TITLE: Prove sequence $a_n=n^{1/n}$ is convergent
QUESTION [18 upvotes]: How to prove that the sequence $a_n=n^{1/n}$ is convergent using definition of convergence?

REPLY [19 votes]: To show $\lim\limits_{n \to \infty} n^{1/n} = 1$, we need to show that for all $\varepsilon > 0$, there exists $N \in \mathbb{N}$ s.t. $\mid n^\frac{1}{n} - 1 \mid < \varepsilon$ for any $n > N$.
Expand 
\begin{equation}
\mid n^\frac{1}{n} - 1 \mid < \varepsilon \Leftrightarrow (1- \varepsilon)^n < n < (1+\varepsilon)^n
\end{equation}
The left part holds as $(1- \varepsilon)^n < 1 \le n$
For the right part, we use the binomial theorem,
$$
(1+\varepsilon)^n = \sum_{i=0}^n \binom{n}{i}\varepsilon^i > \binom{n}{2}\varepsilon^{2} = \frac{n(n-1)}{2} \varepsilon^2
$$
Set $N = \lceil \frac{2}{\varepsilon^2} + 1 \rceil$,
we have 
$$
(1+\varepsilon)^n  > \frac{n(n-1)}{2} \varepsilon^2 > n
$$