# Evaluate $\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^n\left(\left(\frac{3i}{n}\right)^2-\left(\frac{3i}{n}\right)\right)$

How to evaluate the following?

$$\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^n\left(\left(\frac{3i}{n}\right)^2-\left(\frac{3i}{n}\right)\right)$$

I simply expanded but I did not find the answer. I think there might be some trick or clever observation, but now I'm not seeing it.

• Think in terms of a Riemann sum: what definite integral might this correspond to? – colormegone Jul 17 '16 at 8:28
• Would someone please explain why you voted to close this question? I thought it's a valid math question :( – 3x89g2 Jul 17 '16 at 8:30
• @RecklessReckoner I don't get it. – 3x89g2 Jul 17 '16 at 8:36

It is a Riemann sum, so we have $$\lim_{n\rightarrow\infty}\frac{3}{n}\sum_{k=1}^{n}\left(\left(\frac{3k}{n}\right)^{2}-\frac{3k}{n}\right)=3\int_{0}^{1}(9x^{2}-3x)\,dx=\color{red}{\frac{9}{2}}.$$

Since $$\sum_{i=1}^n i=\frac{n(n+1)}{2}$$ and $$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6},$$ we have $$\lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^n\left(\left(\frac{3i}{n}\right)^2-\left(\frac{3i}{n}\right)\right)=\lim_{n\to\infty}\frac{3}{n}(\frac{9}{n^2}\frac{n(n+1)(2n+1)}{6}-\frac{3}{n}\frac{n(n+1)}{2})=\lim_{n\to\infty}\frac{3}{n}\frac{3}{2n}(n+1)^2=\frac{9}{2}$$

prove that your term has the form $$\frac{3}{n}\cdot \frac{3(n+1)^2}{2n}$$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\lim_{n \to \infty}\,{3 \over n}\sum_{i = 1}^{n} \bracks{\pars{3i \over n}^{2} - \pars{3i \over n}}} = 27\lim_{n \to \infty}\pars{{1 \over n^{3}}\sum_{i = 1}^{n}i^{2}} - 9\lim_{n \to \infty}\pars{{1 \over n^{2}}\sum_{i = 1}^{n}i} \end{align}

However, with $\ds{k = 1,2}$ and Stolz-Cesàro Theorem: \begin{align} \fbox{$\ds{\ \lim_{n \to \infty}{1 \over n^{\color{#f00}{k} + 1}} \sum_{n = 1}^{n}i^{\color{#f00}{k}}\ }$} & = \lim_{n \to \infty}{\pars{n + 1}^{k} \over \pars{n + 1}^{k + 1} - n^{k + 1}} = \lim_{n \to \infty}{\sum_{\ell = 0}^{k - 1}{k \choose \ell}n^{\ell}\ +\ n^{k} \over \sum_{\ell = 0}^{k - 1}{k + 1 \choose \ell}n^{\ell}\ +\ \pars{k + 1}n^{k}} \\[4mm] & = \lim_{n \to \infty}{\sum_{\ell = 0}^{k - 1}{k \choose \ell}n^{\ell - k}\ +\ 1 \over \sum_{\ell = 0}^{k - 1}{k + 1 \choose \ell}n^{\ell - k}\ +\ \pars{k + 1}} = \fbox{$\ds{\ {1 \over \color{#f00}{k} + 1}\ }$} \end{align}

Note that $\ds{\left.\quad\lim_{n \to \infty}\,\,n^{\ell - k} \,\,\right\vert_{\ \ell\ <\ k}\,\,\, =\, \color{#f00}{0}}$.

\begin{align} &\color{#f00}{\lim_{n \to \infty}\,{3 \over n}\sum_{i = 1}^{n} \bracks{\pars{3i \over n}^{2} - \pars{3i \over n}}} = 27\,{1 \over \color{#f00}{2} + 1} - 9\,{1 \over \color{#f00}{1} + 1} = \color{#f00}{9 \over 2} \end{align}