Finding Finite Series for Euler's Number (e)

[eddybob123]

well, i kno that e is derived from (1+(1/n))^n, as n goes to infinity, but i want to find a finite series . if i plot a curve like this on a graph, then take its derivative, i can tske a value of e.

 

[disregardthat]

well, i kno that e is derived from (1+(1/n))^n, as n goes to infinity, but i want to find a finite series . if i plot a curve like this on a graph, then take its derivative, i can tske a value of e.

A "finite series"? What do you mean? If you meant an infinite series a common one is $e = \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...$. In fact e^x is sometimes defined as
$$\frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$

 

[TylerH]

Another useful definition of e, especially for calc I derivation of the e^x rule, is $$lim_{n \to 0} (1+n)^{1/n}$$.

$$\frac{d}{dx}e^x=lim_{h \to 0} \frac{e^{x+h}-e^x}{h}$$
$$\frac{d}{dx}e^x=e^xlim_{h \to 0} \frac{e^h-1}{h}$$
$$\frac{d}{dx}e^x=e^xlim_{h \to 0} \frac{\left(\left(1+h\right)^{1/h}\right)^h-1}{h}$$
$$\frac{d}{dx}e^x=e^xlim_{h \to 0} \frac{1+h-1}{h}$$
$$\frac{d}{dx}e^x=e^x$$

 

[HallsofIvy]

well, i kno that e is derived from (1+(1/n))^n, as n goes to infinity, but i want to find a finite series . if i plot a curve like this on a graph, then take its derivative, i can tske a value of e.

Any finite series, involving only integers and fractions of integers, must give a rational number. e is not rational so there can be no such finite series. In fact, because e is transcendental, any such series would have to involve transcendental numbers as coefficients which would be as difficult to work with as e itself.