
Baby Analysis Question
I'm reading through Rudin and got to this proof, which I can't understand: Prove that $\displaystyle \displaystyle \lim_{n \rightarrow \infty} \sqrt[n]{p} = 1$ if $\displaystyle p > 0$.
The proof goes: Assume $\displaystyle p > 1$, then $\displaystyle \sqrt[n]{p}  1 > 0$. Let $\displaystyle x_{n} = \sqrt[n]{p}  1$, then by the binomial theorem $\displaystyle (x_{n} +1)^{n} = p \geq 1 + nx_{n}$. It's this last inequality that I don't get. If what we're doing is dividing each $\displaystyle x_{n}^{ni}$th term by $\displaystyle x_{n}^{ni1}$ to get $\displaystyle x_{n}$, nmany times, then how do we know that we're decreasing the expression, since it's possible that $\displaystyle x_{n} < 1$?

Well the binomial theorem gives you this:
$\displaystyle (a+b)^n = \sum_{k=0}^n \binom{n}{k}a^kb^{nk}$
In our case, we put $\displaystyle a = x_n$ and $\displaystyle b = 1$, giving:
$\displaystyle (x_n+1)^n = \sum_{k=0}^n \binom{n}{k}x_n^k$
Now the RHS of this equation has more than two terms, and all terms are positive (because $\displaystyle p > 1$, so $\displaystyle x_n > 0$ just by its definition). Drop all the terms of the series except when k=0 and k=1. The RHS will be at least as large as this because it's equal to it plus all the terms we dropped. This will give you that:
$\displaystyle (x_n+1)^n \ge \binom{n}{0}x_n^0 + \binom{n}{1}x_n^1 = 1+nx_n$.