# Binomial Theorem?

1. Dec 25, 2008

### samspotting

Rudin's proof of lim n-> inf (p^(1/n)) = 1

1+n*x_n <= (1 + x_n)^n = o

I don't see it from the binomial theorem, which is what he says that is from.

He also does things with the binomial theorem like:

(1+x_n)^n >= ((n(n-1)) / 2) *x_n^2

I'm not sure what he did to get these two inequalities.

2. Dec 25, 2008

### Vid

Those are terms in the expansion given by the binomial theorem. Since there are more terms in the actual expansion, it's an inequality.

3. Dec 26, 2008

### HallsofIvy

Staff Emeritus
1+ n*x_n <= (1+ x_n)^n- o perhaps, not "= o". (1+ x_n)^n, by the binomial theorem that you mention in your title, says that (1+ x_n)^n= 1+ n x_n+ terms of higher order in x_n which will go to 0 as x_n goes to 0: o(x_n) or "small o". Assuming that x_n is positive, all those missing terms in the binomial expansion are positive and so the first two terms are less than or equal to the whole thing.

n(n-1)/2 is the second binomial coefficient nC2= n!(2!(n-2)!)= n(n-1)/2. (1+ x_n)^n= 1+ n x_n+ (n(n-2)/2) x_n^2+ higher terms. Again, assuming that x_n is positive, the 'whole thing' is greater than or equal to just one term.

Last edited: Dec 26, 2008
4. Dec 26, 2008

### samspotting

Thanks. I feel pretty stupid lol, but its been a while since ive seen that thing.