TITLE: How does one transform a sum into a telescoping sum?
QUESTION [7 upvotes]: Very possibly this has been asked before, but as I couldn't find it:
Let $(a_n)_{n\in\mathbb N}$ be a sequence, and 
$$\sum_{n= 0}^\infty a_n $$
an infinite sum.
I'm looking for general methods of transforming this sum into a telescoping sum, i.e. finding a solution to the recurrence formula 
$$a_n= r(n) - r(n-1) $$
This can be seen as an inhomogenous recurrence equation:
$$r(n)=  r(n-1)+a_n $$
One failed attempt would be this:
Using the methods of discrete math, we can show that the generating function of $r(n)$ has the closed form:
$$ \sum_{i=0}^\infty r(i)x^i = \frac{r(0) - \sum_{n=d}^\infty a_n x^n}{1-x}$$
However, to make use of this, we would first need to find a closed form for
$$ \sum_{n=d}^\infty a_n x^n$$
However, if we find a closed form for that, we might as well find a closed form for 
$$\sum_{n=0}^\infty a_n x^n$$
, substitute $x=1$ and be done with it.
TL, DR: In this form, my approach either is missing something, or plain pointless. What are other possible ways of going about solving this recurrence equation, or other general ways to transform sums into telescopic-sum-form?

REPLY [3 votes]: You are looking for
$$
a(n) = r(n) - r(n - 1)
$$
or, what is more standardly adopted
$$
a(n) = s(n + 1) - s(n) = \Delta s(n)
$$
Then $s(n)$ is called the Antidelta or Indefinite Sum of $a(n)$, and same as for integrals
some functions have a closed form for the Antidelta, and some don't.
You can find more details in the indicated link.