<s>[INST] 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 $0$?
 [/INST] Hint: If we can set $a=(n+1)^{\frac 17}$ and $b=n^{\frac 17}$, then the expression becomes $$\lim_{n \to \infty} [a-b]=0.$$ It suffices to prove that $$a-b \to 0$$ as $n \to \infty$. 
The Binomial Theorem at degree $7$ gives us
$$a^7-b^7=(a-b)(a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6). \tag{$\star$}$$
We have
\begin{align}
a-b
&= \frac{a^7-b^7}{a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6} \quad \text{by } (\star)\\
&= \frac 1{a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6} \\
&\to 0
\end{align}
as $n \to \infty$. 
Explanations for the above steps:


*

*$a^7-b^7=[(n+1)^{\frac 17}]^7-[n^{\frac 17}]^7=(n+1)-n=1$

*$a =(n+1)^{\frac 17} \to \infty$ and $b = n^{\frac 17} \to \infty$ 

*The variables $a$ and $b$ are in only the denominator for the penultimate step above.
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