compute $\displaystyle \lim_{n \to \infty}\frac{\sum_{i=1}^{n-2} \sum_{j = 1} ^{n-i-1} ( n-(i + j))}{n^3}$? I was trying to solve the following problem 

If we pick k numbers from the interval (0, 1), what is the probability the sum of those numbers is <1 when ? (k = 6)

for k = 2, got $\displaystyle\lim_{n \to \infty}\frac{\sum_{i = 1}^{n-1} i}{n^2}=0.5$
for k = 3, I got the expression $\displaystyle \lim_{n \to \infty}\frac{\sum_{i=1}^{n-2} \sum_{j = 1} ^{n-i-1} ( n-(i + j))}{n^3}$
how to compute it and is there a way to generalize it for picking k numbers from the interval (0,1)?
 A: Presumably each number is chosen independently and uniformly from $(0,1)$.
A more direct way to compute the probability is is to think of the three-dimensional cube consisting of points $(x,y,z)$ such that $x,y,z \in (0,1)$. Then the probability you seek is the volume of the piece of the cube that lies below the plane $x+y+z=1$.
This will be a pyramid, whose volume you can compute easily.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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$\ds{\lim_{n \to \infty}{\sum_{i = 1}^{n - 2}\sum_{j = 1}^{n - i - 1}
\bracks{n - \pars{i + j}} \over n^{3}}}$

\begin{align}
&\lim_{n \to \infty}{\sum_{i = 1}^{n - 2}\sum_{j = 1}^{n - i - 1}
\bracks{n - \pars{i + j}} \over n^{3}}
\\[5mm] = &\
\lim_{n \to \infty}\braces{{1 \over n^{3}}
\sum_{i = 1}^{n - 2}\bracks{\pars{n - i}\pars{n - i - 1} -
{\pars{n - i - 1}\pars{n - i} \over 2}}}
\\[5mm] = &\
{1 \over 2}\lim_{n \to \infty}\bracks{{1 \over n^{3}}
\sum_{i = 1}^{n - 2}\pars{n - i - 1}\pars{n - i}}
\\[5mm] = &\
{1 \over 2}\lim_{n \to \infty}\braces{{1 \over n^{3}}
\sum_{i = 1}^{n - 2}\bracks{n - \pars{n - 1 - i} - 1}
\bracks{n - \pars{n - 1 - i}}}
\\[5mm] = &\
{1 \over 2}\lim_{n \to \infty}\bracks{{1 \over n^{3}}
\sum_{i = 1}^{n - 2}i\pars{i + 1}} =
{1 \over 2}\lim_{n \to \infty}\pars{{1 \over n^{3}}
\sum_{i = 1}^{n - 2}\pars{i + 1}^{\underline{2}}} =
{1 \over 2}\lim_{n \to \infty}\pars{{1 \over n^{3}}
\left.{\pars{i + 1}^{\underline{3}} \over 3}\right\vert_{\ 1}^{\ n - 1}}
\\[5mm] = &\
{1 \over 6}\lim_{n \to \infty}\bracks{{1 \over n^{3}}\,
\pars{n^{\underline{3}} - 1^{\underline{3}}}} =
{1 \over 6}\lim_{n \to \infty}{\pars{n - 1}\pars{n - 2} \over n^{2}}
=
\bbx{\ds{1 \over 6}}
\end{align}
