<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] $$
a^7-b^7 = (a-b)(a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6)
$$
So for $a=\sqrt [ 7 ]{ n+1 } $ and $b = \sqrt [ 7 ]{n  } $, we get:
$$
\lim _{ n\rightarrow +\infty  }{ \sqrt [ 7 ]{ n+1 } -\sqrt [ 7 ]{ n } \quad  } \quad =\quad \lim _{ n\rightarrow +\infty  }{ \frac { n+1-n }{ (\sqrt [ 7 ]{ n+1 } )^{ 6 }+((\sqrt [ 7 ]{ n+1 } ))^{ 5 }\left( \sqrt [ 7 ]{ n }  \right) +(\sqrt [ 7 ]{ n+1 } )^{ 4 }\left( \sqrt [ 7 ]{ n }  \right) ^{ 2 }+(\sqrt [ 7 ]{ n+1 } )^{ 3 }\left( \sqrt [ 7 ]{ n }  \right) ^{ 3 }+(\sqrt [ 7 ]{ n+1 } )^{ 2 }\left( \sqrt [ 7 ]{ n }  \right) ^{ 4 }+(\sqrt [ 7 ]{ n+1 } )\left( \sqrt [ 7 ]{ n }  \right) ^{ 5 }+\left( \sqrt [ 7 ]{ n }  \right) ^{ 6 } } \quad  } \\ \qquad \qquad \qquad \qquad \qquad =\quad \lim _{ n\rightarrow +\infty  }{ \frac { 1 }{ (\sqrt [ 7 ]{ n+1 } )^{ 6 }+((\sqrt [ 7 ]{ n+1 } ))^{ 5 }\left( \sqrt [ 7 ]{ n }  \right) +(\sqrt [ 7 ]{ n+1 } )^{ 4 }\left( \sqrt [ 7 ]{ n }  \right) ^{ 2 }+(\sqrt [ 7 ]{ n+1 } )^{ 3 }\left( \sqrt [ 7 ]{ n }  \right) ^{ 3 }+(\sqrt [ 7 ]{ n+1 } )^{ 2 }\left( \sqrt [ 7 ]{ n }  \right) ^{ 4 }+(\sqrt [ 7 ]{ n+1 } )\left( \sqrt [ 7 ]{ n }  \right) ^{ 5 }+\left( \sqrt [ 7 ]{ n }  \right) ^{ 6 } } \quad  } \\ \qquad \qquad \qquad \qquad \qquad =\quad 0
$$
</s>