How does x^n/n! approach 0 as n increases to infinity?

[Howers]

x^n/n! -> 0 as n-> +oo

I got 2 texts that explain this in 2 lines. A lot of steps skipped as you can imagine. Can anyone prove this in toddler language?

 

[robert Ihnot]

I see you mean: (x^n)/n!. Let's suppose J is the next largest integer to x, and consider the larger value of (j^n)/n!, now when n=j, we have (j^j)/j!, where only the last term is j/j=1. But, develope a second part of the series (j^j/j!)(j^j)(2j!/j!), this second product is of the form:

\prod_{i=1+j}^{2j}\frac{j}{i} in which every term is less than 1, but we can continue to increase n to 3j producing another product:

\prod_{i=2j+1}^{3j}\frac{j}{i} producing a even smaller product, etc. Multiplying these products all together as n becomes boundless gives us our result.

 

[Gib Z]

Or we could just note that n! grows much faster than x^n for any finite x. We can see this by looking at subsequent terms of the sequence. If x is some finite number, as soon as you get past the x-th term, the denominator is multiplying by numbers larger than x, whilst the denominator is still multiplying by x.