If you have the sums $ (1+2+.. +n) + (1+2+3+.. +n-1)+ (1+2+3+.. +n-2)+(1+2+3+.. +n-3)+... +(1+2+3)+(1+2)+1$for large enough $n$
$$\frac {n^3}{3!} \approx (1+2+.. +n) + (1+2+3+.. +n-1)+ (1+2+3+.. +n-2)+(1+2+3+.. +n-3)+... +(1+2+3)+(1+2)+1$$ if divided the sum by the divisor let's call it $x$ (can be any number $1,2,3.. $) we get $$\frac {n^3}{3!x^2} \approx (1x+2x+.. +\frac {n}{x}x) + (1x+2x+3x+.. +\frac {n-1}{x}x)+ (1x+2x+3x+.. +\frac {n-2}{x}x)+(1x+2x+3x+.. +\frac {n-3}{x}x)+... +(1x+2x+3x)+(1x+2x)+1x$$If the difference between the closest numbers let's call it $d$, $d=\frac {1}{10^k}$, we get $$\frac {n^3}{3!x^2d^2} \approx (1dx+2dx+3dx.. +\frac {n}{x}x) + (1dx+2dx+3dx+.. +\frac {n-d}{x}x)+ (1dx+2dx+3dx+.. +\frac {n-2d}{x}x)+(1dx+2dx+3dx.. +\frac {n-3d}{x}x)+... +(1dx+2dx+3dx)+(1dx+2dx)+1dx$$ if we assume $k\to\infty$ we get $$\frac {n^3}{3!x^2d^2} = (1dx+2dx+3dx.. +\frac {n}{x}x) + (1dx+2dx+3dx+.. +\frac {n-d}{x}x)+ (1dx+2dx+3dx+.. +\frac {n-2d}{x}x)+(1dx+2dx+3dx.. +\frac {n-3d}{x}x)+... +(1dx+2dx+3dx)+(1dx+2dx)+1dx$$ OR $$\frac {n^3}{3!x^2} = (1d^3x+2d^3x+3d^3x.. +\frac {n}{x}dx) + (1d^3x+2d^3x+3d^3x+.. +\frac {n-d}{x}d^2x)+ (1d^3x+2d^3x+3d^3x+.. +\frac {n-2d}{x}d^2x)+(1d^3x+2d^3x+3d^3x.. +\frac {n-3d}{x}d^2x)+... +(1d^3x+2d^3x+3d^3x)+(1d^3x+2d^3x)+1d^3x$$ Now if we assume that $n$ value of circle arc and $x$ value of diameter, we get $n - \frac {n^3}{3!x^2} \approx$ value of chord, So we got the first condition of Taylor series.The question is my calculations are correct?