Proving limit of difference is $0$: $\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$ I am trying to show that
$$\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$$

and I know intuitively this must be so since the "+1" contribution in the first term becomes negligible as $n$ becomes sufficiently large. I am wondering whether showing the ratio between the two terms approaching $1$ is sufficient to show the difference is $0$; if so, how can this be formalized? If not, how can it more directly be shown that the difference approaches $0$?
Given $$\displaystyle \lim_{n\rightarrow \infty}\left[(n+1)^{\frac{1}{7}}-n^{\frac{1}{7}}\right]\;,$$ Now let $$\displaystyle n= \frac{1}{y}\;,$$ Then $y\rightarrow 0$

So $$\displaystyle \lim_{y\rightarrow 0}\frac{(1+y)^{\frac{1}{7}}-1}{y^{\frac{1}{7}}} = \lim_{y\rightarrow 0}\frac{(1+y)^{\frac{1}{7}}-1}{(1+y)-1}\cdot \frac{y}{y^{\frac{1}{7}}} = \lim_{y\rightarrow 0}\frac{(1+y)^{\frac{1}{7}}-1}{(1+y)-1}\cdot \lim_{y\rightarrow 0}y^{\frac{6}{7}} = \frac{1}{7}\times 0 = 0$$

Above we used the formula $$\displaystyle \lim_{x\rightarrow a}\frac{x^n-a^n}{x-a} = na^{n-1}$$
Divide by $(n)^{1/7}$:

$$\begin{align}\\\lim_{n\to \infty}\left[\left(\frac {n+1}{n}\right)^{1/7} -1\right]&amp;=0\\1-1&amp;=0\\0&amp;=0\end{align}$$