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 $0$?
On
$$ a^7-b^7 = (a-b)(a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6) $$
So for $a=\sqrt [ 7 ]{ n+1 } $ and $b = \sqrt [ 7 ]{n } $, we get:
$$ \lim _{ n\rightarrow +\infty }{ \sqrt [ 7 ]{ n+1 } -\sqrt [ 7 ]{ n } \quad } \quad =\quad \lim _{ n\rightarrow +\infty }{ \frac { n+1-n }{ (\sqrt [ 7 ]{ n+1 } )^{ 6 }+((\sqrt [ 7 ]{ n+1 } ))^{ 5 }\left( \sqrt [ 7 ]{ n } \right) +(\sqrt [ 7 ]{ n+1 } )^{ 4 }\left( \sqrt [ 7 ]{ n } \right) ^{ 2 }+(\sqrt [ 7 ]{ n+1 } )^{ 3 }\left( \sqrt [ 7 ]{ n } \right) ^{ 3 }+(\sqrt [ 7 ]{ n+1 } )^{ 2 }\left( \sqrt [ 7 ]{ n } \right) ^{ 4 }+(\sqrt [ 7 ]{ n+1 } )\left( \sqrt [ 7 ]{ n } \right) ^{ 5 }+\left( \sqrt [ 7 ]{ n } \right) ^{ 6 } } \quad } \\ \qquad \qquad \qquad \qquad \qquad =\quad \lim _{ n\rightarrow +\infty }{ \frac { 1 }{ (\sqrt [ 7 ]{ n+1 } )^{ 6 }+((\sqrt [ 7 ]{ n+1 } ))^{ 5 }\left( \sqrt [ 7 ]{ n } \right) +(\sqrt [ 7 ]{ n+1 } )^{ 4 }\left( \sqrt [ 7 ]{ n } \right) ^{ 2 }+(\sqrt [ 7 ]{ n+1 } )^{ 3 }\left( \sqrt [ 7 ]{ n } \right) ^{ 3 }+(\sqrt [ 7 ]{ n+1 } )^{ 2 }\left( \sqrt [ 7 ]{ n } \right) ^{ 4 }+(\sqrt [ 7 ]{ n+1 } )\left( \sqrt [ 7 ]{ n } \right) ^{ 5 }+\left( \sqrt [ 7 ]{ n } \right) ^{ 6 } } \quad } \\ \qquad \qquad \qquad \qquad \qquad =\quad 0 $$
On
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.
On
Notice that
$$\left(\sqrt[7]{n+1}\right)^7-\left(\sqrt[7]{n}\right)^7=\left(\sqrt[7]{n+1}-\sqrt[7]{n}\right)\left(\sqrt[7]{(n+1)^6}+\cdots\sqrt[7]{n^6}\right)=1,$$
where we have used the identity
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right),$$
which holds for odd $n$. Then,
$$\lim_{n\to\infty} \sqrt[7]{n+1}-\sqrt[7]{n}=\lim_{n\to\infty}\frac{1}{\sqrt[7]{(n+1)^6}+\cdots\sqrt[7]{n^6}}=0,$$
since the denominator is going to $\infty$ and the numerator is $0$.
On
I am wondering whether showing the ratio between the two terms approaching 1 is sufficient to show the difference is 0"
$\lim\limits_{x\to\infty} \frac{x + 1}{x} = 1$, but $\lim\limits_{x\to\infty}{ (x + 1) - (x)} \ne 0$
One approach to proving your limit would be to let $n = 1/x$ and let $x\to 0^+$. Then apply L'Hopital's rule.
On
Divide by $(n)^{1/7}$:
$$\begin{align}\\\lim_{n\to \infty}\left[\left(\frac {n+1}{n}\right)^{1/7} -1\right]&=0\\1-1&=0\\0&=0\end{align}$$
On
Use the Taylor expansion of $(1 + x)^\alpha = 1 + \alpha x + o(x)$: \begin{align*} & (n + 1)^{1/7} - n^{1/7} \\ = & n^{1/7}\left[\left(1 + \frac{1}{n}\right)^{1/7} - 1\right] \\ = & n^{1/7}\left(1 + \frac{1}{7}\frac{1}{n} + o\left(\frac{1}{n}\right) - 1\right) \\ = & \frac{1}{7}\frac{1}{n^{6/7}} + o\left(\frac{1}{n^{6/7}}\right)\\ \to & 0 \end{align*} as $n \to \infty$.
On
Notice, we have $$\lim_{n\to \infty}[(n+1)^{1/7}-n^{1/7}]$$ $$=\lim_{n\to \infty}\left[n^{1/7}\left(1+\frac{1}{n}\right)^{1/7}-n^{1/7}\right]$$ Using binomial expansion of $\left(1+\frac{1}{n}\right)^{1/7}$ & neglecting higher power terms $$=\lim_{n\to \infty}\left[n^{1/7}\left(1+\frac{1}{7}\frac{1}{n}\right)-n^{1/7}\right]$$ $$=\lim_{n\to \infty}\left[n^{1/7}+\frac{1}{7}\frac{n^{1/7}}{n}-n^{1/7}\right]$$ $$=\lim_{n\to \infty}\left[\frac{1}{7}\frac{1}{n^{6/7}}\right]$$ $$=\frac{1}{7}\lim_{n\to \infty}\left[\frac{1}{n^{6/7}}\right]=0$$
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 $$\displaystyle \lim_{x\rightarrow a}\frac{x^n-a^n}{x-a} = na^{n-1}$$