Let $p_n$ be the probability that the figure formed by connecting the four points $O,P,Q,R $ in this order is a convex square. (Problem)
Let rhombus OABC be a rhombus on the plane with one side of length $1$. $\angle A=60^{\circ}  $. Let $n$ be a positive integer and $i, j, k $ are integers between $1$ and $n$. Three points P,Q,R are defined by
$\overrightarrow{OP}=\dfrac{i}{n}\overrightarrow{OA}$ ,
$\overrightarrow{OQ}=\dfrac{j}{n}\overrightarrow{OB}$ ,
$\overrightarrow{OR}=\dfrac{k}{n}\overrightarrow{OC}$ .Let T be the intersection of PR and OB.
Let $p_n$ be the probability that the figure formed by connecting the four points $O,P,Q,R $ in this order is a convex quadrilateral. Find $\lim_{n \to \infty}p_n$.
(MY idea)
I've calculated $|\overrightarrow{OT}|$ which is $|\overrightarrow{OT}|=\dfrac{2ik}{n(i+k)}$
And when $O,P,Q,R $ in this order is a convex quadrilateral, $|\overrightarrow{OT}|<|\overrightarrow{OQ}|$ holds.
Although my teacher gave me a hint to use $\int_{0}^{n}\dfrac{ix}{i+x}dx<\sum_{k=1}^n \dfrac{ik}{i+k}<\int_{0}^{n+1}\dfrac{ix}{i+x}dx$ when calculating $\lim_{n \to \infty}p_n$.
But I can't find $p_n$ yet.
Sorry for my poor English.(I'm not a native speaker)
 A: Hints:

*

*$\displaystyle \big|\,\vec{OT}\,\big|<\big|\,\vec{OQ}\,\big|\ \Leftrightarrow\ \frac{2ik}{i+k}<j$

*$\displaystyle
  \begin{align}
  \\\\\therefore\ p_n&=\frac{1}{n^3}\left|\left\{(i,j,k)\,\left|\,1\le i\le n,1\le k\le
   n,\frac{2ik}{i+k}<j\le n\right.\right\}\right|\\
&=\frac{1}{n^3}\sum_{i=1}^n\sum_{k=1}^n\left|\left\{j\,\left|\,\frac{2ik}{i+k}<j\le n\right.\right\}\right|\end{align}$

*$\displaystyle n-\frac{2ik}{i+k}-1\le\left|\left\{j\,\left|\,\frac{2ik}{i+k}<j\le n\right.\right\}\right|\le n-\frac{2ik}{i+k}$

*$\displaystyle \therefore\ 1-\frac{1}{n^3}\sum_{i=1}^n\sum_{k=1}^n\frac{2ik}{i+k}-\frac{1}{n}\le p_n\le1-\frac{1}{n^3}\sum_{i=1}^n\sum_{k=1}^n\frac{2ik}{i+k} $

*Therefore, if $\displaystyle\ \lim_{n\rightarrow\infty}\frac{1}{n^3}\sum_{i=1}^n\sum_{k=1}^n\frac{2ik}{i+k}=\ell\ $ then $\displaystyle\ \lim_{n\rightarrow\infty}p_n=1-\ell\ $.

*Using a two-dimensional version your teacher's  hint,$$\int_0^n\int_0^n\frac{xy}{x+y}\,dydx\le \sum_{i=1}^n\sum_{k=1}^n\frac{ik}{i+k}\le\int_0^{n+1}\int_0^{n+1}\frac{xy}{x+y}\,dydx$$

*Using the substitutions $\ u=\frac{x}{m}, v=\frac{y}{m}\ $ in the integral $\ \displaystyle\int_0^m\int_0^m\frac{xy}{x+y}\,dydx\ $ gives $$\displaystyle\int_0^m\int_0^m\frac{xy}{x+y}\,dydx=m^3 \displaystyle\int_0^1\int_0^1\frac{uv}{u+v}\,dvdu$$

*Putting all this together you should be able to get $$\lim_{n\rightarrow\infty}p_n=1-2\int_0^1\int_0^1\frac{uv}{u+v}\,dvdu$$
The integral $\ \displaystyle\int_0^1\int_0^1\frac{uv}{u+v}\,dvdu\ $ isn't difficult to evaluate, but requires a fair amount of tedious calculation.  When I tried to do it, I somewhere lost the factor of $\ 2\ $ multiplying the logarithm in Wolfram alpha's answer of $\ \displaystyle\frac{2-2\ln2}{3}\approx0.204569\ $. I couldn't be bothered trying to locate the presumable error in my calculation.
