Limit of $\frac{1}{\sqrt[n] n}$ 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.
 A: 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
$$ \lim_{n\to\infty}\frac1{\sqrt[n]n}=1. $$
A: An old standby:
By Bernoulli's inequality,
$(1+n^{-1/2})^n
\ge 1+n^{1/2}
\gt n^{1/2}
$.
Raising to the $2/n$ power,
$n^{1/n}
\lt (1+n^{-1/2})^2
=1+2n^{-1/2}+n^{-1}
\lt 1+3n^{-1/2}
\to 1
$.
A: Not sure if that would help
your equation $= \dfrac{1}{n^\frac{1}{n}} = n^\frac{-1}{n} = \exp\left(\dfrac{-1}{n}*\ln(n)\right)$
$\dfrac{-1}{n}*\ln(n)$ goes to $0$ when $N$ goes to $\infty$
Therefore $\exp(0) = 1$
A: Let's do this just using straightforward algebra and some crude (but clever) bounds.
For $n\ge1$ we have, using the algebraic identity $(1-a^n)=(1-a)(1+a+a^2+\cdots+a^{n-1})$ and the inequality $\sqrt[n]n\ge1$,
$$\begin{align}
0\le1-{1\over\sqrt[n]n}&=1-n^{-1/n}\\
&=(1-n^{-1/n}){1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}\over1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}}\\
&={1-n^{-1}\over1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}}\\
&\lt{1\over(n^{-1/2}+n^{-1/2}+\cdots+n^{-1/2})+(0+0+\cdots+0)}\qquad\text{(see note below)}\\
&\lt{1\over(n/3)n^{-1/2}}\\
&={3\over\sqrt n}\\
&\lt\epsilon\qquad\text{if $n\gt{9\over\epsilon^2}$}
\end{align}$$
The key step is the replacement of the (roughly) first third of the terms in $1+n^{-1/n}+n^{-2/n}+\cdots+n^{-(n-1)/n}$ with the smaller value $n^{-1/2}$ and the rest with $0$; one can, of course, go up close to the first half and replace the $3$ here with something close to $2$, but there's not much to be gained in doing so.
A: Pick $\varepsilon>0$. We have
$$\sqrt[n]{n}-1 < \varepsilon \iff \sqrt[n]{n} < 1+\varepsilon \iff n < (1+\varepsilon)^n$$
Now, we have $$(1+\varepsilon)^n = 1+n\varepsilon + {n \choose 2}\varepsilon^2 + \cdots > {n \choose 2}\varepsilon^2 = \frac12(n^2-n)\varepsilon^2$$
so if we pick $$n \ge \sqrt{1+\frac2{\varepsilon^2}} \implies n \le \frac12(n^2-n)\varepsilon^2 < (1+\varepsilon)^n$$
and hence $\lim_{n\to\infty} \sqrt[n]{n} = 1$. Now therefore also
$$\lim_{n\to\infty} \frac1{\sqrt[n]{n}} = 1.$$
A: I think the following method can help much. For each $n$, there exists an integer $k$ such that $2^{k-1}\le n\le2^k$ and we have either
$$
2^{k-1\over 2^{k-1}}\le n^{1\over n}\le 2^{k\over 2^k}
$$
or
$$
2^{k\over 2^{k}}\le n^{1\over n}\le 2^{k-1\over 2^{k-1}}
$$
Now prove $k\over 2^k$ tends to $0$.
