Plz Get This One Right

Consider the lines $L_n$ given by the equation $x = \left( n + \frac12 \right) y$ for $n=0,1,2,\ldots.$ The probability that the closest integer to $x/y$ is $n$ is precisely the area enclosed by $L_{n-1}, L_n,$ and the unit square--except when $n=0,$ when that probability is the area of the triangle enclosed by the left and top sides of the square and $L_0,$ which is $1/4.$

For $n = 2k,$ $k \ge 1,$ the area enclosed by $L_{n-1},L_n,$ and the unit square is the area of a triangle with vertices at the origin and the points $\left( 1, \frac2{4k-1}\right), \left( 1, \frac2{4k+1}\right).$ This is a triangle with height $1$ and base $\frac2{4k-1} - \frac2{4k+1},$ so its area is $\frac1{4k-1} - \frac1{4k+1}.$

Hence the probability that $x/y$ is even is $\frac14 + \sum_{k=1}^\infty \left( \frac1{4k-1} - \frac1{4k+1} \right) = \frac14 + \left( \frac13 - \frac15 + \frac17 - \frac19 + \cdots \right).$ Now it's well-known (the Leibniz formula for $\pi$ ) that $\pi/4 = 1 - \frac13 + \frac15 - \frac17 + \cdots,$ so $1-\pi/4 = \frac13 - \frac15 + \frac17 - \frac19 + \cdots.$ Substituting into the probability calculation above, we get $\frac14 + (1-\pi/4) = \frac{5-\pi}4.$

Let $X,Y$ be independent random variables, $X,Y\sim U(0,1)$ . So

$\mathbb{P} (\frac{X}{Y} \le t)=\intop_0^{min\{t,1\}} \int_{\frac{x}{t}}^1 dydx = \begin{cases} 1-\frac{1}{2t} & t>1\\ \frac{t}{2} & 0<t\le 1\\ \end{cases}$

So for all $n\in \mathbb N$ ,

$\mathbb{P} (2n-\frac{1}{2} < \frac{X}{Y} \le 2n+\frac{1}{2}) = \frac{1}{4n-1} -\frac{1}{4n+1}$

Finally,

$\mathbb{P} (\frac{X}{Y}<\frac{1}{2}) + \sum_{n=1}^\infty \mathbb{P} (2n-\frac{1}{2} < \frac{X}{Y} < 2n + \frac{1}{2}) = \frac{1}{4} + \sum_{n=1}^\infty (\frac{1}{4n-1} - \frac{1}{4n+1}) = \frac{1}{4} \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2k+1} = \frac{1}{4} + 1-\frac{\pi}{4} = \boxed { \frac{5-\pi}{4} }$