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.
Best Answer
You do not really need the binomial theorem. For any $n\geq 2$ we have
$$ n = \frac{n}{n-1}\cdot\frac{n-1}{n-2}\cdot\ldots\cdot\frac{2}{1}\cdot \frac{1}{1}\tag{1}$$ hence $n^{1/n}$ is the geometric mean of $1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}$, which by the AM-GM inequality is less than $$ \frac{1}{n}\left[1+\sum_{k=1}^{n-1}\left(1+\frac{1}{k}\right)\right]=1+\frac{H_{n-1}}{n}\leq 1+\frac{1+\int_{1}^{n}\frac{dx}{x}}{n}=1+\frac{1+\log(n)}{n}\tag{2} $$ but clearly greater than one. Squeezing then leads to $\lim_{n\to +\infty} n^{1/n}=1$, as expected from the fact that this limit equals $\exp\lim_{n\to +\infty}\frac{\log n}{n}$.