# Easy Sequence

1. Sep 28, 2007

### 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?

2. Sep 29, 2007

### 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.

Last edited: Sep 29, 2007
3. Sep 29, 2007

### 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.