Prove sequence $a_n=n^{1/n}$ is convergent How to prove that the sequence $a_n=n^{1/n}$ is convergent using definition of convergence?

Well, the easiest proof is that the sequence is decreasing and bounded below (by 1); thus it converges by the Monotone Convergence Theorem...
The proof from definition of convergence goes like this:
A sequence $a_{n}$ converges to a limit L in $\mathbb{R}$ if and only if $\forall \epsilon > 0 $, $\exists N\in\mathbb{N}$ such that $\left | L - a_{n} \right | < \epsilon$ for all $n \geq N $.
The proposition: $\lim_{n\to\infty} n^{1/n} = 1 $
Proof:
Let $\epsilon > 0$ be given. Then by Archimedean property of the real numbers, there exists $M \in \mathbb{N}$ such that $M < \epsilon + 1$ then find $x\in\mathbb{R}; x>2$ such that $1+M>x^{1/x}$ and let $P = \left \lceil x \right \rceil$. Then, since $f(x)=x^{1/x}$ is decreasing (for $x>e$) (trivial and left to the reader :D) take any $x\in\mathbb{N}$ such that $x>P$ and observe that (because of our choice and $M$ and $P$) we have $n^{1/n} \leq P^{1/P} \leq M \le 1 + \epsilon$ whenever $n\geq P$ and so $\left | 1 - a_{n} \right | < \epsilon$ whenever $n\geq P$. Thus $a_{n}$ converges (to 1).
Edit: We can not always find a natural number M such that $M < \epsilon$ (what if $0 < \epsilon < 1$)? But we can always find a natural number M such that $M < \epsilon + 1$.
