Abel's Summation (memorizing the theorem) I'm trying to intuitively remember and understand Abel's Summation by Parts (https://en.wikipedia.org/wiki/Summation_by_parts):
$$\sum_{k=m}^n f_k(g_{k+1}-g_k)=[f_{n+1}g_{n+1}-f_mg_m]-\sum_{k=m}^n g_{k+1}(f_{k+1}-f_k)$$
Are there any tips to understand and memorize this formula? I know it is the discrete analogue of Integration by Parts, but I am having problem with the subscripts (e.g. remembering whether it is $k+1$ vs $k$).
My current method is to memorize the form $\int f\,dg=[fg]-\int g\,df$, and then perform some guess and check to verify the subscripts. This probably reflects some lack of understanding of Abel's lemma on my part. (The only proof I know is the direct verification by expanding out everything).
Any more intuitive ways of understanding and remembering Abel's Summation by Parts? 
Thanks. Any comments will be appreciated.
 A: This is easier to remember:
$$\sum f\Delta g=fg-\sum {\rm E}g\Delta f\tag{1}$$
where $\Delta f(k)=f(k+1)-f(k)$ and ${\rm E}f(k)=f(k+1)$. For some functions $f(k)$ and $g(k)$ the above is written as
$$\sum f(k)\Delta g(k)\delta k=f(k)g(k)-\sum {\rm E}g(k)\Delta f(k)\delta k\tag{2}$$
where $\delta k$ is similar to ${\rm d}x$ in infinitesimal calculus, it is used to show the variable that is "summed". The above formulas $(1)$ and $(2)$ have no limits, i.e. are indefinite sums, analogous to indefinite integrals of infinitesimal calculus.
This is the summation by parts for finite calculus. Taking limits in $(1)$ or $(2)$ must be done following this rule of finite calculus:
$$\sum_{k=a}^b f(k)=\sum\nolimits_a^{b+1} f(k)\delta k$$
Then notice that
$$f_{n+1}g_{n+1}-f_mg_m=f(k)g(k)\Big|_m^{n+1}$$
i.e. taking limits on $(2)$ we write
$$\sum\nolimits_a^{b+1} f(k)\Delta g(k)\delta k=f(k)g(k)\Big|_a^{b+1}-\sum\nolimits_a^{b+1} {\rm E}g(k)\Delta f(k)\delta k$$
what is equivalent to write
$$\sum_{k=a}^{b} f(k)\Delta g(k)=f(k)g(k)\Big|_a^{b+1}-\sum_{k=a}^{b} {\rm E}g(k)\Delta f(k)$$
In short: the Abel summation is the discrete analog of
$$\int_a^b f(x)g'(x){\rm d}x=f(x)g(x)\Big|_a^b-\int_a^b f'(x)g(x){\rm d}x$$
A: Personally, understanding/proving/remembering this formula is especially easy when you refer to a graphical representation, as can be found for example in (http://mathbmt.com/sumofparts). Of course, there is a limitation at first to positive numbers, but this limitation can be overcome.
