The discrete derivative with respect to n is defined as ## \Delta_n f(n) = f(n+1) - f(n) ## in the document. The discrete integral is defined as ##\sum_{n: a \rightarrow b} f(n) = f(a)+f(a+1)+ ...+f(b-1) ##.

It had been a few months since I had read this article so when I started exploring this math, I defined the discrete integral as the regular sum, from a to b ( ##\sum_{n= a}^{b} f(n) = f(a)+f(a+1)+ ...+f(b) ## ). For this reason, I found it best to define the discrete integral as ## \Delta_n f(n) = f(n) - f(n-1) ##

The Fundamental Theorem:

## \Delta_n F(n) = f(n) \leftrightarrow \sum_{n=a}^{b} f(n) = F(b)-F(a-1) ##

I did a few problems and everything worked out well when I didn't make any errors. However, just now I was trying to prove

##\sum_{n} a^{n} = \frac{a^{n+1}-1}{a-1}##

but couldn't seem to. My work was as follows:

##\sum_{n} a^{n} ##

## u = a^{n} ##

##\Delta_u = (a^{n}-a^{n-1})*\Delta_n##

## \Delta_u = a^{n-1}(a-1) * \Delta_n ##

## \frac{\Delta_u*a^{1-n}}{a-1} = \Delta_n ##

## \frac{a}{a^{n}(a-1)}*\Delta_u = \Delta_n ##

So I then make the substitutions

##\sum_{u} u*\frac{a}{a^{n}(a-1)} ##

##\sum_{u} u*\frac{a}{u(a-1)} ##

##\sum_{u} \frac{a}{(a-1)} ##

I can then bring out that fraction

##\frac{a}{(a-1)} \sum_{u} 1 ##

##***## ##\sum_{k} 1 = k ## ## ***##

##\frac{a}{(a-1)} u ##

##\frac{a}{(a-1)} a^{n} ##

And in the end:

##\frac{a^{k+1}}{a-1} ##

so

##\sum_{n} a^{n} = \frac{a^{k+1}}{a-1} ##

but

##\frac{a^{k+1}}{a-1} \not= \frac{a^{n+1}-1}{a-1}##

Somewhere I've made a mistake. If someone could either point it out for me or guide me to it, it would be greatly appreciated. Also, if my reasoning in someplace isn't clear, please tell me and I'll explain.