it's possible to invert summation/ series limits?

Question

If the summation just sum every term i was thinking that for instance 1+2+3+4 = 4+3+2+1

so why this $$\sum\limits_{i=1}^{n} (3i)\ = \sum\limits_{i=n}^{1} (3i) $$ is not true ?

And how i can invert the limits of the summation without changing the result ?

Answer 1

as others pointed out it is common (not convention) that indexes start from low values to high in that order, but when it is obvious it is undertood

$\sum\limits_{i=1}^{n} (3i)\ = \sum\limits_{i=n}^{1} (3i) $ is true, not sure why you think $3+6+9+\cdots +3n \neq 3n+\cdots+9+6+3$?

Just remember it is a shorthand, it is meaningless without defining what it means, you coud even define it with additional properties $\sum\limits_{k \text{ is prime} \le10000} k $ means $2+3+5+7+11+\cdots$ only primes less than 10000.

Answer 2

As a matter of convention, indices rise. If you want to reverse the order of the terms, you need to write $\sum_{i=1}^na_i=\sum_{i=1}^na_{n+1-i}$.

Answer 3

To be consistent with the summation identity $\sum\limits_a^{b-1}+\sum\limits_b^c=\sum\limits_a^c$ when $a<b<c$

You need to set $\displaystyle \sum\limits_{i=M}^m x_i=-\sum\limits_{i=m+1}^{M-1} x_i$ for the case $m<M\ $ (and $0$ if $|M-m|\le 1$).

Have a look at this paper for a detailed approach.

As for integrals when bounds are reversed, a negative sign appears before the sum.