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The summation operator $\sum$ is often used to study sequences whose terms are given by a regular pattern:

$$ \sum_{n=n_0}^{\infty}a_n = a_0+a_1+a_2+\ldots $$

We also have various tools that can be used to study the convergence of sums and their properties. Why is it that we do not have similar tools and definitions for a difference of terms in a sequence, i.e.:

$$ \overset{\infty}{\Delta}_{n=n_0} a_n = a_0-a_1-a_2-\ldots $$

where $\forall n: n>0$ (in order to restrict alternating sequences), for the sake of simplicity.

This could be useful in studying progressions that start at a given initial value and decrease gradually - perhaps even converge, from above!

Of course, it is easy to show that:

\begin{align} \overset{\infty}{\Delta}_{n=n_0} a_n = a_0-a_1-a_2-\ldots &= 2a_0 - \sum_{n=n_0}^{\infty}a_n \\ &= a_0 - \sum_{n=n_1}^{\infty}a_n \end{align}

But wouldn't it be useful to formally define a $\Delta$ for such purposes?

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closed as primarily opinion-based by Hans Lundmark, Martin R, Claude Leibovici, Shaun, abiessu Dec 7 '17 at 20:28

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ There is no difference operator because there is no need for it. Introducing a new operator for something that is basically the same as the $\sum$ operator doesn't really bring much to the table, except for one more operator to confuse matters. $\endgroup$ – 5xum Dec 7 '17 at 12:42
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    $\begingroup$ The one you are thinking of is pretty useless. What you want is a difference operator that acts on a sequence and produces a new sequence of differences between adjacent terms. The inverse of this operator is the sequence of partial sums. They are direct analogues of the differentiation and integration operators for continuous functions, and they indeed are used and nice rules can be told in this form (building on an analogy with differentiation) $\endgroup$ – orion Dec 7 '17 at 12:42
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    $\begingroup$ On the left we have $n = n_0$ but on the right the indexes seem to start at $n = 0$. Also, where does $2 a_0$ come from? $\endgroup$ – eternalGoldenBraid Dec 7 '17 at 12:43
  • $\begingroup$ Indeed, check out the calculus of finite differences: en.wikipedia.org/wiki/Finite_difference $\endgroup$ – orion Dec 7 '17 at 12:43
  • $\begingroup$ Because mathematicians don't introduce notations that are not interesting enough. $\endgroup$ – Cave Johnson Dec 7 '17 at 12:58
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There is a difference operator. It doesn't subtract all the terms of a sequence, it just subtracts adjacent pairs. But it's very useful for analyzing sequences and series. More precisely, define a sequence in $k$ as

$$(\Delta \{a_n\}_n)_k=a_{k+1}-a_k. $$

So for example, a way to characterize arithmetic sequences is as sequence $a_n$ for which $\Delta a_n=r$ for $r$ constant. A quadratic sequence is one whose difference of differences is constant. Etc.

As for the operation you suggest, well that's still a sum.

$$a_0-a_1-a_2-\dotsb=2a_0+\sum -a_n = 2a_0-\sum a_n $$ as you pointed out yourself. So there's no need for a new notation for it.

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To add an entirely new operator for something that can be performed with an existing operator in a reasonably simple way does not really make sense. It just adds confusing elements, especially since we generally work with sums.

Furthermore, addition is a basic operation. Subtraction is not: subtraction is generally defined in terms of addition. Indeed, you have to define the difference operator here in terms of the summation operator.

Additionally, the operator is strange because it always starts with the first term as the term itself and then it subtracts each term from that. Finally, in order to show that it is reasonable to add this new operator to our repertoire, you should have to show that there are multiple cases where the operator would simplify existing results.

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