How can the following limits be rigorously proved using the definition of limits？

Question

How can I prove with the $\epsilon - N$ formalism (that is, by definition) the following?

$$\lim_{n\to+\infty} \sqrt[n]{n^k} = 1 \qquad \qquad \qquad k\in\mathbb{N}$$

Thank you for your help!

Answer 1

To prove this limit we need to show that $\forall \epsilon > 0$ there exists $N>0$, with $N\in\mathbb{N}$, such that

$$|n^{k/n} - 1| < \epsilon$$

$\forall n > N$.

Then we can rewrite it as

$$1 - \epsilon < n^{k/n} < 1 + \epsilon$$

Arranging the exponents we get

$$(1 - \epsilon)^{n/k} < n < (1 + \epsilon)^{n/k}$$

Now the left part is true since

$$(1 - \epsilon)^{n/k} < 1 \leq n$$

For what concerns the right part we can use the binomial theorem:

\begin{equation*} \begin{split} (1 + \epsilon)^{n/k} = \left((1 + \epsilon)^n\right)^{1/k} & = \left(\sum_{j = 0}^n \binom{n}{j} \epsilon^j \right)^{1/k}\\\\ & > \left(1 + n\epsilon + \dfrac{n(n-1)}{2}\epsilon^2 \right)^{1/k}\\\\ & > \left(\dfrac{n(n-1)}{2}\epsilon^2\right)^{1/k} \end{split} \end{equation*}

So that

$$(1 + \epsilon)^{n/k} > \left(\dfrac{n(n-1)}{2}\epsilon^2\right)^{1/k} > n$$

Choose $N = \dfrac{2}{\epsilon^2}$ and you're done.