$\sum_{j=3n}^{2n}a_{j}=\sum_{j=2n}^{3n}a_{j}$: true or false? $$\sum_{j=0}^{n}a_{3n-j}=\sum_{j=2n}^{3n}a_{j}$$
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 $-\sum_{j=2n}^{3n}a_{j}$?
 A: 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
$$
  \sum\nolimits_a^bf(x)\delta x = -\sum\nolimits_b^af(x)\delta x
\quad\text{and}\quad
  \sum\nolimits_a^bf(x)\delta x + \sum\nolimits_b^cf(x)\delta x = \sum\nolimits_a^cf(x)\delta x.
$$
A: 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.
A: 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:
$$
\sum_{i=n}^m a_i := \begin{cases} a_n+\cdots +a_m &\text{, if } n\leq m\\
0 &\text{, otherwise.}
\end{cases}
$$
