 Ask Remember Me? abver Posts: 34, Reputation: 1 Junior Member #1 Mar 20, 2016, 11:18 AM
Infinity sum
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 $$\frac {n^3}{3!x^2} \approx (1x+2x+.. +\frac {n}{x}x) + (1x+2x+3x+.. +(\frac {n}{x}-1)x)+ (1x+2x+3x+.. +(\frac {n}{x}-2)x)+(1x+2x+3x+.. +(\frac {n}{x}-3)x)+... +(1x+2x+3x)+(1x+2x)+1x$$If the difference between the closest numbers let's call it $d$, $d=\frac {1}{10^k}$, we get $$\frac {n^3}{3!x^2d^2} \approx (1dx+2dx+3dx.. +\frac {n}{x}x) + (1dx+2dx+3dx+.. +(\frac {n}{x}-d)x)+ (1dx+2dx+3dx+.. +(\frac {n}{x}-2d)x)+(1dx+2dx+3dx.. +(\frac {n}{x}-3d)x)+... +(1dx+2dx+3dx)+(1dx+2dx)+1dx$$ if we assume $k\to\infty$ we get $$\frac {n^3}{3!x^2d^2} = (1dx+2dx+3dx.. +\frac {n}{x}x) + (1dx+2dx+3dx+.. +(\frac {n}{x}-d)x)+ (1dx+2dx+3dx+.. +(\frac {n}{x}-2d)x)+(1dx+2dx+3dx.. +(\frac {n}{x}-3d)x)+... +(1dx+2dx+3dx)+(1dx+2dx)+1dx$$ The question is, my calculations are correct? RickJ Posts: 7,762, Reputation: 864 Uber Member #2 Mar 20, 2016, 06:24 PM
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