A minimalist proof of

$1+2+3+...=-1/12$

$$\begin{array}{rl}{S}_1& =1-1+1-1+\dots \\ 1-S_1& =1-(1-1+1-1+\dots )\\ & =1-1+1-1+\dots =S_1\\ S_1& =\frac{1}{2}\end{array}$$

$$\begin{array}{rl}{S}_2& =1-2+3-4+\dots \\ 2S_2& =1-2+3-4+\dots \\ & +1-2+3-4+\dots \\ & =1-1+1-1+\dots =S_1\\ S_2& =\frac{1}{4}\end{array}$$

$$\begin{array}{rl}S& =1+2+3+4+\dots \\ S-S_2& =1+2+3+4+\dots \\ & -(1-2+3-4+\dots )\\ & =4(1+2+3+4\dots )=4S\\ S& =-\frac{1}{3}{S}_2=-\frac{1}{12}\end{array}$$