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What is the limit when $n \to \infty$?

$$\lim_{n \to \infty} \frac{1}{n^4} \sum_{J=0}^{2n-1} J^3=?$$

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    There's a closed-form expression for that inner summation. You can do some research to find that. Then just evaluate the integral of that expression and you should be able to take it from there. – 2017-01-21
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    The integral is confusing. What are you integrating and with respect to what? – 2017-01-21
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    Sorry there was no integral sign – 2017-01-21
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    This is a uniform Riemann sum for $\int_0^2 x^3 \, dx$, ergo... – 2017-01-21
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    You've [the tutorial of this MSE](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). Is a quick reference and with it you can edit nice formulas. Also put the mouse over a formula and with the right clik you can select Show Math as> Tex commands, and then copy and modify the code. Good luck. – 2017-01-21
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    Erick Wong how did you get the limits of the integratio.? – 2017-01-21
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    Nice one thanks user243301 7 – 2017-01-21

2 Answers 2

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Hint: $$1^3+2^3+3^3+\cdots+k^3=\left(\frac{k(k+1)}{2}\right)^2.$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\bracks{{1 \over n^{4}}\sum_{J = 0}^{2n - 1}J^{3}} &\ =\ \overbrace{\lim_{n \to \infty}\bracks{{1 \over \pars{n + 1}^{4} - n^{4}} \pars{\sum_{J = 0}^{2n + 1}J^{3} - \sum_{J = 0}^{2n - 1}J^{3}}}} ^{\ds{\mbox{Stolz-Ces$\mrm{\grave{a}}$ro Theorem}}} \\[5mm] & = \lim_{n \to \infty}\,\, {\pars{2n + 1}^{3} + \pars{2n}^{3} \over 4n^{3} + 6n^{2} + 4n + 1} = \lim_{n \to \infty}\,\, {16n^{3} + 12n^{2} + 6n + 1\over 4n^{3} + 6n^{2} + 4n + 1} = \bbx{\ds{4}} \end{align}

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    Isn't it easier to type `\mbox{Stolz-Cesàro Theorem}` instead of `\ds{\mbox{Stolz-Ces$\mrm{\grave{a}}$ro Theorem}}`? By the way, both in TeX/LaTeX and MathJax, the command `\displaystyle` doesn't take an argument: it is a declaration that affects everything in the current group. – 2017-01-21
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    @egreg Thanks for your advice. $\texttt{\grave}$ is browser independent. It doesn't depend on keyboard layout or/and language. – 2017-01-21
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    I guess that every browser nowadays is able to understand U+00E0. `;-)` – 2017-01-21