Chain Reaction 1!

Let's look what $f(x, y)$ is. $\begin{aligned}f(x,y) &= \lim_{n\to\infty}\left(\lfloor x\rfloor^n + y^n\right)^{\frac 1n} \text{ for } x>y\Rightarrow \lfloor x\rfloor\ge y\\&= L(\text{say})\\\ln L& = \dfrac1n\ln\left(\lfloor x\rfloor^n + y^n\right)\\&= \dfrac1n\ln\left(\lfloor x\rfloor^n\left( 1+ \left(\dfrac y{\lfloor x\rfloor}\right)^n\right)\right)\\&= \dfrac1n\left[n\ln\lfloor x\rfloor + \ln\left( 1+ \left(\dfrac y{\lfloor x\rfloor}\right)^n\right)\right]&\color{#3D99F6}{ \left(\dfrac y{\lfloor x\rfloor}\right) < 1 \Rightarrow \left(\dfrac y{\lfloor x\rfloor}\right)^\infty = 0}\\&= \ln \lfloor x\rfloor \\\Rightarrow L &= \lfloor x\rfloor \end{aligned}$

So that, we can take $g(x, y) = f(x,y) = \max\left(\lfloor x\rfloor ,\lfloor y\rfloor\right)$

Given $m = e\Rightarrow\lfloor m\rfloor = 2$

$\dfrac{(\lfloor m\rfloor-1)\lfloor m\rfloor(\lfloor m\rfloor+1)}3 = 2$ $\begin{aligned}\lim_{n\to \infty} j(n, e) &= \dfrac1{2\cdot2} + \dfrac1{1\cdot2+2\cdot 3} + \dfrac1{1\cdot2+2\cdot 3+3\cdot 4} + \cdots \\ &=\lim_{n\to \infty}\dfrac14+\sum_{s=2}^n \dfrac1{\displaystyle\sum_{r=1}^sr(r+1)}\\ &=\lim_{n\to \infty}\dfrac14+3\sum_{s=2}^n\dfrac1{s(s+1)(s+2)}\\ &= \lim_{n\to \infty}\dfrac14 + \dfrac32\cdot\left(\dfrac1{2\cdot3} - \dfrac1{n(n+1)}\right)\\ &= \dfrac12 = \boxed{0.5}\end{aligned}$