Two part question that I want to make sure I did correctly.

a) Let $x\_n = \sqrt[n]{n} - 1$. Use the fact that $(1 + x\_n)^n = n$ to show that $x\_n^2 \leq \frac{2}{n}$.
Hint given to use the binomial theorem and throw away most terms.

Expanding $(1+x\_n)^n$ using the binomial theorem gives us ${{n}\choose{0}}x\_n^{n-0}1^0 + {{n}\choose{1}}x\_n^{n-1}1^1 + \cdots + {{n}\choose{n}}x\_n^{0}1^n = n$.

Want to show $x\_n^2 \leq 2/n$, that is $(\sqrt[n]{n} - 1)^2 \leq \frac{2}{{{n}\choose{0}}x\_n^{n-0}1^0 + {{n}\choose{1}}x\_n^{n-1}1^1 + \cdots + {{n}\choose{n}}x\_n^{0}1^n}$ or $n^{2/n} - 2\sqrt[n]{n} + 1\leq {{n}\choose{0}}x\_n^{n-0}1^0 + {{n}\choose{1}}x\_n^{n-1}1^1 + \cdots + {{n}\choose{n}}x\_n^{0}1^n$ which we can see is true by getting rid of the right terms

b) Compute $\lim \limits\_{n\to \infty} n^{1/n}$

Using our inequality in part a), we can use Squeeze Theorem to compute the limit.

We have $1^{1/n} \leq n^{1/n} \leq 2/n + 1$
Since $1^{1/n} \rightarrow 1$ and $2/n +1 \rightarrow 1$, we have that $ n^{1/n} \rightarrow 1$.
So $\lim \limits\_{n\to \infty} n^{1/n} = 1$.

Just wanted to double check this.