No Squares

Assume the probability distribution mentioned follows a uniform distribution .

Define the notation $[ a,b]_c$ as the probability of randomly choosing 2 real numbers between $0$ and $c$ such that they fall in between $a$ and $b$ , where $0\leq a\leq b\leq c$ .

For example, $f(1) = [0,1]_1 = 1$ and $f(2) = [0,1]_4 + [1,4]_4 = \left(\frac 14\right)^2 + \left(\frac34\right)^2 = \frac58$ .

In general,

$\begin{array} { l c l } f(n) &=& [0,1]_{n^2} + [1,4]_{n^2} + [4,9]_{n^2} + \cdots + [(n-1)^2 , n^2]_{n^2} \\ &=& \displaystyle \left(\frac1{n^2} \right)^2 + \left(\frac3{n^2} \right)^2 + \left(\frac5{n^2} \right)^2 + \cdots + \left(\frac{2n-1}{n^2} \right)^2 \\ &=& \displaystyle \frac1{n^4} \left [ 1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2 \right ] \end{array}$

We have $\displaystyle \lim_{n\to\infty} n f(n) = \lim_{n\to\infty} \dfrac{1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2 }{n^3}$ .

Let $a_n = 1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2$ and $b_n = n^3$ . By Stolz–Cesàro theorem , if $\displaystyle \lim_{n\to\infty}\dfrac{a_{n+1} - a_n}{b_{n+1} - b_n}$ is finite and is equal to $l$ , then so does the limit in question.

$\lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim_{n\to\infty} \dfrac{(2n+1)^2}{ (n+1)^3 - n^3} = \dfrac43$

Hence, the limit in question is equal to $\frac43$ . Our answer is $4 + 3 = \boxed7$ .