Is 0! Really the Same as Dividing by Zero in Series Homework?

[Kidphysics]

Homework Statement

Stumbled onto this picture..

Homework Equations

The Attempt at a Solution

I see the first term in the series has in the denominator 0! but isn't that the same thing as dividing by 0 or do we treat the first term as just the number 1?

 

[Dick]

0! is by convention 1. You want n!=n*(n-1)!. So 1!=1 should be 1*0!. Better set 0! to one.

 

[Mentallic]

e^0=1 and using that formula, we get e^0=\lim_{n\to\infty}\left(\frac{1}{0!}+\frac{0}{1!}+\frac{0}{2!}+...+\frac{0}{n!}\right)=\frac{1}{0!} so you can then conclude that since we have 1=\frac{1}{0!} then 0!=1

 

[Juanriq]

You can also use the gamma function for a quick way of seeing this.
\Gamma (x) = \int_0 ^{\infty} t^{x-1} e^{-t} dt for x>0. Plugging in x = 1, you see that \Gamma (1) = 1. For integers, the gamma function has the recursion \Gamma (n+1) = n!, so for n = 0 we have 1 = \Gamma (1) = 0!.

 

[trautlein]

I little unrelated, but I was going through all of the homework threads to give myself some much needed practice.
But none of them made me smile like this : )