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Can the index of summation of a sum be anything other than an integer? What is the reasoning behind the answer?

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  • $\begingroup$ Sure, you can sum over any set you like. It turns out that if you sum over something uncountable, the sum will diverge unless all but countably many of the terms are zero, but you can still sum over them. $\endgroup$ Jan 4, 2019 at 21:24

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You will have obvious issues if you are attempting an uncountable sum of non-zero values, but otherwise there is no greater difficulty than sums over an integer index

Here is an example:

$$\sum_{\frac{a}{b} \in \mathbb{Q} \cap (0,1]} \left(\frac{1}{b}\right)^k = \frac{\zeta(k-1)}{\zeta(k)}$$ where $\frac a b$ is a rational expressed in lowest terms and real $k \gt 2$

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  • $\begingroup$ I get $\zeta(k-1)$ for the sum. Where is my mistake? $\begin{array}\\ \sum_{\frac{a}{b} \in \mathbb{Q} \cap (0,1]} \left(\frac{1}{b}\right)^k &=\sum_{n=1}^{\infty} \sum_{j=1}^n (1/n)^k\\ &=\sum_{n=1}^{\infty} (1/n)^k\sum_{j=1}^n 1\\ &=\sum_{n=1}^{\infty} (1/n)^k\ n\\ &=\sum_{n=1}^{\infty} (1/n)^{k-1}\\ &=\zeta(k-1)\\ \end{array} $ $\endgroup$ Jan 4, 2019 at 20:27
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    $\begingroup$ @martycohen Have you included four cases of $(1/4)^k$ in your sum? There should only be two for $\frac14$ and $\frac34$, as $\frac24=\frac12$ and $\frac44=\frac11$ and so have already been included as $(1/2)^k$ and $(1/1)^k$ $\endgroup$
    – Henry
    Jan 4, 2019 at 20:35
  • $\begingroup$ Aha! I included all fractions, not just lowest terms. So that $n$ in the third line should be $\phi(n)$. Thanks. $\endgroup$ Jan 4, 2019 at 20:36
  • $\begingroup$ "there is no greater difficulty" Well, defining the sum as a limit of the partial sums doesn't work if the set of the indices is not ordered, and the order of a traditional infinite sum does matter if there are both positive and negative terms. $\endgroup$
    – JiK
    Jan 5, 2019 at 1:56
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Normally yes, although it can be generalized to members of enumerable sets, for example, instead of writing

$$\sum_{i=0}^N f(x_i)$$ you can simply write $$\sum_{x\in A} f(x)$$ where $A$ is the set over which summation is done. This is very common simplification of notation.

What you are probably asking is, can they be noninteger real numbers... the answer is: those are not sums in the same sense, but people have generalized summation to noninteger indices. You can look up summability calculus, it's a beautiful theory, although I do not know what your mathematical background is.

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The "index" of summation, the way I'm interpreting your question, corresponds to an enumeration of the elements of the set over which you are summing. For example, if you want to add the numbers $x_{1},...,x_{n}$, then you would write $$\sum_{k=1}^{n}x_{k}$$ You can also think of this as $$\sum_{x\in S}x$$ where $S=\{x_{1},...,x_{n}\}$. However, there are sets which are not countable - this means the elements of the set are not in one-to-one correspondence with the set of integers. For example, you can take the sum of every real number in $[0,1]$, which you could write as $$\sum_{x\in [0,1]}x$$ The problem is that because this set is uncountable, so there's no natural ordering for which to add these numbers. And in this case the sum is infinite. However if you know that a sum $$\sum_{x\in S}x<\infty$$ then, even if $S$ is uncountable, this can be rewritten as a sum over a countable set.

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Sums over non-integer indices are actually quite common. Of course the notation must make it clear what indices are being used. But you sometimes see expressions such as

$$ \sum_{\alpha \in A} f(\alpha) $$

where $A$ is some set (not necessarily of integers).

It should be noted that if $A$ is infinite, conditionally convergent sums will depend on the order in which the terms are taken, and that would have to be specified. For an absolutely convergent sum there is no problem. If $A$ is uncountable, the sum can't be well-defined and finite unless all but countably many $f(\alpha)$ are $0$.

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Absolutely! See https://en.wikipedia.org/wiki/Summation#Capital-sigma_notation.

Really, the main purpose of using integers to index is because it's fairly intuitive but really any way of distinguishing different elements of a list will work (especially since addition will (generally) be commutative over the set of numbers so even the ordering of the elements is irrelevant).

An index set (while describing ways to index sets in a union rather than elements to be summed) provides a nice analogy
(see https://planetmath.org/indexingset)

Also, it's really often inconvenient to limit ourselves to plain old integer indexing. More abstract notation is generally more useful.
Say you wanted the sum of all primes under 4000. You could do a lot of work to first figure out how many primes there are under 4000 and use that to index them all but it'd be far easier to just say something like

$\sum_{p:p=prime,p<4000} p$

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    $\begingroup$ For anyone who's wondering, there are 550 primes under 4000. $\endgroup$
    – Dan
    Jan 5, 2019 at 0:06
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As a more practical answer, if you want the index of summation to increment by a constant non-integer value, you could write

$\sum_{x=0\ by\ .01}^1 f(x) $.

However, both symbolically and numerically I would write this as $\sum_{n=0}^{100} f(n/100) $.

The numerical reason is that if the increment is not representable exactly ($0.01$ is not in binary floating point), the loop may execute an extra or one less time than expected.

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