Just a long problem

The value of $x$ simplifies to

$\huge x = \sum _{n=1}^{\infty }\left(\frac{1}{\frac{n\left(n+1\right)}{2}}\right)$

$\large x =\sum _{n=1}^{\infty }\frac{2}{n\left(n+1\right)}$

$\large x =\sum _{n=1}^{\infty }\:2\left(\frac{1}{n}-\frac{1}{n+1}\right)\:\:$

Giving us $\boxed {\large x = 2}$ .

The value of $y$ , on the other hand, simplifies to

$y = \huge \sum _{n=1}^{\infty }\left(\frac{1}{n^2}\right)$

which simplifies to $\boxed{ \large y = \frac {\pi^{2}}{6} }$ .

Thus, $z = \huge \frac {2 \times \frac{\pi^{2}}{6}}{ \pi^{2}}$

$\boxed {z = \frac {1}{3}}$

Let $w$ be the exact value of the nested square root.

$\large w=\sqrt{z+2z\sqrt{z+2z\sqrt{z+2z\sqrt{z+...}}}}$

$\large w = \sqrt{z+2zw}$

$\large w^{2} = \frac {1}{3} + \frac{2}{3}w$

$\large w^{2} - \frac{2}{3}w - \frac {1}{3} = 0$

$\large 3w^{2} - 2w - 1 = 0$

$\large (3w + 1)(w-1) = 0$

since $w$ is positive, we take $\large{ \boxed { w = 1 } }.$