\sum_{k\in\Bbb{Z}} a_k$ is the limit of the net $ \{ \sum_{k \in F} a_k : F \subset \Bbb{Z} \text{ and $F$ is finite} \}$.

  2. $\sum_{k\in\Bbb{Z}} a_k = \sum_{k=-\infty}^{\infty} a_k$ when the series is absolutely convergent, i.e., $\sum_{k=-\infty}^{\infty} |a_k| < \infty$.

As you can see from the second definition, the sum $\sum_{k\in\Bbb{Z}} a_k$ is a strictly stronger notion than the doubly infinite sum $\sum_{k=-\infty}^{\infty} a_k$.

Example. Let us consider $a_k = (-1)^k /k$ for $k \neq 0$ and $a_0 = 0$, then $\sum_{k=-\infty}^{\infty} a_k = 0$ is easy to check. On the other hand, $\sum_{k\in\Bbb{Z}} a_k$ is simply undefined.

And the first definition is exactly a special case of the definition introduced in your link.

Sangchul Lee
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  • Do you have a reference for the equivalence of these two notions? Also, am I right if I say that a Fourier series is a series of the form $\displaystyle\sum_{k=-\infty}^{\infty}\hat{f}(k)e^{ikt}$ (not $\displaystyle\sum_{k\in\mathbb{Z}}\hat{f}(k)e^{ikt}$), but that Parseval's formula can be written equivalently as $$\displaystyle\sum_{k=-\infty}^{\infty}|\hat{f}(k)|^2=|f|2^2$$ or $$\displaystyle\sum{k\in\mathbb{Z}}|\hat{f}(k)|^2=|f|_2^2$$ – NeedForHelp Feb 02 '17 at 00:56
  • @NeedForHelp, I am not sure about exact references, but it is a common exercise for those who learn the concept of the net. – Sangchul Lee Feb 02 '17 at 01:10
  • For your second question, I guess it depends on which notion of convergence you are interested in. Certainly $\sum_{k=-\infty}^{\infty} \hat{f}(k)e^{ikt} $ covers more general situation under pointwise convergence, but it is the same as $\sum_{k\in\Bbb{Z}} \hat{f}(k)e^{ikt} $ under $L^2$-convergence. – Sangchul Lee Feb 02 '17 at 01:22
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Its the integral w.r.t. counting measure when $\sum_{k \in \mathbb Z} \left| a_k \right| < \infty$, in which case equivalence is given by the Dominated Convergence Theorem.

If $(a_k)$ is not absolutely summable, the definitions might not agree. For example, if $a_k = k$, the first definition gives a sum of $0$ while the second one is undefined. (If the definition is taken to be the limit of symmetric sums)

(2) is true but does not necessarily holds for non-absolutely summable series. In which case the order of summation does matter and the sum can converge to any value in $\mathbb R$ if all $a_k$ are real.

Henricus V.
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