Question on Rudin sequences? In baby Rudin, Rudin shows that $$\lim_{n \to \infty}\sqrt[n]{p} = 1.$$
In the proof of limit he tries to prove that the limit is $1$. So he takes $x_n = \sqrt[n]{p} - 1$.
I have never noticed this before, but I could have tried to prove the limit for any $n \neq 1$ (and obviously it wouldn't hold). But in the process of proving that the limit is $1$, did he have some special insight that $\lim_{n \to \infty} = 1$? My question becomes clearer below: He now says that by the binomial theorem, $$0 < 1 + nx_n \leq (1 + x_n)^n = p$$
But how in the world, as an undergraduate, am I supposed to pull that out of my nearly-empty math toolbox? I never even knew such a technique existed. In fact, with Rudin, he uses many esoteric proofs (that all work out, anyway) and I have no idea what could have caused him to realize that is the way (in this case, using the binomial theorem) to prove such a sequence.
I noticed indeed that he just selected a sequence that is smaller than $p$ and show by the squeeze theorem of upper and lower limits that $x_n \to 0$. This is a red herring.
But here is what he specifically wrote:
$$0 < x_n < \frac{p - 1}{n}$$
But $p = (1 + x_n)^n$, so how do we know indeed that $\lim_{n\to\infty}\sqrt[n]{p} = 1$??
My main concern is, how can I learn the methods to prove sequences and series without being inefficient? Should I just look at many solutions to limits of sequences, go hard and kill myself at the problems anyway, or keep reading (which is very slow for me, at perhaps an hour a page)?
 A: The log trick is always helpful, when your base is non-negative:
$p^{1/n}=e^{(1/n)lnp}$. Now, $lnp$ is fixed, and $(1/n) \rightarrow 0$ , so you can find
an  $M$ , so that for $n>M$ , $\frac {1}{n}<\frac{1}{\epsilon(lnp)}$ ....
EDIT: Another useful trick is that of sequential continuity, which works on the Reals:
the idea is : given a sequence $x_n $ with $x_n \rightarrow x$ , and $f$ is continuous, it follows that $f(x_n)\rightarrow f(x)$ . In this particular case, the function $x^a$ is continuous. You can then say, for $x^{1/n}:$ $Lim_{n\rightarrow \infty}x^{1/n}=e^{Lim_{n\rightarrow \infty}{(1/n)}=e^0=1}.$ You can also get a lot of mileage out of this one.
A: Here is another avenue.  Suppose $p \ge 1$. Then by the Geometric Series Theorem,
$$\root{n}\of{p} - 1 = {p- 1\over \sum_{k=0}^{n-1} p^{k/n}}\le {p-1\over n}. $$
Hence $\root{n}\of {p} \to 1$ as $n\to\infty$.  If $0 < p < 1$, you can apply this result to $1/p$.
A: Another way to look at this is the following: recall that 
$$
\liminf_{n \to \infty} \frac{c_{n+1}}{c_n} \leq \liminf_{n \to \infty} \sqrt[n]{c_n} \leq \limsup_{n \to \infty} \sqrt[n]{c_n} \leq \limsup_{n \to \infty} \frac{c_{n+1}}{c_n},$$
where $c_n > 0.$ Then you obtain the desired result for the constant sequence $c_n \equiv p.$ 
