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yes

Woah. You confused on here?

\((n + 1)! = n! \times (n + 1)\)

That's an axiom, isn't it?

\( \color{Black}{\Rightarrow (4 + 1)! = 4! \times 5 = 1 \times 2 \times 3 \times 4 \times5 = 5!}\)

That is an axiom, and that's also a way people prove that \(0! = 1\). :)

I suppose that makes sense, it just seems odd to me

\[n! = n (n-1)(n-2)...\]
SO,
\[(n+1)! = (n+1)n(n-1)(n-2)... = (n+1)n!\]

\[\sum_{n=1}^{\infty} n!(2x-1)^n\] for example, is one of the easier ones

\(n! = n(n - 1)(n - 2) \cdots 1\)
Correction* :)

At first I couldn't tell what they were doing to get rid of the n!'s on the ratio test

I believe that a factorial may be expressed as \(\prod\).

\[THANKS\]
@ParthKohli

\[Correct\]

\[\prod_{i = 1}^{n}i = n!\]

sometimes we take 0!=1 as a convention :)

Alright so that's the trick, anything else I should know about factoring n! out?

Well......... Nothing so important.

It's actually a simple fact.

Simple, but not intuitive at first glance, at least not to me

a small execise : re-write \[\frac{1.3.5.7.........(2n+1)}{2.4.6.8.............2n}\]
using "!"

|dw:1343146818151:dw|

|dw:1343146998874:dw|

Woah! That's some hardcore Mathematics!

Yeah I'm getting an interval of convergence of -0.5 < x < 0.5
And a radius of convergence of 0.5

Would you agree with that @experimentX ?
The limit for this series goes to infinity, it diverges.

wait .. something went wrong!!

I'm using the ratio test... I think that's what you did too *looks back at what your wrote*

What a change!
\(\mathbf{Factorials \Longrightarrow Calculus}\)

well .. factorials are the basics!!

lol ... i did the opposite!!

Using the axiom Parth pointed out I can then cancel...

But the radius? 0.5? 0? o_O

Oh no! Not THAT type of limits!

PS: People please give @experimentX a medal...

Done.

|dw:1343147733699:dw|

no ... i enjoy weird problems!!