Find the limit of $(1^7 + 2^7 + .......+ n^7)^{1/n}$ as $n \rightarrow \infty$

Question

The question tells me find the limit of $(1^7 + 2^7 + .......+ n^7)^{1/n}$.

I thought that I would use an idea similar to the one given below:

Exercise 1.8.4. $a_n=\sqrt[n]n.$

Solution. If we take $b=\sqrt[n]n$ and apply $(1.14)$, we obtain $$n>\frac{n(n-1)}2(\sqrt[n]n-1)^2.$$ It follows that $$0<\sqrt[n]n-1<\sqrt{\frac2{n-1}}\to0,$$ so $\lim{\sqrt[n]n}=1$.

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but with $b = n^{14/n}$ and using the $14^{\text{th}}$ term of the binomial theorem which is $$\frac{n(n-1)(n-2)\cdots(n-13)}{14!} (n^{14/n} - 1)^{14}$$ but I got stucked.

Could anyone help me please?

Answer 1

We have that $n^{\frac{8}{n}} = (n^{\frac{1}{n}})^{8} \to 1^{8}=1$.

Now $ 1^{1/n} \leq (1+\ldots+n^7)^{1/n} \leq (n^8)^{1/n}$.

Since both limits are $1$, we have that our limit is $1$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{% \lim_{n \to \infty}\pars{1^{7} + 2^{7} + \cdots + n^{7}}^{1/n}} = \exp\pars{\lim_{n \to \infty}{\ln\pars{\sum_{k = 1}^{n}k^{7}} \over n}} \\[5mm] = &\ \exp\pars{\lim_{n \to \infty}{\ln\pars{\sum_{k = 1}^{n + 1}k^{7}} - \ln\pars{\sum_{k = 1}^{n}k^{7}} \over \pars{n + 1} - n}} \\[5mm] = &\ \exp\pars{\lim_{n \to \infty}{\ln\pars{1 + {\bracks{n + 1}^{7} \over \sum_{k = 1}^{n + 1}k^{7}}}}} \\[5mm] = &\ \exp\pars{\ln\pars{1 + \lim_{n \to \infty}{\bracks{n + 2}^{7} - \bracks{n + 1}^{7} \over \bracks{n + 2}^{7}}}} = \exp\pars{\ln\pars{1}} = \bbx{1} \end{align}

We were using the Stolz-Ces$\mrm{\grave{a}}$ro Theorem in the last two lines.

Answer 3

Assuming that you could be interested by more than the limit itself.

Considering Faulhaber polynomials $$S_n=\sum_{i=1}^n i^7=\frac {a^2(6a^2-4a+1)} 3 \qquad \text{where} \qquad a=\frac 12 n (n+1)$$ you can write $$S_n=\frac{n^8}8 \left(1+\frac{4}{n}+\frac{14}{3 n^2}-\frac{7}{3 n^4}+\frac{2}{3 n^6} \right)$$ $$\log(S_n)=\log\left(\frac{n^8}8\right)+\log \left(1+\frac{4}{n}+\frac{14}{3 n^2}-\frac{7}{3 n^4}+\frac{2}{3 n^6} \right)$$ Now, using Taylor expansion $$\log(S_n)=\left(8 \log \left({n}\right)-\log (8)\right)+\frac{4}{n}-\frac{10}{3 n^2}+O\left(\frac{1}{n^3}\right)$$

$$\frac 1 n \log(S_n)=\frac{8 \log \left({n}\right)-\log (8) }n+\frac{4}{n^2}+O\left(\frac{1}{n^3}\right)$$

Using Taylor again $$S_n^{\frac 1n}=e^{\frac 1 n \log(S_n)}=1+\frac{8 \log \left({n}\right)-\log (8) }n+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.