I would like to show that the limit of the sequence $(\frac{1}{\sqrt[n] n})_{n=1}^{\infty}$ is equal to 1 using the definition of the limit of a sequence. (Another way to say this is that it converges to 1).
Definition: A sequence $(a_n)_{n=1}^{\infty}$ converges to a real number A iff for each $\epsilon > 0$ there is a positive integer N such that for all $n \geq N$ we have $|a_n – A| < \epsilon$.
Edit: Here is my work.
Let $\epsilon > 0$. There exists $N$ such that for $n \geq N$,
$$|\frac{1}{\sqrt[n] n}-1| = |\frac{n^{\frac{n-1}{n}}}{n} – 1| = |\frac{n^{\frac{n-1}{n}} – n}{n}| = \frac{n-n^{\frac{n-1}{n}}}{n}$$
Here is where I am stuck.
Best Answer
Let $$ 1-\frac{1}{\sqrt[n]{n}}=a_n. \tag{1} $$ Clearly $a_n\in(0,1)$. From (1), one has $$ \frac{1}{\sqrt[n]{n}}=1-a_n. $$ So $$ n=\frac1{(1-a_n)^n}=(1+a_n+a_n^2+\cdots)^n\ge(1+a_n)^n\ge\binom{n}{2}a_n^2 $$ from which one has $$ 0<a_n\le\sqrt{\frac{2}{n-1}}. $$ Thus, for $\forall \epsilon>0$, letting $\sqrt{\frac{2}{n-1}}<\epsilon$ gives $n>\frac{2}{\epsilon^2}+1$. Define $N=\lfloor\frac{2}{\epsilon^2}+1\rfloor+1$ and then when $n\ge N$, one has $0<a_n<\epsilon$ or $$ \left|\frac1{\sqrt[n]n}-1\right|<\epsilon,$$ namely $$ \lim_{n\to\infty}\frac1{\sqrt[n]n}=1. $$