Find the limit of $(1^7 + 2^7 + .......+ n^7)^{1/n}$ as $n \rightarrow \infty$ 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?
 A: 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 $1$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\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. 

A: 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.
