How can I prove that the following limit is equal to $ \frac{1}{3} $

Question

$$ \lim_{n \to \infty}\frac{1}{n^3}\sum_{k=1}^nk^2 $$

I have noticed a similarity to $ \frac{1}{n^3}\int_{0}^{n}k^2dk=\frac{1}{3} $.

Answer 1

HINT:

Use that $$\sum_{k=1}^nk^2=\frac{ n \left( n+1 \right) \left( 2\,n+1 \right)}{6} $$

Answer 2

Using

$$\frac{1}{n^3}\int_{0}^{n}x^2~\mathrm dx=\frac{1}{3}$$

We can squeeze,

$$\frac{1}{n^3}\int_{0}^{n}x^2~\mathrm dx-\frac1n<\frac1{n^3}\sum_{k=0}^nk^2<\frac{1}{n^3}\int_{0}^{n}x^2~\mathrm dx$$

Answer 3

Consider this $$k^3-(k-1)^3=3k^2-3k+1.$$ Adding this from $1$ to $n$, the sum telescopes and $$n^3=3\sum_{k=1}^nk^2-3\sum_{k=1}^nk+\sum_{k=1}^n1.$$ Alternatively $$1=\frac3{n^3}\sum_{k=1}^nk^2-\frac3{n^3}\sum_{k=1}^nk+ \frac1{n^3}\sum_{k=1}^n1.\tag{*}$$ Obviously $$0\le\sum_{k=1}^n k\le n^2$$ so $$0\le\frac1{n^2}\sum_{k=1}^n k\le \frac1n.$$ Therefore $$\frac1{n^2}\sum_{k=1}^n k\to 0.$$ Similarly, $$\frac1{n^2}\sum_{k=1}^n 1\to 0.$$ From $(*)$ then $$\frac3{n^3}\sum_{k=1}^nk^2\to 1.$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\LARGE\left.a\right)}$

\begin{align} \lim_{n \to \infty}\pars{{1 \over n^{3}}\sum_{k = 1}^{n}k^{2}} & = \lim_{n \to \infty}\bracks{{1 \over n^{3}} \sum_{k = 1}^{n}\pars{k^{\overline{2}} + k^{\overline{1}}}} \\[5mm] & = \lim_{n \to \infty}{\pars{n + 1}^{\overline{3}}/3 + \pars{n + 1}^{\overline{2}}/2 - 1^{\overline{3}}/2 - 1^{\overline{2}}/2 \over n^{3}} \\[5mm] & = \lim_{n \to \infty}{\pars{n + 1}n\pars{n - 1}/3 + \pars{n + 1}n/2 \over n^{3}} = \bbx{1 \over 3} \end{align}

$\ds{\LARGE\left.b\right)}$. With Stolz-Ces$\mrm{\grave{a}}$ro Theorem:

\begin{align} \lim_{n \to \infty}\pars{{1 \over n^{3}}\sum_{k = 1}^{n}k^{2}} & = \lim_{n \to \infty}{\sum_{k = 1}^{n + 1}k^{2} - \sum_{k = 1}^{n}k^{2} \over \pars{n + 1}^{3} - n^{3}} = \lim_{n \to \infty}{\pars{n + 1}^{2} \over 3n^{2} + 3n + 1} \\[5mm] & = {1 \over 3}\,\lim_{n \to \infty}{\pars{1 + 1/n}^{2} \over 1 + 1/n + 1/\pars{3n^{2}}} = \bbx{1 \over 3} \end{align}