prove the limit of $k^{1/k}$ is $1$

Question

I want to prove that the limit of the sequence $k^{1/k}$ is $1$ as $k$ tends to infinity without using advanced rules such as L'Hospital's Rule and just using the basic rules in real analysis. How would I go about doing this?

Answer 1

Consider the sequence $\ \ x_k = k^{\frac1k} - 1 \geq 0$

$k ^{\frac1k} \to 1 \iff x_k \to 0 \ as\ k \to \infty $

Now, $\ k = (1 + x_k)^k$

Apply binomial theorem : $$(1+x_k)^k = 1 + kx_k + \frac{(k)(k-1)}{2}x_k^2 + \ ... \ + x_k^k = k$$

Remove all the extra positive terms except $\frac{(k)(k-1)}{2}x_k^2$ and convert it into an inequality

$$ 0 \leq \frac{(k)(k-1)x_k^2}{2} < k^{\frac1k} \ \ \ \ \ \ \ \ \ \ \ \ \ \ for \ k > 1$$

So $$0 \leq x_k^2 < \frac2{k-1}$$

$$ 0 \leq x_k < \sqrt{\frac2{k-1}} \ \ \ \ \ \ \ \ \ \ \ since \ \ x_k \geq 0$$ Now apply squeeze theorem, $ x_k \to 0 \ as \ \ k \to \infty $

Answer 2

Hints:

Let $y(k)=k^{\frac1k}$. Then $\ln y=\frac1k \ln k$. So we just compute the limit of $\frac1k \ln k$. Let $t=\ln k$, then $\lim_{k\to \infty} \frac1k \ln k=\lim_{t\to \infty} \frac{t}{e^t}$.