Limit by $\epsilon-N$ method Question is to prove that $\lim_{n\rightarrow \infty}n^{\frac{1}{n}}=1$.
This question is discussed in this site so many times and in all of the answers i have seen they have not considered the basic definition case :
Given $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\left|n^{\frac{1}{n}}-1\right|<\epsilon$ for all $n\geq N$.
We want to see if $(1+\epsilon)^n>n$
I have tried the following.
Let $\epsilon\geq 1$ then $(1+\epsilon)^n=1+n\epsilon+*\epsilon^2+\cdots+\epsilon^n>1+1+\cdots+1=n$.
So, we do have $(1+\epsilon)^n>n$ for all $n$ irrespective of $\epsilon$.
Suppose $\epsilon<1$ first of all i want one $N$ such that $(1+\epsilon)^N>N$. I was looking for $N$ such that $(1+\epsilon)^N>1+N\epsilon>N$ i.e., $\dfrac{1}{1-\epsilon}>N$.. As  $\epsilon<1$ it is possible to get such $N$.. But this argument does not make much sense.
Help me to solve this.
 A: If you define $x_n=\sqrt[n]{n}-1$, then
$$
n = (1+x_n)^n \geq \frac{n(n-1)}{2}x_n^2.
$$
Hence
$$
0 \leq x_n \leq \sqrt{\frac{2}{n-1}}
$$
for all $n \geq 2$. Now you can play with $\epsilon$.
Bibliography
W. Rudin. Principles of mathematical analysis.
A: Take
$$n>N=\frac1{\ln^2(1+\epsilon)}.$$
Then
$$\frac{\ln(n)}n<\frac1{\sqrt n}<\ln(1+\epsilon)$$
and
$$|n^{1/n}-1|<\epsilon.$$

Note that for $n>4$,
$$\ln(n)<\sqrt n$$ because $$\ln(4)<\sqrt4$$and after derivation
$$\frac1n<\frac1{2\sqrt n}.$$
A: We have to prove that $\lim_{n\to \infty}n^{1/n}=1$. Consider $a_n = n^{1/n}-1$. Since $n\geq 1$ we have that $n^{1/n}\geq 1$ and therefore $a_n\geq 0$. On there other hand, $n = (a_n+1)^n$. Using the binomial theorem (https://en.wikipedia.org/wiki/Binomial_theorem) we conclude that $$(a_n+1)^n=\binom{n}{0} + \binom{n}{1}a_n+\binom{n}{2}a_n^2 + \cdots  + \binom{n}{n-1}a_n^{n-1} +\binom{n}{n}a_n^{n}.$$
From that equality we obtain $$(a_n+1)^n \geq \binom{n}{2}a_n^2 = \frac{n(n-1)}{2}a_n^2.$$
So, since $n = (a_n+1)^n \geq (n(n-1)/2)a_n^2$, for every $n>1$ we have
$$\frac{n(n-1)}{2}a_n^2 \leq n,$$
$$a_n^2 \leq \frac{2}{n-1},$$
$$a_n \leq \frac{\sqrt 2}{\sqrt{n-1}}.$$
Fix $\epsilon >0$ and take $N\in \mathbb N$ larger than $2/\epsilon^2+1$. If $n> N$ then $n>2/\epsilon^2+1$ and therefore $$\frac{\sqrt{2}}{\sqrt{n-1}}<\epsilon.$$
So for every $n>N$ we have $a_n < \epsilon$ and therefore $|n^{1/n}-1|=|a_n|<\epsilon$ because $a_n = |a_n|$.
A: If you notice that $n^{1/n}>1$, for any $n$, you can forget about the absolute value, and
$$n^{\frac{1}{n}}-1<\epsilon\to n^{\frac{1}{n}}<\epsilon+1$$
take logarithms on both sides (you can do that since the logarithm is a monotonously increasing function)
$$\log(n)/n<\log(1+\epsilon)$$
This equation has solution, however I am worried it is not analytic. Then, for any positive $\epsilon$ you can find an $n$ that fulfills the convergence expression.
I hpoe this will be useful for you!
A: I think it can be done as follows :
We have $(1+\epsilon)^n=1+n\epsilon+***$ 
Suppose $\epsilon>1$ then we have $(1+\epsilon)^n\geq n\epsilon>n$..
I have done seperately for $\epsilon\leq 1$.
So,  given $\epsilon>0$ we have $(1+\epsilon)^n>n$ for all $n$ and so we are done
