Algebra behind the ratio test?

[cdotter]

Homework Statement

$$\sum\limits_{i=0}^\infty \frac{i!}{5^i}$$

Homework Equations

The Attempt at a Solution

(1) $$\sum\limits_{i=0}^\infty \frac{i!}{5^i}$$
(2) $$= \lim_{i\rightarrow \infty}|{\frac{(i+1)!}{(5+1)^i} \cdot \frac{5^i}{i!}|$$
(3) $$= \lim_{i\rightarrow \infty}{\frac{(i+1)!}{5i!}$$

Could someone explain to me the jump from (2) to (3)? I can't make sense of it and my lecture notes don't mention anything in particular. Do the properties of factorials come into play here (I never learned factorials)?

 

[wisvuze]

nothing happened, you should have

$$= \lim_{i\rightarrow \infty}|{\frac{(i+1)!}{(5)^{i+1}} \cdot \frac{5^i}{i!}|$$

and the 5^i / 5^i+1 gives you 1/5

 

[HallsofIvy]

There is no jump from (2) to (3)- Your (2) is simply wrong.

As wizvuze said, you should have $5^{i+1}$, not $(5+ 1)^i$.

$$\frac{(i+ 1)!}{i!}= \frac{(i+1)i!}{i!}= i+ 1$$
and
$$\frac{5^i}{5^{i+1}}= \frac{5^i}{(5^i)(5)}= \frac{1}{5}$$

 

[cdotter]

Ok, what about:

$$\frac{1}{(i+1)!} \cdot \frac{i!}{1} = \frac{i!}{(i+1)!}$$

My notes says it simplifies to this:

$$\frac{1}{i+1}$$

How?

 

[Bohrok]

$$\frac{i!}{(i + 1)!} = \frac{i!}{(i + 1)i!} = \frac{1}{i + 1}$$

 

[rakalakalili]

Because you said that you never learned factorials let me further explain Bohrok's reply.
5! = 5*4*3*2*1, 6! = 6*4*3*2*1 and so on.
Think of n! as multiplying all of integers from 1 up to n together.
i! = i*(i-1)*(i-2)...*3*2*1. The product of all the integers from 1 up to i.
(i+1)! = (i+1)*(i)*(i-1)*(i-2)*...*3*2*1. The product of all integers from 1 up to (i+1).
so then we have that (i+1)! = (i+1)*(i)*(i-1)*(i-2)...*3*2*1 = (i+1)*(i)!. Make this substitution on the bottom and you see that the i! on top and bottom cancel out.