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$$\sum_{j=0}^{n}a_{3n-j}=\sum_{j=2n}^{3n}a_{j}$$ I know this is true because you can calculate some terms and are the exact same terms written in the reverse order.

But if you substitute $t=3n-j$, you get: $$\sum_{t=3n}^{2n}a_{t}=\sum_{j=2n}^{3n}a_{j}$$ $$\sum_{j=3n}^{2n}a_{j}=\sum_{j=2n}^{3n}a_{j}$$

Now I would say it is still true, but I'm not sure, because if it was an integral (instead of a sum) there would be a minus!

I mean, is $\sum_{j=3n}^{2n}a_{j}$ equal to $\sum_{j=2n}^{3n}a_{j}$ or equal to $-\sum_{j=2n}^{3n}a_{j}$?

collimarco
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    The convention is the lower limit is by definition not higher than the upper limit in summation. Even if we want to be flexible with this, there would be no minus sign involved, as the intention is clear, you want to sum for all indices in the set $[2n, 3n]$. – Macavity Sep 22 '13 at 10:14
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    The left sum equals $0$. –  Sep 22 '13 at 10:32
  • Related. – Did Sep 22 '13 at 10:36

3 Answers3

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Finite sums and (commutative) products can be defined over finite sets of indices, not just over intervals of the integers. Since it is very common to rearrange the order of summation freely, this set approach is usually taken as the basis of defining the $\sum$ and $\prod$ operators, and $\sum_{i=a}^bx_i$ and $\prod_{i=a}^bx_i$ are just interpreted as alternative ways of writing $\sum_{a\leq i\leq b}x_i$ (or more formally $\sum_{i\in\{\, k\in\Bbb Z\mid a\leq k\leq b\,\}}x_i$) respectively $\prod_{a\leq i\leq b}x_i$. It follows that it is legal to have $b<a$, in which case the sum or product is empty (and has value $0$ respectively $1$).

It is on the other hand useful to have a calculus of summations over intervals of integers, where it is legal to glue together intervals just like one can do for integrals. The only place where I have seen this described is in Concrete Mathematics, pages 48,49, where one defines the notation $$ \sum\nolimits_a^bf(x)\delta x = \begin{cases}\sum_{x=a}^{b-1}f(x)&\text{when $a\leq b$}\\-\sum_{x=b}^{a-1}f(x)&\text{when $b\leq a$}\end{cases} $$ (that case $a=b$ gives $0$, twice), so that in all cases $$ \sum\nolimits_a^bf(x)\delta x = -\sum\nolimits_b^af(x)\delta x \quad\text{and}\quad \sum\nolimits_a^bf(x)\delta x + \sum\nolimits_b^cf(x)\delta x = \sum\nolimits_a^cf(x)\delta x. $$

Marc van Leeuwen
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You need to be careful about the way you define the summation symbol. To be most useful, we like to have $$ \sum_{j=k}^{m-1}a_j+\sum_{j=m}^{n-1}a_j=\sum_{j=k}^{n-1}a_j\tag{1} $$ In this case, $$ \sum_{j=k}^ma_j=a_k+a_m-\sum_{j=m}^ka_j\tag{2} $$ and $$ \sum_{j=k}^{k-1}a_j=0\tag{3} $$ whereas $$ \sum_{j=k-1}^ka_j=a_{k-1}+a_k\tag{4} $$ Just as the definite integral depends on the order of its limits, a summation also depends on the order of its limits.

robjohn
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  • This is not the convention most used to define summations. However you will find relatively few summutations $\sum_{i=a}^bx_i$ used at all with $b<a-1$ (but an empty sum for $b=a-1$ is quite common). So the way summations are mostly used, it doesn't matter how they are exactly defined. – Marc van Leeuwen Sep 23 '13 at 13:32
  • @MarcvanLeeuwen: You mention the convention used in Concrete Mathematics in your answer. $(1)$ is exactly the same, without the special notation. The rest of what I have above follows from $(1)$. Certainly, summations are usually written so that the upper index is at least the lower index (possibly one less), but since the question asked specifically about reversed indices, I addressed that. – robjohn Sep 23 '13 at 14:37
  • I was saying summation is usually defined so that (1) does not always hold, but only if $k\leq m\leq n$. In any case (1) does not supply a definition for summations; but defining it so that $\sum_{i=a}^bx_i=\sum\nolimits_a^{b+1}a_i\delta i$ would be the only reasonable option. However, I don't think that is done anywhere. Instead $\sum_{i=a}^bx_i=\sum_{a\leq i\leq b}x_i$ is taken as definition, which gives $0$ whenever $b<a$; the comment to the question by André confirms this point of view. In manipulating multiple sums one may get conflicting inequalities, which just means "no terms here". – Marc van Leeuwen Sep 23 '13 at 15:05
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It is a matter of notation, I guess.

In the theory of Riemann integrals, one has to set: $$ \int_\beta^\alpha f(x)\ \text{d} x := - \int_\alpha^\beta f(x)\ \text{d} x $$ in order to use the additive property of the integral in any possible case. In fact, with the definition above you have: $$ \int_\beta^\alpha f(x)\ \text{d} x + \int_\alpha^\beta f(x)\ \text{d} x = 0 = \int_\beta^\beta f(x)\ \text{d} x $$ and: $$ \int_\alpha^\beta f(x)\ \text{d} x + \int_\beta^\alpha f(x)\ \text{d} x = 0 = \int_\alpha^\alpha f(x)\ \text{d} x $$ and everything works fine (formally).

On the other hand, when using sums, one always sets by definition: $$ \sum_{i=n}^m a_i := \begin{cases} a_n+\cdots +a_m &\text{, if } n\leq m\\ 0 &\text{, otherwise.} \end{cases} $$

Pacciu
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